DAV Class 8 Maths Chapter 8 Practice Worksheet

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DAV Class 8 Maths Chapter 8 Practice Worksheet

Polynomials Practice Worksheet


Section - A

1. If \((x + 2)(x + a) = x^2 + 5x + 6\), then the value of \(a\) is

\( \begin{aligned} (a)\,& 2 \\ (b)\,& 3 \\ (c)\,& 5 \\ (d)\,& 6 \\ \end{aligned} \)

Solution

\[ \begin{aligned} & = x^2 + 5x + 6 \\[4pt] & = x^2 + 2x + 3x + 6 \\[4pt] & = x(x + 2) + 3(x + 2) \\[4pt] & = (x + 2)(x + 3) \\[4pt] a & =\color{green}{3} \end{aligned} \]

Answer \(\color{orange}{(b)}\ \color{red}{3}\)

2. The degree of a quartic (biquadratic) polynomial is

\( \begin{aligned} (a)\,& 1 \\ (b)\,& 2 \\ (c)\,& 3 \\ (d)\,& 4 \\ \end{aligned} \)

Solution

Quartic (biquadratic) = degree is 4

Answer \(\color{orange}{(d)}\ \color{red}{4}\)

3. The factors of \(p^2 + 3p + 2p + 6\) are

\( \begin{aligned} (a)\,& (p + 1)(p + 3) \\ (b)\,& (p + 1)(p + 5) \\ (c)\,& (p + 2)(p + 3) \\ (d)\,& (p + 1)(p + 6) \\ \end{aligned} \)

Solution

\[ \begin{aligned} &= p^2 + 3p + 2p + 6 \\ &= p(p+3) + 2(p+3) \\ &= (p+2)(p+3) \end{aligned} \]

Answer \(\color{orange}{(c)}\ \color{red}{(p+2)(p+3)}\)

4. The degree of the polynomial \((x^2 + 3)(x^5 - 3x + 2)\) is

\( \begin{aligned} (a)\,& 10 \\ (b)\,& 6 \\ (c)\,& 7 \\ (d)\,& 12 \\ \end{aligned} \)

Solution

\[ \begin{aligned} & = (x^2 + 3)(x^5 - 3x + 2) \\ & = x^2 (x^5 - 3x + 2) +3(x^5 - 3x + 2) \\ & = x^7 - 3x^3 + 2x^2 +3x^5 - 9x + 6 \\ & = x^7 + 3x^5- 3x^3 + 2x^2 - 9x + 6 \\ Degree & = 7 \end{aligned} \]

Answer \(\color{orange}{(c)}\ \color{red}{7}\)

5. If \(9x^2 + kx + 6 = (3x + 2)(3x + 3)\), then the value of \(k\) is

\( \begin{aligned} (a)\,& 5 \\ (b)\,& 9 \\ (c)\,& 15 \\ (d)\,& 6 \\ \end{aligned} \)

Solution

\[ \begin{aligned} & = (3x+2)(3x+3) \\ &=9x^2+(2 + 3)3x+(2)(3)\\ &=9x^2+ 15x + 6 \\ k&=\color{green}{15} \end{aligned} \]

Answer \(\color{orange}{(c)}\ \color{red}{15}\)

6. The number to be added to \((x^2 - 5x + 4)\) so that \((x - 2)\) is a factor of the resulting polynomial is

\( \begin{aligned} (a)\,& 5 \\ (b)\,& 10 \\ (c)\,& 15 \\ (d)\,& 2 \\ \end{aligned} \)

Solution

\[ \begin{array}{l} \hspace{1.7cm}\color{green}{x \ - \ 3} \\ x-2\ \enclose{longdiv}{\ x^2 - 5x + 4 + k}\\ \hspace{1cm}\overset{\color{red}\scriptsize(-)}{+}x^2 \overset{\color{red}\scriptsize(+)}{-}2x\\ \hline \hspace{1.5cm}0 -3x + 4 + k\\ \hspace{2cm}\overset{\color{red}\scriptsize(+)}{-}3x \overset{\color{red}\scriptsize(-)}{+}6\\ \hline \hspace{3.cm}-2+k \end{array} \] \[ \begin{aligned} \text{Remainder } \implies & -2 + k=0 \\ \Rightarrow\ k & = \color{green}{2} \end{aligned} \]

Answer \(\color{orange}{(d)}\ \color{red}{2}\)

7. The value of \(p\) for which \( (x^2 + 3x + p) \) divided by \( (x - 2) \) is

\( \begin{aligned} (a)\,& 10 \\ (b)\,& -10 \\ (c)\,& 5 \\ (d)\,& -5 \\ \end{aligned} \)

Solution

\[ \begin{array}{l} \hspace{1.7cm}\color{green}{x \ + \ 5} \\ x-2\ \enclose{longdiv}{\ x^2 + 3x + p}\\ \hspace{1cm}\overset{\color{red}\scriptsize(-)}{+}x^2 \overset{\color{red}\scriptsize(+)}{-}2x\\ \hline \hspace{1.5cm}0 + 5x + p\\ \hspace{2cm}\overset{\color{red}\scriptsize(-)}{+}5x \overset{\color{red}\scriptsize(+)}{-}10\\ \hline \hspace{3.25cm}p+10 \end{array} \] \[ \begin{aligned} \text{Remainder } \implies & p+10=0 \\ \Rightarrow\ p & = \color{green}{-10} \end{aligned} \]

Answer \(\color{orange}{(b)}\ \color{red}{-10}\)

8. Assertion (A): The sum of the degrees of polynomials \(x^4 + 5\) and \(x^4 - 3x^7 + 2\) is \(11\).
Reason (R): The highest power of the variable is the degree of the polynomial.

\( \begin{aligned} (a)\,& \text{Both A and R are true, and R is the correct explanation of A} \\ (b)\,& \text{Both A and R are true, but R is not the correct explanation of A} \\ (c)\,& \text{A is true, but R is false} \\ (d)\,& \text{A is false, but R is true} \\ \end{aligned} \)

Solution

\[ \begin{aligned} x^4+5=Degree \implies 4 \\ x^4-3x^7+2=Degree \implies 7 \\ \text{Sum}=4+7=\color{green}{11} \end{aligned} \]

Answer \(\color{orange}{(a)}\ \color{red}{\text{Both true, and R explains A}}\)

9. Assertion (A): A polynomial of degree \(5\) divided by a quadratic polynomial will have a remainder of degree at most \(1\).
Reason (R): The degree of the remainder is always equal to the degree of the divisor.

\( \begin{aligned} (a)\,& \text{Both A and R are true, and R is the correct explanation of A} \\ (b)\,& \text{Both A and R are true, but R is not the correct explanation of A} \\ (c)\,& \text{A is true, but R is false} \\ (d)\,& \text{A is false, but R is true} \\ \end{aligned} \)

Answer \(\color{orange}{(c)}\ \color{red}{\text{A true, R false}}\)

Section - B

1. Divide the following polynomials.

(a) \(6x^4 - 24x^3 + 15x^2 + 9 \ \text{ by } \ (-3x^2)\)

Solution

\[ \begin{aligned} &= \frac{6x^4 - 24x^3 + 15x^2 + 9}{-3x^2} \\[6pt] &= \frac{6x^4}{-3x^2}\;-\;\frac{24x^3}{-3x^2}\;+\;\frac{15x^2}{-3x^2}\;+\;\frac{9}{-3x^2} \\[8pt] &= -\frac{\cancel{6}^{\,2}}{\cancel{3}_{\,1}}\,x^{4-2} \;+\;\frac{\cancel{24}^{\,8}}{\cancel{3}_{\,1}}\,x^{3-2} \;-\;\frac{\cancel{15}^{5}\,\cancel{x^2}}{\cancel{3}_{1}\,\cancel{x^2}} \;-\;\frac{\cancel 9^{3}}{\cancel 3 \, x^2} \\[8pt] &= -2x^2 \;+\; 8x \;-\; 5 \;-\; \frac{3}{x^2} \end{aligned} \]

Answer \(\color{red}{-2x^2 + 8x - 5 - \dfrac{3}{x^2}}\)

(b) \(\dfrac{2}{3}z^4 - \dfrac{1}{3}z^2 - 1 \ \text{ by } \ \dfrac{1}{3}z\)

Solution

\[ \begin{aligned} &= \frac{\dfrac{2}{3}z^4 - \dfrac{1}{3}z^2 - 1}{\dfrac{1}{3}z} \\[6pt] &= \frac{\dfrac{2}{3}z^4}{\dfrac{1}{3}z} \;-\;\frac{\dfrac{1}{3}z^2}{\dfrac{1}{3}z} \;-\;\frac{1}{\dfrac{1}{3}z} \\[8pt] &= \left(\frac{2}{\cancel 3}\times\frac{\cancel 3}{1}\right) z^{4-1} \;-\; z^{2-1} \;-\; 1\times \frac{3}{z} \\[8pt] &= 2z^3 \;-\; z \;-\; \frac{3}{z} \end{aligned} \]

Answer \(\color{red}{2z^3 - z - \dfrac{3}{z}}\)

(c) \( 56x^5 - 49x^4 + 63x^2 - 14x \ \text{ by } \ (7\sqrt{7}\,x^2)\)

Solution

\[ \begin{aligned} &= \left(\frac{56x^5 - 49x^4 + 63x^2 - 14x}{7\sqrt{7}\,x^2} \right) \times \color{green} \frac{\sqrt{7}}{\sqrt{7}}\\[8pt] &= \frac{56\sqrt{7}x^5 - 49\sqrt{7}x^4 + 63\sqrt{7}x^2 - 14\sqrt{7}x}{49x^2} \\[8pt] &= \frac{\cancel{56}^{ \ 8} \sqrt{7}x^5}{\cancel{49}_7 \ x^2} - \frac{\cancel{49}\sqrt{7}x^4}{\cancel{49}x^2} + \frac{\cancel{63}^{ \ 9 }\sqrt{7}x^2}{\cancel{49}_7 \ x^2} - \frac{\cancel{14}^{\ 2}\sqrt{7}x}{\cancel{49}_7 \ x^2} \\[8pt] & = \frac{8\sqrt{7}}{7}x^3 - \sqrt{7}x^2 + \frac{9\sqrt{7}}{7} - \frac{2\sqrt{7}}{7x} \end{aligned} \]

Answer \(\color{red} \dfrac{8\sqrt{7}}{7}x^3 - \sqrt{7}x^2 + \dfrac{9\sqrt{7}}{7} - \dfrac{2\sqrt{7}}{7x} \)

2. Using factor method, divide the following polynomials.

(a) \( (x^2 + 19x + 84) \ \text{ by } \ (x + 7) \)

Solution

\[ \begin{aligned} &= \frac{x^2 \ {\color{magenta}+ \ 19x} + 84}{(x+7)}\\[6pt] &= \frac{x^2 \ {\color{magenta}+ \ 12x + 7x} + 84}{(x+7)}\\[6pt] &= \frac{x(x+12) + 7(x+12)}{(x+7)}\\[6pt] & = \frac{(x+12)\cancel{(x+7)}}{\cancel{(x+7)}}\\[4pt] &= x+12 \end{aligned} \]

Answer \( \color{red}{x+12} \)

(b) \( (x^2 + 5x - 104) \ \text{ by } \ (x - 8) \)

Solution

\[ \begin{aligned} &= \frac{x^2 \ {\color{magenta}+ \ 5x} - 104}{(x-8)}\\[6pt] &= \frac{x^2 \ {\color{magenta}+ \ 13x - 8x} - 104}{(x-8)}\\[6pt] &= \frac{x(x+13) - 8(x+13)}{(x-8)}\\[6pt] &= \frac{\cancel{(x-8)}(x+13)}{\cancel{(x-8)}}\\[4pt] &= x+13 \end{aligned} \]

Answer \( \color{red}{x+13} \)

(c) \( (z^2 - 4z - 77) \ \text{ by } \ (z - 11) \)

Solution

\[ \begin{aligned} &= \frac{z^2 \ {\color{magenta}- \ 4z} - 77}{(z-11)}\\[6pt] &= \frac{z^2 \ {\color{magenta}- \ 11z + 7z} - 77}{(z-11)}\\[6pt] &= \frac{z(z-11) + 7(z-11)}{(z-11)}\\[6pt] &= \frac{(z+7)\cancel{(z-11)}}{\cancel{(z-11)}}\\[4pt] &= z+7 \end{aligned} \]

Answer \( \color{red}{z+7} \)

(d) \( (p^2 - 14p + 45) \ \text{ by } \ (p - 9) \)

Solution

\[ \begin{aligned} &= \frac{p^2 \ {\color{magenta}- \ 14p} + 45}{(p-9)}\\[6pt] &= \frac{p^2 \ {\color{magenta}- \ 9p - 5p} + 45}{(p-9)}\\[6pt] &= \frac{p(p-9) - 5(p-9)}{(p-9)}\\[6pt] &= \frac{(p-5)\cancel{(p-9)}}{\cancel{(p-9)}}\\[4pt] &= p-5 \end{aligned} \]

Answer \( \color{red}{p-5} \)

3. What should be subtracted from the polynomial \((12x^3-2x^2+x+5)\) so that \((3x+1)\) is a factor of the resulting polynomial?

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{12x^3-2x^2+x+5} \\[6pt] \text{Divisor} &= \color{magenta}{3x+1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{\ 4x^2 \ - \ 2x \ + \ 1\ } \\[2pt] 3x+1\ \enclose{longdiv}{\ 12x^3 - 2x^2 + x + 5}\\ \hspace{1.3cm}\overset{\color{red}\scriptsize(-)}{+}12x^3 \overset{\color{red}\scriptsize(-)}{+}4x^2\\ \hline \hspace{2.2cm}0 - 6x^2 + x + 5 \\ \hspace{2.6cm}\overset{\color{red}\scriptsize(+)}{-}6x^2 \overset{\color{red}\scriptsize(+)}{-}2x\\ \hline \hspace{3.8cm}3x + 5\\ \hspace{3.5cm}\overset{\color{red}\scriptsize(-)}{+}3x \overset{\color{red}\scriptsize(-)}{+}1\\ \hline \hspace{4.6cm}4 \end{array} \] \[ \begin{aligned} \text{Quotient }&=4x^2-2x+1 \\ \text{Remainder } &=4 \end{aligned} \] Subtract \(\boxed{\color{green}4} \) from the polynomial \( (12x^3-2x^2+x+5) \) so that \((3x+1)\) is the factor.

Answer \( \color{red}{4} \)

4. Find whether \((x^2+1)\) is a factor of \((x^4-3x^3-4x^2+3x+2)\).

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{x^4 - 3x^3 - 4x^2 + 3x + 2}\\ \text{Divisor} &= \color{magenta}{x^2 + 1} \end{aligned} \]\[ \begin{array}{l} \hspace{1.7cm}\color{green}{x^2 \ - \ 3x \ - \ 5} \\ x^2+1\ \enclose{longdiv}{\ x^4 - 3x^3 - 4x^2 + 3x + 2}\\ \hspace{1.1cm}\overset{\color{red}\scriptsize(-)}{+}x^4 \hspace{1.3cm} \overset{\color{red}\scriptsize(-)}{+}x^2 \\ \hline \hspace{1.8cm}0 - 3x^3 - 5x^2 + 3x + 2\\ \hspace{2.2cm}\overset{\color{red}\scriptsize(+)}{-}3x^3 \hspace{1.1cm} \overset{\color{red}\scriptsize(+)}{-}3x \\ \hline \hspace{3.0cm}0 - 5x^2 + 6x + 2\\ \hspace{3.35cm}\overset{\color{red}\scriptsize(+)}{-}5x^2 \hspace{1.05cm} \overset{\color{red}\scriptsize(+)}{-}5 \\ \hline \hspace{4.75cm}6x + 7 \end{array} \]\[ \begin{aligned} \color{green}{\text{Quotient}}&=x^2-3x-5 \\ \color{green}{\text{Remainder}} &=6x+7 \end{aligned} \] \(\therefore\ (x^2+1) \) is not a factor of \( (x^4-3x^3-4x^2+3x+2) \)

Answer \(\color{red} \text{Not a factor }\)

5. If \((z^4 - z^3 + 3z^2 - 2z + k)\) is divisible by \((z+1)\), find the value of \(k\).

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{z^4 - z^3 + 3z^2 - 2z + k} \\ \text{Divisor} &= \color{magenta}{z + 1} \end{aligned} \]\[ \begin{array}{l} \hspace{1.4cm}\color{green}{z^3 \ - \ 2z^2 \ + \ 5z \ - \ 7} \\ z+1\ \enclose{longdiv}{\ z^4 - z^3 + 3z^2 - 2z + k}\\ \hspace{1.1cm}\overset{\color{red}\scriptsize(-)}{+}z^4 \overset{\color{red}\scriptsize(-)}{+}z^3 \\ \hline \hspace{1.4cm}0 - 2z^3 + 3z^2 - 2z + k \\ \hspace{1.8cm}\overset{\color{red}\scriptsize(+)}{-}2z^3 \overset{\color{red}\scriptsize(+)}{-}2z^2 \\ \hline \hspace{2.5cm}0 + 5z^2 - 2z + k \\ \hspace{3cm}\overset{\color{red}\scriptsize(-)}{+}5z^2 \overset{\color{red}\scriptsize(-)}{+}5z \\ \hline \hspace{3.7cm}0 - 7z + k \\ \hspace{4.2cm}\overset{\color{red}\scriptsize(+)}{-}7z \overset{\color{red}\scriptsize(+)}{-}7 \\ \hline \hspace{5cm}k + 7 \end{array} \]\[ \begin{aligned} \color{green}{\text{Quotient}} &= z^3 - 2z^2 + 5z - 7 \\ \color{green}{\text{Remainder}} &= k + 7 \end{aligned} \]\[ \begin{aligned} k+7&=0 \\ k &=-7 \end{aligned} \]

Answer \(\color{red}{-7}\)

6. Find the difference between the quotient and remainder, when \((8x^2-2-3x+12x^3)\) is divided by \((4x^2-1)\).

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{12x^3+8x^2-3x-2}\\ \text{Divisor} &= \color{magenta}{4x^2-1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{3x \ + \ 2} \\ 4x^2-1\ \enclose{longdiv}{\ 12x^3 + 8x^2 - 3x - 2}\\ \hspace{1.5cm}\overset{\color{red}\scriptsize(-)}{+}12x^3 \hspace{1.05cm} \overset{\color{red}\scriptsize(+)}{-}3x \\ \hline \hspace{2.2cm}0 + 8x^2 + 0x - 2 \\ \hspace{2.6cm}\overset{\color{red}\scriptsize(-)}{+}8x^2 \hspace{1.05cm} \overset{\color{red}\scriptsize(+)}{-}2 \\ \hline \hspace{3.9cm}0 \end{array} \] \[ \begin{aligned} \text{Quotient} &= \color{green}{3x+2}\\ \text{Remainder} &= \color{green}{0}\\[4pt] \text{Difference} &= \text{Quotient}-\text{Remainder}\\ &=(3x+2)-0\\ &=\color{green}{3x+2} \end{aligned} \]

Answer \( \color{red}{3x+2} \)

7. Divide and check: \(\ (3y^2 + 30y + 6y^5 - 28y^3 - 9)\ \text{by}\ (2y^2 - 6)\).

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{6y^5 + 0y^4- 28y^3 + 3y^2 + 30y - 9} \\ \text{Divisor} &= \color{magenta}{2y^2 - 6} \end{aligned} \] \[ \begin{array}{l} \hspace{1.6cm}\color{green}{3y^3 \ - \ 5y \ + \ \dfrac{3}{2}} \\ 2y^2 - 6\ \enclose{longdiv}{\ 6y^5 + 0y^4 - 28y^3 + 3y^2 + 30y - 9}\\ \hspace{1.5cm}\overset{\color{red}\scriptsize(-)}{+}6y^5 \hspace{1.0cm} \overset{\color{red}\scriptsize(+)}{-}18y^3 \\ \hline \hspace{3.2cm}0 - 10y^3 + 3y^2 + 30y - 9\\ \hspace{3.7cm}\overset{\color{red}\scriptsize(+)}{-}10y^3 \hspace{1cm} \overset{\color{red}\scriptsize(-)}{+}30y \\ \hline \hspace{4.5cm} 0 + 3y^2 + 0y - 9\\ \hspace{4.9cm}\overset{\color{red}\scriptsize(-)}{+}3y^2 \hspace{1cm} \overset{\color{red}\scriptsize(+)}{-}9 \\ \hline \hspace{6.4cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 3y^3 - 5y + \dfrac{3}{2} \\ \color{green}{\text{Remainder}} & = 0 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} 6y^5 - 28y^3 + 3y^2 + 30y - 9 \\[6pt] RHS&= (2y^2-6) \left(3y^3 - 5y + \dfrac{3}{2}\right) + 0 \\[6pt] &= 2y^2 \left(3y^3 - 5y + \dfrac{3}{2}\right) -6\left(3y^3 - 5y + \dfrac{3}{2}\right) \\[6pt] &= 6y^5 - 10y^3 + 3y^2 - 18y^3 + 30y - 9 \\[6pt] &= 6y^5 - 10y^3 - 18y^3 + 3y^2 + 30y - 9 \\[6pt] RHS &\implies \color{green} 6y^5 - 28y^3 + 3y^2 + 30y - 9 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

8. Divide \((z^5 - 9z)\) by \((z^2 + 3)\) and check your answer.

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{z^5 + 0z^4 + 0z^3 + 0z^2 - 9z + 0} \\ \text{Divisor} &= \color{magenta}{z^2 + 3} \end{aligned} \] \[ \begin{array}{l} \hspace{1.55cm}\color{green}{z^3 \ - \ 3z} \\ z^2 + 3\ \enclose{longdiv}{\ z^5 + 0z^4 + 0z^3 + 0z^2 - 9z + 0}\\ \hspace{1.25cm}\overset{\color{red}\scriptsize(-)}{+}z^5 \hspace{1.1cm} \overset{\color{red}\scriptsize(-)}{+}3z^3 \\ \hline \hspace{2.8cm}0 - 3z^3 + 0z^2 - 9z + 0\\ \hspace{3.2cm}\overset{\color{red}\scriptsize(+)}{-}3z^3 \hspace{1.2cm} \overset{\color{red}\scriptsize(+)}{-}9z \\ \hline \hspace{5cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= z^3 - 3z \\ \color{green}{\text{Remainder}} &= 0 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} z^5 - 9z \\[6pt] RHS &= (z^2+3)\,(z^3-3z) + 0 \\[4pt] &= z^2(z^3-3z) + 3(z^3-3z) \\[4pt] &= z^5 - 3z^3 + 3z^3 - 9z \\[4pt] RHS &\implies \color{green} z^5 - 9z \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

9. Divide \(4x(2x^3+3)+5(2x^3+3)\) by \((5+4x)\) and check your answer.

Solution

\[ \begin{aligned} &= 4x(2x^3+3)+5(2x^3+3) \\[4pt] &= 8x^4 + 12x + 10x^3 + 15 \\[4pt] \text{Dividend} &= \color{magenta}{8x^4 + 10x^3 + 0x^2 + 12x + 15} \\ \text{Divisor} &= \color{magenta}{4x + 5} \end{aligned} \] \[ \begin{array}{l} \hspace{1.6cm}\color{green}{2x^3 \ + \ 3} \\ 4x + 5 \ \enclose{longdiv}{\ 8x^4 + 10x^3 + 0x^2 + 12x + 15}\\ \hspace{1.3cm}\overset{\color{red}\scriptsize(-)}{+}8x^4 \overset{\color{red}\scriptsize(-)}{+}10x^3 \\ \hline \hspace{4.5cm}0 + 12x + 15\\ \hspace{4.8cm}\overset{\color{red}\scriptsize(-)}{+}12x \overset{\color{red}\scriptsize(-)}{+}15 \\ \hline \hspace{6cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 2x^3 + 3 \\ \color{green}{\text{Remainder}} &= 0 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} 8x^4 + 10x^3 + 12x + 15 \\[6pt] &= (4x + 5)(2x^3+3) + 0 \\[4pt] &= 4x(2x^3+3) + 5(2x^3+3) \\[4pt] &= 8x^4 + 12x + 10x^3 + 15 \\[4pt] RHS &\implies \color{green} 8x^4 + 10x^3 + 12x + 15 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

10. Find the quotient and remainder when \((p^4 + p^3 - p^2 + 1)\) is divided by \((p - 1)\). Also write the degree of the quotient.

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{p^4 + p^3 - p^2 + 0p + 1} \\ \text{Divisor} &= \color{magenta}{p - 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.45cm}\color{green}{p^3 \ + \ 2p^2 \ + \ p \ + \ 1} \\ p-1\ \enclose{longdiv}{\ p^4 + p^3 - p^2 + 0p + 1}\\ \hspace{1cm}\overset{\color{red}\scriptsize(-)}{+}p^4 \overset{\color{red}\scriptsize(+)}{-}p^3 \\ \hline \hspace{1.5cm}0 + 2p^3 - p^2 + 0p + 1 \\ \hspace{1.8cm}\overset{\color{red}\scriptsize(-)}{+}2p^3 \overset{\color{red}\scriptsize(+)}{-}2p^2 \\ \hline \hspace{2.5cm}0 + p^2 + 0p + 1 \\ \hspace{3.cm}\overset{\color{red}\scriptsize(-)}{+}p^2 \overset{\color{red}\scriptsize(+)}{-}p \\ \hline \hspace{3.5cm}0 + p + 1 \\ \hspace{3.9cm}\overset{\color{red}\scriptsize(-)}{+}p \overset{\color{red}\scriptsize(+)}{-}1 \\ \hline \hspace{4.9cm}2 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= p^3 + 2p^2 + p + 1 \\ degree & = 3 \\ \color{green}{\text{Remainder}} &= 2 \end{aligned} \]

Answer Quotient \(=\color{red}{p^3+2p^2+p+1}\) (degree \(3\)), Remainder \(=\color{red}{2}\)

11. Divide: \(3x(5x^2 + 3x^3 + 2) - (2x^2 + 8 - x)\) by \((-2 + 3x)\). Also, write the quotient and remainder.

Solution

\[ \begin{aligned} &= 3x(5x^2 + 3x^3 + 2) - (2x^2 + 8 - x) \\ &= 15x^3 + 9x^4 + 6x - 2x^2 + x - 8 \\ &= 9x^4 + 15x^3 - 2x^2 + 6x + x - 8 \\ \text{Dividend}&= \color{magenta}{9x^4 + 15x^3 - 2x^2 + 7x - 8} \\ \text{Divisor} & = \color{magenta}{3x - 2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.65cm}\color{green}{3x^3 \ + \ 7x^2 \ + \ 4x \ + \ 5} \\[2pt] 3x - 2 \ \enclose{longdiv}{\ 9x^4 + 15x^3 - 2x^2 + 7x - 8}\\ \hspace{1.4cm}\overset{\color{red}\scriptsize(-)}{+}9x^4 \overset{\color{red}\scriptsize(+)}{-}6x^3\\ \hline \hspace{2cm}0 + 21x^3 - 2x^2 + 7x - 8 \\ \hspace{2.4cm}\overset{\color{red}\scriptsize(-)}{+}21x^3 \overset{\color{red}\scriptsize(+)}{-}14x^2\\ \hline \hspace{3.3cm}0 + 12x^2 + 7x - 8 \\ \hspace{3.6cm}\overset{\color{red}\scriptsize(-)}{+}12x^2 \overset{\color{red}\scriptsize(+)}{-}8x\\ \hline \hspace{4.4cm}0 + 15x - 8 \\ \hspace{4.7cm}\overset{\color{red}\scriptsize(-)}{+}15x \overset{\color{red}\scriptsize(+)}{-}10 \\ \hline \hspace{5.8cm}2 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 3x^3 + 7x^2 + 4x + 5 \\ \color{green}{\text{Remainder}} &= 2 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} 9x^4 + 15x^3 - 2x^2 + 7x - 8 \\[6pt] RHS &= (3x - 2)(3x^3 + 7x^2 + 4x + 5) + 2 \\[4pt] &= 3x(3x^3 + 7x^2 + 4x + 5) - 2(3x^3 + 7x^2 + 4x + 5) + 2 \\[4pt] &= 9x^4 + 21x^3 + 12x^2 + 15x - 6x^3 - 14x^2 - 8x - 10 + 2 \\[4pt] RHS &\implies \color{green} 9x^4 + 15x^3 - 2x^2 + 7x - 8 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer Quotient \(=\color{red}{3x^3 + 7x^2 + 4x + 5}\), Remainder \(=\color{red}{2}\)

12. Divide and check whether \((3y^2 + 5)\) is a factor of \((6y^5 + 15y^4 + 16y^3 + 4y^2 + 10y - 35)\) or not?

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{6y^5 + 15y^4 + 16y^3 + 4y^2 + 10y - 35} \\ \text{Divisor} &= \color{magenta}{3y^2 + 5} \end{aligned} \] \[ \begin{array}{l} \hspace{1.6cm}\color{green}{2y^3 \ + \ 5y^2 \ + \ 2y \ - \ 7} \\ 3y^2 + 5\ \enclose{longdiv}{\ 6y^5 + 15y^4 + 16y^3 + 4y^2 + 10y - 35}\\ \hspace{1.5cm}\overset{\color{red}\scriptsize(-)}{+}6y^5 \hspace{1.2cm} \overset{\color{red}\scriptsize(-)}{+}10y^3 \\ \hline \hspace{2.2cm}0 + 15y^4 + 6y^3 + 4y^2 + 10y - 35 \\ \hspace{2.5cm}\overset{\color{red}\scriptsize(-)}{+}15y^4 \hspace{1.2cm} \overset{\color{red}\scriptsize(-)}{+}25y^2 \\ \hline \hspace{3.5cm}0 + 6y^3 - 21y^2 + 10y - 35 \\ \hspace{3.9cm}\overset{\color{red}\scriptsize(-)}{+}6y^3 \hspace{1.3cm} \overset{\color{red}\scriptsize(-)}{+}10y \\ \hline \hspace{4.8cm}0 - 21y^2 + 0 - 35 \\ \hspace{5.2cm}\overset{\color{red}\scriptsize(+)}{-}21y^2 \hspace{.8cm} \overset{\color{red}\scriptsize(+)}{-}35 \\ \hline \hspace{6.8cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 2y^3 + 5y^2 + 2y - 7 \\ \color{green}{\text{Remainder}} &= 0 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} 6y^5 + 15y^4 + 16y^3 + 4y^2 + 10y - 35 \\[6pt] RHS &= (3y^2 + 5)(2y^3 + 5y^2 + 2y - 7) + 0 \\[4pt] &= 3y^2(2y^3 + 5y^2 + 2y - 7) + 5(2y^3 + 5y^2 + 2y - 7) \\[4pt] &= 6y^5 + 15y^4 + 6y^3 - 21y^2 + 10y^3 + 25y^2 + 10y - 35 \\[4pt] RHS &\implies \color{green} 6y^5 + 15y^4 + 16y^3 + 4y^2 + 10y - 35 \\[6pt] LHS &= RHS \\ & \text{Hence verified}\\ & \text{Since Remainder is 0, it is a factor.} \end{aligned} \]

Answer Yes, \(\color{red} (3y^2 + 5) \) is a factor of \(\color{red} (6y^5 + 15y^4 + 16y^3 + 4y^2 + 10y - 35)\)

13. Divide and check: \(( -3x^3 + 11x + 1 - 6x^4 + 5x^2 )\) by \((1 - 3x)\). Find the degree of the quotient.

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{-6x^4 - 3x^3 + 5x^2 + 11x + 1} \\ \text{Divisor} & = \color{magenta}{-3x + 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{2x^3 \ + \ \dfrac{5}{3}x^2 \ - \ \dfrac{10}{9}x \ - \ \dfrac{109}{27}} \\[6pt] -3x + 1 \ \enclose{longdiv}{\ -6x^4 - 3x^3 + 5x^2 + 11x + 1}\\ \hspace{2.1cm}\overset{\color{red}\scriptsize(+)}{-}6x^4 \overset{\color{red}\scriptsize(-)}{+}2x^3\\ \hline \hspace{2.8cm}0 - 5x^3 + 5x^2 + 11x + 1 \\ \hspace{3.1cm}\overset{\color{red}\scriptsize(+)}{-}5x^3 \overset{\color{red}\scriptsize(-)}{+}\dfrac{5}{3}x^2\\ \hline \hspace{3.7cm} 0 +\dfrac{10}{3}x^2 + 11x + 1 \\ \hspace{3.8cm}\overset{\color{red}\scriptsize(-)}{+}\dfrac{10}{3}x^2 \overset{\color{red}\scriptsize(+)}{-}\dfrac{10}{9}x\\ \hline \hspace{5cm}0 + \dfrac{109}{9}x + 1 \\ \hspace{5.2cm}\overset{\color{red}\scriptsize(-)}{+}\dfrac{109}{9}x \overset{\color{red}\scriptsize(+)}{+}\dfrac{109}{27}\\ \hline \hspace{6cm}0 + \dfrac{136}{27} \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 2x^3 + \dfrac{5}{3}x^2 - \dfrac{10}{9}x - \dfrac{109}{27} \\ \text{Degree} &= \color{green}{3} \\[4pt] \color{green}{\text{Remainder}} &= \dfrac{136}{27} \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} -6x^4 - 3x^3 + 5x^2 + 11x + 1 \\[6pt] RHS &= (-3x + 1)\left(2x^3 + \dfrac{5}{3}x^2 - \dfrac{10}{9}x - \dfrac{109}{27}\right) + \dfrac{136}{27} \\[4pt] &= -3x\left(2x^3 + \dfrac{5}{3}x^2 - \dfrac{10}{9}x - \dfrac{109}{27}\right) + \left(2x^3 + \dfrac{5}{3}x^2 - \dfrac{10}{9}x - \dfrac{109}{27}\right) + \dfrac{136}{27} \\[4pt] &= -6x^4 - 5x^3 + \dfrac{10}{3}x^2 + \dfrac{109}{9}x + 2x^3 + \dfrac{5}{3}x^2 - \dfrac{10}{9}x - \dfrac{109}{27} + \frac{136}{27} \\[4pt] &= -6x^4 - 5x^3 + 2x^3 + \dfrac{10}{3}x^2 + \dfrac{5}{3}x^2 + \dfrac{109}{9}x - \dfrac{10}{9}x - \dfrac{109}{27} + \frac{136}{27} \\[4pt] &= -6x^4 - 3x^3 + \dfrac{15}{3}x^2 + \dfrac{99}{9}x + \frac{27}{27} \\[4pt] &= -6x^4 - 3x^3 + 5x^2 + 11x + 1 \\[4pt] RHS &\implies \color{green} -6x^4 - 3x^3 + 5x^2 + 11x + 1 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer Quotient \(=\color{red}{2x^3 + \dfrac{5}{3}x^2 - \dfrac{10}{9}x - \dfrac{109}{27}}\), Degree of Quotient \(=\color{red}{3} \), Remainder \(=\color{red}{\dfrac{136}{27}}\)

14. Divide \((-1 + x^4)\) by \((-1 + x)\) and check your answer.

Solution

\[ \begin{aligned} \text{Dividend} & = \color{magenta}{x^4 + 0x^3 + 0x^2 + 0x - 1} \\ \text{Divisor} & = \color{magenta}{x - 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{x^3 \ + \ x^2 \ + \ x \ + \ 1} \\ x - 1\ \enclose{longdiv}{\ x^4 + 0x^3 + 0x^2 + 0x - 1}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+}x^4 \overset{\color{red}\scriptsize(+)}{-}x^3\\ \hline \hspace{1.8cm}0 + x^3 + 0x^2 + 0x - 1 \\ \hspace{2.1cm}\overset{\color{red}\scriptsize(-)}{+}x^3 \overset{\color{red}\scriptsize(+)}{-}x^2 \\ \hline \hspace{2.8cm}0 + x^2 + 0x - 1 \\ \hspace{3.1cm}\overset{\color{red}\scriptsize(-)}{+}x^2 \overset{\color{red}\scriptsize(+)}{-}x \\ \hline \hspace{3.8cm}0 + x - 1 \\ \hspace{4.1cm}\overset{\color{red}\scriptsize(-)}{+}x \overset{\color{red}\scriptsize(+)}{-}1 \\ \hline \hspace{4.9cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= x^3 + x^2 + x + 1 \\ \color{green}{\text{Remainder}} &= 0 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} x^4 - 1 \\[6pt] RHS &= (x - 1)(x^3 + x^2 + x + 1) + 0 \\[4pt] &= x(x^3 + x^2 + x + 1) - 1(x^3 + x^2 + x + 1) \\[4pt] &= x^4 + x^3 + x^2 + x - x^3 - x^2 - x - 1 \\[4pt] &= x^4 + x^3 - x^3 + x^2 - x^2 + x - x - 1 \\[4pt] RHS &\implies \color{green} x^4 - 1 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer Quotient \(=\color{red}{x^3 + x^2 + x + 1}\), Remainder \(=\color{red}{0}\)

15. Find the quotient when \((p^2 + 13p + 42)\) is divided by \((p + 6)\) using factor method. Check your answer by long division method.

Solution

Factor Method:

\[ \begin{aligned} &= \frac{p^2 \ {\color{magenta}+ \ 13p} + 42}{(p+6)}\\[6pt] &= \frac{p^2 \ {\color{magenta}+ \ 6p + 7p} + 42}{(p+6)}\\[6pt] &= \frac{p(p+6) + 7(p+6)}{(p+6)}\\[6pt] & = \frac{(p+7)\cancel{(p+6)}}{\cancel{(p+6)}}\\[4pt] &= p+7 \end{aligned} \]

Check (Long Division Method):

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{p^2 + 13p + 42} \\ \text{Divisor} &= \color{magenta}{p + 6} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{p \ + \ 7} \\ p + 6\ \enclose{longdiv}{\ p^2 + 13p + 42}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+}p^2 \overset{\color{red}\scriptsize(-)}{+}6p\\ \hline \hspace{1.8cm}0 + 7p + 42 \\ \hspace{2.1cm}\overset{\color{red}\scriptsize(-)}{+}7p \overset{\color{red}\scriptsize(-)}{+}42 \\ \hline \hspace{3.1cm}0 \end{array} \] \[ \begin{aligned} \text{Quotient} &= p + 7 \\ \text{Remainder} &= 0 \\ \end{aligned} \] The quotient matches the result from the factor method.

Answer \( \color{red}{p+7} \)

16. Divide and check your answer for the following:

(a) \( (34x - 22x^3 - 12x^4 - 10x^2 - 75) \ \text{ by } \ (3x + 7) \)

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{-12x^4 - 22x^3 - 10x^2 + 34x - 75} \\ \text{Divisor} &= \color{magenta}{3x + 7} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{-4x^3 \ + \ 2x^2 \ - \ 8x \ + \ 30} \\ 3x + 7\ \enclose{longdiv}{\ -12x^4 - 22x^3 - 10x^2 + 34x - 75}\\ \hspace{1.3cm}\overset{\color{red}\scriptsize(+)}{-}12x^4 \overset{\color{red}\scriptsize(+)}{-}28x^3\\ \hline \hspace{2.8cm}0 + 6x^3 - 10x^2 + 34x - 75 \\ \hspace{3.1cm}\overset{\color{red}\scriptsize(-)}{+}6x^3 \overset{\color{red}\scriptsize(-)}{+}14x^2\\ \hline \hspace{4.cm}0 - 24x^2 + 34x - 75 \\ \hspace{4.3cm}\overset{\color{red}\scriptsize(+)}{-}24x^2 \overset{\color{red}\scriptsize(+)}{-}56x\\ \hline \hspace{5.4cm}0 + 90x - 75 \\ \hspace{5.7cm}\overset{\color{red}\scriptsize(-)}{+}90x \overset{\color{red}\scriptsize(-)}{+}210 \\ \hline \hspace{6.8cm}-285 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= -4x^3 + 2x^2 - 8x + 30 \\ \color{green}{\text{Remainder}} &= -285 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} -12x^4 - 22x^3 - 10x^2 + 34x - 75 \\[6pt] RHS &= (3x + 7)(-4x^3 + 2x^2 - 8x + 30) - 285 \\[4pt] &= 3x(-4x^3 + 2x^2 - 8x + 30) + 7(-4x^3 + 2x^2 - 8x + 30) - 285 \\[4pt] &= -12x^4 + 6x^3 - 24x^2 + 90x -28x^3 + 14x^2 - 56x + 210 - 285 \\[4pt] &= -12x^4 + 6x^3 -28x^3 - 24x^2 + 14x^2 + 90x - 56x + 210 - 285 \\[4pt] RHS &\implies \color{green} -12x^4 - 22x^3 - 10x^2 + 34x - 75 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer Quotient \(=\color{red}{-4x^3 + 2x^2 - 8x + 30}\), Remainder \(=\color{red}{-285}\)

(b) \( (8x + 20 + 16x^4 + 12x^3 - 10x^2) \ \text{ by } \ (4x - 3) \)

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{16x^4 + 12x^3 - 10x^2 + 8x + 20} \\ \text{Divisor} &= \color{magenta}{4x - 3} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{4x^3 \ + \ 6x^2 \ + \ 2x \ + \ \dfrac{7}{2}} \\ 4x - 3\ \enclose{longdiv}{\ 16x^4 + 12x^3 - 10x^2 + 8x + 20}\\ \hspace{1.3cm}\overset{\color{red}\scriptsize(-)}{+}16x^4 \overset{\color{red}\scriptsize(+)}{-}12x^3\\ \hline \hspace{2.2cm}0 + 24x^3 - 10x^2 + 8x + 20 \\ \hspace{2.5cm}\overset{\color{red}\scriptsize(-)}{+}24x^3 \overset{\color{red}\scriptsize(+)}{-}18x^2\\ \hline \hspace{3.7cm}0 + 8x^2 + 8x + 20 \\ \hspace{4.cm}\overset{\color{red}\scriptsize(-)}{+}8x^2 \overset{\color{red}\scriptsize(+)}{-}6x\\ \hline \hspace{5.2cm}0 + 14x + 20 \\ \hspace{5.5cm}\overset{\color{red}\scriptsize(-)}{+}14x \overset{\color{red}\scriptsize(+)}{-}\dfrac{21}{2}\\ \hline \hspace{7cm} \dfrac{61}{2} \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 4x^3 + 6x^2 + 2x + \dfrac{7}{2} \\ \color{green}{\text{Remainder}} &= \dfrac{61}{2} \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} 16x^4 + 12x^3 - 10x^2 + 8x + 20 \\[6pt] RHS &= (4x - 3)\left(4x^3 + 6x^2 + 2x + \dfrac{7}{2}\right) + \dfrac{61}{2} \\[4pt] &= 4x\left(4x^3 + 6x^2 + 2x + \dfrac{7}{2}\right) - 3\left(4x^3 + 6x^2 + 2x + \dfrac{7}{2}\right) + \dfrac{61}{2} \\[4pt] &= 16x^4 + 24x^3 + 8x^2 + 14x - 12x^3 - 18x^2 - 6x - \dfrac{21}{2} + \dfrac{61}{2} \\[4pt] &= 16x^4 + 24 x^3 -12x^3 + 8x^2-18x^2 + 14x-6x + \dfrac{40}{2} \\[4pt] &= 16x^4 + 12x^3 - 10x^2 + 8x + 20 \\[4pt] RHS &\implies \color{green} 16x^4 + 12x^3 - 10x^2 + 8x + 20 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer Quotient \(=\color{red}{4x^3 + 6x^2 + 2x + \dfrac{7}{2}}\), Remainder \(=\color{red}{\dfrac{61}{2}}\)

(c) \( (4p^3 - 12p^2 - 37p - 15) \ \text{ by } \ (2p + 1) \)

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{4p^3 - 12p^2 - 37p - 15} \\ \text{Divisor} &= \color{magenta}{2p + 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{2p^2 \ - \ 7p \ - \ 15} \\ 2p + 1\ \enclose{longdiv}{\ 4p^3 - 12p^2 - 37p - 15}\\ \hspace{1.3cm}\overset{\color{red}\scriptsize(-)}{+}4p^3 \overset{\color{red}\scriptsize(-)}{+}2p^2\\ \hline \hspace{2.2cm}0 - 14p^2 - 37p - 15 \\ \hspace{2.5cm}\overset{\color{red}\scriptsize(+)}{-}14p^2 \overset{\color{red}\scriptsize(+)}{-}7p\\ \hline \hspace{3.4cm}0 - 30p - 15 \\ \hspace{3.7cm}\overset{\color{red}\scriptsize(+)}{-}30p \overset{\color{red}\scriptsize(+)}{-}15\\ \hline \hspace{4.9cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 2p^2 - 7p - 15 \\ \color{green}{\text{Remainder}} &= 0 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} 4p^3 - 12p^2 - 37p - 15 \\[6pt] RHS &= (2p + 1)(2p^2 - 7p - 15) + 0 \\[4pt] &= 2p(2p^2 - 7p - 15) + 1(2p^2 - 7p - 15) \\[4pt] &= 4p^3 - 14p^2 - 30p + 2p^2 - 7p - 15 \\[4pt] RHS &\implies \color{green} 4p^3 - 12p^2 - 37p - 15 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer Quotient \(=\color{red}{2p^2 - 7p - 15}\), Remainder \(=\color{red}{0}\)

17. Divide: \((25p^3 - 5p^2 + 5\sqrt{5}p - \sqrt{5})\) by \((-\sqrt{5}p)\)

Solution

\[ \begin{aligned} &= \frac{25p^3 - 5p^2 + 5\sqrt{5}p - \sqrt{5}}{-\sqrt{5}p} \\[6pt] &= \frac{25p^3}{-\sqrt{5}p} - \frac{5p^2}{-\sqrt{5}p} + \frac{5\sqrt{5}p}{-\sqrt{5}p} - \frac{\sqrt{5}}{-\sqrt{5}p} \\[8pt] &= -\frac{25}{\sqrt{5}}p^2 + \frac{5}{\sqrt{5}}p - 5 + \frac{1}{p} \\[8pt] &= -\frac{25 \color{red} \times \sqrt{5}}{\sqrt{5} \color{red} \times \sqrt{5} }p^2 + \frac{5 \color{red} \times \sqrt{5} }{\sqrt{5} \color{red} \times \sqrt{5}}p - 5 + \frac{1}{p} \\[8pt] &= -\frac{25 \sqrt{5}}{5}p^2 + \frac{5\sqrt{5} }{5}p - 5 + \frac{1}{p} \\[8pt] &= -5\sqrt{5}p^2 + \sqrt{5}p - 5 + \frac{1}{p} \end{aligned} \]

Answer \(\color{red}{-5\sqrt{5}p^2 + \sqrt{5}p - 5 + \dfrac{1}{p}}\)

18. Is \((3y - 2)\) a factor of \((15y^4 - 16y^3 + 9y^2 - \dfrac{10}{3}y)\)?

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{15y^4 - 16y^3 + 9y^2 - \dfrac{10}{3}y + 0}\\ \text{Divisor} &= \color{magenta}{3y - 2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{5y^3 \ - \ 2y^2 \ + \ \dfrac{5}{3}y} \\ 3y - 2\ \enclose{longdiv}{\ 15y^4 - 16y^3 + 9y^2 - \dfrac{10}{3}y + 0}\\ \hspace{1.1cm}\overset{\color{red}\scriptsize(-)}{+}15y^4 \overset{\color{red}\scriptsize(+)}{-}10y^3 \\ \hline \hspace{2.2cm}0 - 6y^3 + 9y^2 - \dfrac{10}{3}y + 0\\ \hspace{2.5cm}\overset{\color{red}\scriptsize(+)}{-}6y^3 \overset{\color{red}\scriptsize(-)}{+}4y^2 \\ \hline \hspace{3.6cm}0 + 5y^2 - \dfrac{10}{3}y + 0\\ \hspace{3.9cm}\overset{\color{red}\scriptsize(-)}{+}5y^2 \overset{\color{red}\scriptsize(+)}{-}\dfrac{10}{3}y \\ \hline \hspace{5.8cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}}&=5y^3 - 2y^2 + \dfrac{5}{3}y \\ \color{green}{\text{Remainder}} &= 0 \end{aligned} \] Since the remainder is 0 \( \color{red} (3y-2) \) is a factor of \( \color{red} \left(15y^4 - 16y^3 + 9y^2 - \dfrac{10}{3}y\right) \)

Answer \(\color{red} \text{Yes, it is a factor.}\)

19. Divide and check your answer for the following polynomials.

(a) \( 6x(2x^3 + x^2 + x + 1) - x(8x^2 + 5x + 2) + 5 \ \text{ by } \ (1 + 3x) \)

Solution

\[ \begin{aligned} &= 6x(2x^3 + x^2 + x + 1) - x(8x^2 + 5x + 2) + 5 \\ &= 12x^4 + 6x^3 + 6x^2 + 6x - 8x^3 - 5x^2 - 2x + 5 \\ &= 12x^4 + 6x^3-8x^3 + 6x^2-5x^2 + 6x-2x + 5 \\ \text{Dividend} &= \color{magenta}{12x^4 - 2x^3 + x^2 + 4x + 5} \\ \text{Divisor} &= \color{magenta}{3x + 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{4x^3 \ - \ 2x^2 \ + \ x \ + \ 1} \\ 3x + 1\ \enclose{longdiv}{\ 12x^4 - 2x^3 + x^2 + 4x + 5}\\ \hspace{1.3cm}\overset{\color{red}\scriptsize(-)}{+}12x^4 \overset{\color{red}\scriptsize(-)}{+}4x^3\\ \hline \hspace{2.2cm}0 - 6x^3 + x^2 + 4x + 5 \\ \hspace{2.5cm}\overset{\color{red}\scriptsize(+)}{-}6x^3 \overset{\color{red}\scriptsize(+)}{-}2x^2\\ \hline \hspace{3.4cm}0 + 3x^2 + 4x + 5 \\ \hspace{3.7cm}\overset{\color{red}\scriptsize(-)}{+}3x^2 \overset{\color{red}\scriptsize(-)}{+}x\\ \hline \hspace{4.6cm}0 + 3x + 5 \\ \hspace{4.9cm}\overset{\color{red}\scriptsize(-)}{+}3x \overset{\color{red}\scriptsize(-)}{+}1\\ \hline \hspace{5.8cm}4 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 4x^3 - 2x^2 + x + 1 \\ \color{green}{\text{Remainder}} &= 4 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} 12x^4 - 2x^3 + x^2 + 4x + 5 \\[6pt] RHS &= (3x + 1)(4x^3 - 2x^2 + x + 1) + 4 \\[4pt] &= 3x(4x^3 - 2x^2 + x + 1) + 1(4x^3 - 2x^2 + x + 1) + 4 \\[4pt] &= 12x^4 - 6x^3 + 3x^2 + 3x + 4x^3 - 2x^2 + x + 1 + 4 \\[4pt] &= 12x^4 -6x^3+4x^3 + 3x^2-2x^2 + 3x+1x + 1+4 \\[4pt] RHS &\implies \color{green} 12x^4 - 2x^3 + x^2 + 4x + 5 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer Quotient \(=\color{red}{4x^3 - 2x^2 + x + 1}\), Remainder \(=\color{red}{4}\)

(b) \( [3x(5x^2 + 3x^3 + 2) - (2x^2 - x + 8)] \ \text{ by } \ (-2 + 3x) \)

Solution

\[ \begin{aligned} &= [3x(5x^2 + 3x^3 + 2) - (2x^2 - x + 8)] \\ &= 15x^3 + 9x^4 + 6x - 2x^2 + x - 8 \\ &= 9x^4 + 15x^3 - 2x^2 + 6x + x - 8 \\ \text{Dividend} &= \color{magenta}{9x^4 + 15x^3 - 2x^2 + 7x - 8} \\ \text{Divisor} &= \color{magenta}{3x - 2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.65cm}\color{green}{3x^3 \ + \ 7x^2 \ + \ 4x \ + \ 5} \\[2pt] 3x - 2 \ \enclose{longdiv}{\ 9x^4 + 15x^3 - 2x^2 + 7x - 8}\\ \hspace{1.4cm}\overset{\color{red}\scriptsize(-)}{+}9x^4 \overset{\color{red}\scriptsize(+)}{-}6x^3\\ \hline \hspace{2.2cm}0 + 21x^3 - 2x^2 + 7x - 8 \\ \hspace{2.5cm}\overset{\color{red}\scriptsize(-)}{+}21x^3 \overset{\color{red}\scriptsize(+)}{-}14x^2\\ \hline \hspace{3.3cm}0 + 12x^2 + 7x - 8 \\ \hspace{3.6cm}\overset{\color{red}\scriptsize(-)}{+}12x^2 \overset{\color{red}\scriptsize(+)}{-}8x\\ \hline \hspace{4.4cm}0 + 15x - 8 \\ \hspace{4.7cm}\overset{\color{red}\scriptsize(-)}{+}15x \overset{\color{red}\scriptsize(+)}{-}10 \\ \hline \hspace{6.3cm}2 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 3x^3 + 7x^2 + 4x + 5 \\ \color{green}{\text{Remainder}} &= 2 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} 9x^4 + 15x^3 - 2x^2 + 7x - 8 \\[6pt] RHS &= (3x - 2)(3x^3 + 7x^2 + 4x + 5) + 2 \\[4pt] &= 3x(3x^3 + 7x^2 + 4x + 5) - 2(3x^3 + 7x^2 + 4x + 5) + 2 \\[4pt] &= 9x^4 + 21x^3 + 12x^2 + 15x - 6x^3 - 14x^2 - 8x - 10 + 2 \\[4pt] RHS &\implies \color{green} 9x^4 + 15x^3 - 2x^2 + 7x - 8 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer Quotient \(=\color{red}{3x^3 + 7x^2 + 4x + 5}\), Remainder \(=\color{red}{2}\)

(c) \( 3x(5x^2 + 3x^3 + 2) - (2x^2 - x + 8) \ \text{ by } \ (2x + 3) \)

Solution

\[ \begin{aligned} \text{Dividend} &= (15x^3 + 9x^4 + 6x) - (2x^2 - x + 8) \\ &= \color{magenta}{9x^4 + 15x^3 - 2x^2 + 7x - 8} \\ \text{Divisor} &= \color{magenta}{2x + 3} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{\dfrac{9}{2}x^3 + \dfrac{3}{4}x^2 - \dfrac{17}{8}x + \dfrac{107}{16}} \\ 2x + 3\ \enclose{longdiv}{\ 9x^4 + 15x^3 - 2x^2 + 7x - 8}\\ \hspace{1.3cm}\overset{\color{red}\scriptsize(-)}{+}9x^4 \overset{\color{red}\scriptsize(-)}{+}\dfrac{27}{2}x^3\\ \hline \hspace{2.2cm}0 + \dfrac{3}{2}x^3 - 2x^2 + 7x - 8 \\ \hspace{2.5cm}\overset{\color{red}\scriptsize(-)}{+}\dfrac{3}{2}x^3 \overset{\color{red}\scriptsize(-)}{+}\dfrac{9}{4}x^2\\ \hline \hspace{3.2cm}0 - \dfrac{17}{4}x^2 + 7x - 8 \\ \hspace{3.5cm}\overset{\color{red}\scriptsize(+)}{-}\dfrac{17}{4}x^2 \overset{\color{red}\scriptsize(+)}{-}\dfrac{51}{8}x\\ \hline \hspace{4.4cm}0 + \dfrac{107}{8}x - 8 \\ \hspace{4.6cm}\overset{\color{red}\scriptsize(-)}{+}\dfrac{107}{8}x \overset{\color{red}\scriptsize(-)}{+}\dfrac{321}{16}\\ \hline \hspace{5.8cm} -\dfrac{449}{16} \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= \dfrac{9}{2}x^3 + \dfrac{3}{4}x^2 - \dfrac{17}{8}x + \dfrac{107}{16} \\ \color{green}{\text{Remainder}} &= -\dfrac{449}{16} \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} 9x^4 + 15x^3 - 2x^2 + 7x - 8 \\[6pt] RHS &= (2x + 3)\left(\dfrac{9}{2}x^3 + \dfrac{3}{4}x^2 - \dfrac{17}{8}x + \dfrac{107}{16}\right) - \dfrac{449}{16} \\[6pt] &= 2x \left(\dfrac{9}{2}x^3 + \dfrac{3}{4}x^2 - \dfrac{17}{8}x + \dfrac{107}{16}\right) +3 \left(\dfrac{9}{2}x^3 + \dfrac{3}{4}x^2 - \dfrac{17}{8}x + \dfrac{107}{16}\right) - \dfrac{449}{16} \\[6pt] &= 9x^4 + \dfrac{3}{2}x^3 - \dfrac{17}{4}x^2 + \dfrac{107}{8}x + \dfrac{27}{2}x^3 + \dfrac{9}{4}x^2 - \dfrac{51}{8}x + \dfrac{321}{16} - \dfrac{449}{16} \\[6pt] &= 9x^4 + \left(\dfrac{3}{2}+\dfrac{27}{2}\right)x^3 + \left(\dfrac{9}{4} -\dfrac{17}{4}\right)x^2 + \left(\dfrac{107}{8}-\dfrac{51}{8}\right)x + \left(\dfrac{321}{16} - \dfrac{449}{16}\right) \\[6pt] &= 9x^4 + \left(\dfrac{30}{2}\right)x^3 + \left(-\dfrac{8}{4}\right)x^2 + \left(\dfrac{56}{8}\right)x + \left(-\dfrac{128}{16}\right) \\[6pt] RHS &\implies \color{green} 9x^4 + 15x^3 - 2x^2 + 7x - 8 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer Quotient \(=\color{red}{\dfrac{9}{2}x^3 + \dfrac{3}{4}x^2 - \dfrac{17}{8}x + \dfrac{107}{16}}\), Remainder \(=\color{red}{-\dfrac{449}{16}}\)

20. Using long division method, check whether the second polynomial is a factor of first polynomial. Check the answer using division algorithm.

(a) \( (x^4 - 16) \ \text{ by } \ (x - 2) \)

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{x^4 + 0x^3 + 0x^2 + 0x - 16} \\ \text{Divisor} &= \color{magenta}{x - 2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{x^3 \ + \ 2x^2 \ + \ 4x \ + \ 8} \\ x - 2\ \enclose{longdiv}{\ x^4 + 0x^3 + 0x^2 + 0x - 16}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+}x^4 \overset{\color{red}\scriptsize(+)}{-}2x^3\\ \hline \hspace{1.8cm}0 + 2x^3 + 0x^2 + 0x - 16 \\ \hspace{2.1cm}\overset{\color{red}\scriptsize(-)}{+}2x^3 \overset{\color{red}\scriptsize(+)}{-}4x^2 \\ \hline \hspace{2.8cm}0 + 4x^2 + 0x - 16 \\ \hspace{3.1cm}\overset{\color{red}\scriptsize(-)}{+}4x^2 \overset{\color{red}\scriptsize(+)}{-}8x \\ \hline \hspace{3.8cm}0 + 8x - 16 \\ \hspace{4.1cm}\overset{\color{red}\scriptsize(-)}{+}8x \overset{\color{red}\scriptsize(+)}{-}16 \\ \hline \hspace{5.2cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= x^3 + 2x^2 + 4x + 8 \\ \color{green}{\text{Remainder}} &= 0 \\ (x-2) & \text{ is a factor of } (x^4 - 16) \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} x^4 - 16 \\[6pt] RHS &= (x - 2)(x^3 + 2x^2 + 4x + 8) + 0 \\[4pt] &= x(x^3 + 2x^2 + 4x + 8) - 2(x^3 + 2x^2 + 4x + 8) \\[4pt] &= x^4 + 2x^3 + 4x^2 + 8x - 2x^3 - 4x^2 - 8x - 16 \\[4pt] RHS &\implies \color{green} x^4 - 16 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer \(\color{red} \text{Yes, it is a factor (Remainder = 0).}\)

(b) \( (y^4 - 1) \ \text{ by } \ (y + 1) \)

Solution

\[ \begin{aligned} \text{Dividend} &= \color{magenta}{y^4 + 0y^3 + 0y^2 + 0y - 1} \\ \text{Divisor} &= \color{magenta}{y + 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.5cm}\color{green}{y^3 \ - \ y^2 \ + \ y \ - \ 1} \\ y + 1\ \enclose{longdiv}{\ y^4 + 0y^3 + 0y^2 + 0y - 1}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+}y^4 \overset{\color{red}\scriptsize(-)}{+}y^3\\ \hline \hspace{1.8cm}0 - y^3 + 0y^2 + 0y - 1 \\ \hspace{2.1cm}\overset{\color{red}\scriptsize(+)}{-}y^3 \overset{\color{red}\scriptsize(+)}{-}y^2 \\ \hline \hspace{2.8cm}0 + y^2 + 0y - 1 \\ \hspace{3.1cm}\overset{\color{red}\scriptsize(-)}{+}y^2 \overset{\color{red}\scriptsize(-)}{+}y \\ \hline \hspace{3.8cm}0 - y - 1 \\ \hspace{4.1cm}\overset{\color{red}\scriptsize(+)}{-}y \overset{\color{red}\scriptsize(+)}{-}1 \\ \hline \hspace{5.2cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= y^3 - y^2 + y - 1 \\ \color{green}{\text{Remainder}} &= 0 \\ (y+1) & \text{ is a factor of } (y^4 - 1) \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green} y^4 - 1 \\[6pt] RHS &= (y + 1)(y^3 - y^2 + y - 1) + 0 \\[4pt] &= y(y^3 - y^2 + y - 1) + 1(y^3 - y^2 + y - 1) \\[4pt] &= y^4 - y^3 + y^2 - y + y^3 - y^2 + y - 1 \\[4pt] RHS &\implies \color{green} y^4 - 1 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]

Answer \(\color{red} \text{Yes, it is a factor (Remainder = 0).}\)

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