DAV Class 8 Maths Chapter 8 Brain Teasers
Polynomials Brain Teasers
1. A. Tick (✓) the correct option.
(a)
Value of \(p\) for which \((x^2 + 3x + p)\) is divisible by \((x-2)\) is— \(
\begin{aligned}
(i)&\ 10 \\
(ii)&\ -10 \\
(iii)&\ 5 \\
(iv)&\ -5 \\
\end{aligned}
\)
Solution
\[ \begin{array}{l} \hspace{1.7cm}\color{green}{x \ + \ 5} \\ x-2\ \enclose{longdiv}{\ x^2 + 3x + p}\\ \hspace{1cm}\overset{\color{red}\scriptsize(-)}{+}x^2 \overset{\color{red}\scriptsize(+)}{-}2x\\ \hline \hspace{1.5cm}0 + 5x + p\\ \hspace{2cm}\overset{\color{red}\scriptsize(-)}{+}5x \overset{\color{red}\scriptsize(+)}{-}10\\ \hline \hspace{3.25cm}p+10 \end{array} \] \[ \begin{aligned} \text{Remainder } \implies & p+10=0 \\ p &=-10 \end{aligned}\]Answer \( {\color{orange}(ii)}\ \color{red}{-10} \)
(b)
Quotient obtained when \((x^2 + 7x - 4)\) is divided by \((x+7)\) is— \(
\begin{aligned}
(i)&\ x \\
(ii)&\ x-4 \\
(iii)&\ x+4 \\
(iv)&\ x+2
\end{aligned}
\)
Solution
\[ \begin{array}{l} \hspace{1.7cm}\color{green}{x} \\ x+7\ \enclose{longdiv}{\ x^2+7x-4}\\ \hspace{1cm}\overset{\color{red}\scriptsize(-)}{+}x^2 \overset{\color{red}\scriptsize(-)}{+}7x\\ \hline \hspace{2.5cm}0 -4 \end{array} \] \[ \begin{aligned} {\text{Quotient}}&=\color{green}x \\ \text{Remainder}&=\color{green}-4 \end{aligned} \]Answer \( {\color{orange}(i)}\ \color{red}{x} \)
(c)
Polynomial which when divided by \((-x^2 + x - 1)\) gives a quotient \((x - 2)\) and remainder \(3\) is— \(
\begin{aligned}
(i)&\ x^3-3x^2+3x-5\\
(ii)&\ -x^3+3x^2-3x+5\\
(iii)&\ -x^3+3x^2-3x-5\\
(iv)&\ x^3-3x^2-3x+5
\end{aligned}
\)
Solution
\[ \begin{aligned} \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor } \times \text{ Quotient } + \text{ Remainder}} \\[6pt]&=\left[(-x^2 + x - 1)(x - 2)\right] + 3 \\[4pt] &=\left[-x^2(x-2) + x(x - 2) - 1(x - 2)\right] + 3 \\[4pt] &= -x^3 + 2x^2 + x^2 - 2x - x + 2 + 3 \\[4pt] \color{magenta}{\text{Dividend}} &= -x^3 + 3x^2 - 3x +5 \\[4pt] \end{aligned} \]Answer \( {\color{orange}(ii)}\ \color{red}{-x^3+3x^2-3x+5} \)
(d)
What should be added to \(x^2-5x+4\) so that \((x-3)\) is a factor? \(
\begin{aligned}
(i)&\ 1 \\
(ii)&\ 2 \\
(iii)&\ 4 \\
(iv)&\ 5
\end{aligned}
\)
Solution
\[ \begin{array}{l} \hspace{1.7cm}\color{green}{x \ - \ 2} \\ x-3\ \enclose{longdiv}{\ x^2-5x+4}\\ \hspace{1cm}\overset{\color{red}\scriptsize(-)}{+}x^2 \overset{\color{red}\scriptsize(+)}{-}3x\\ \hline \hspace{1.5cm}0 - 2x + 4\\ \hspace{1.9cm}\overset{\color{red}\scriptsize(+)}{-}2x \overset{\color{red}\scriptsize(-)}{+}6\\ \hline \hspace{2.8cm}-2 \end{array} \] \[ \begin{aligned} \text{Remainder}=-2 \\ \text{add } \boxed{2}\ \text{to make remainder }0 \end{aligned} \]Answer \( {\color{orange}(ii)}\ \color{red}{2} \)
(e)
What should be subtracted from \(x^2-16x+30\) so that \((x-15)\) is a factor? \(
\begin{aligned}
(i)&\ 30 \\
(ii)&\ 14 \\
(iii)&\ 15 \\
(iv)&\ 16
\end{aligned}
\)
Solution
\[ \begin{array}{l} \hspace{1.7cm}\color{green}{x \ - \ 1} \\ x-15\ \enclose{longdiv}{\ x^2-16x+30}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+}x^2\ \ \overset{\color{red}\scriptsize(+)}{-}15x\\ \hline \hspace{2.05cm}0\ -\ x\ +\ 30\\ \hspace{2.4cm}\overset{\color{red}\scriptsize(+)}{-}x\ \ \overset{\color{red}\scriptsize(-)}{+}15\\ \hline \hspace{3.35cm}15 \end{array} \] \[ \begin{aligned} \text{Remainder}=15\\ \boxed{\color{magenta}15} \text{ must be subtracted} \end{aligned} \]Answer \( {\color{orange}(iii)}\ \color{red}{15} \)
B. Answer the following questions.
(a) Find the difference between the quotient and remainder when \((x^3-6x^2+11x-6)\) is divided by \((x+1)\).
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{x^3-6x^2+11x-6}\\ \text{Divisor} &= \color{magenta}{x+1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.6cm}\color{green}{x^2 \ - \ 7x \ + \ 18} \\ x+1\ \enclose{longdiv}{\ x^3 - 6x^2 + 11x - 6}\\ \hspace{1.05cm}\overset{\color{red}\scriptsize(-)}{+}x^3 \overset{\color{red}\scriptsize(-)}{+}x^2 \\ \hline \hspace{1.7cm}0 - 7x^2 + 11x - 6 \\ \hspace{2.1cm}\overset{\color{red}\scriptsize(+)}{-} 7x^2 \overset{\color{red}\scriptsize(+)}{-}7x \\ \hline \hspace{2.9cm}0 + 18x - 6\\ \hspace{3.25cm}\overset{\color{red}\scriptsize(-)}{+} 18x \overset{\color{red}\scriptsize(-)}{+}18 \\ \hline \hspace{4.35cm}-24 \end{array} \] \[ \begin{aligned} \text{Quotient} &= \color{green}{x^2-7x+18}\\ \text{Remainder} &= \color{green}{-24} \\[6pt] &\text{Difference} \\ & = \text{Quotient} - \text{Remainder} \\ & = (x^2-7x+18)-(-24) \\ &= x^2-7x+18 + 24 \\ &= x^2 - 7x + 42 \\ \end{aligned} \]Answer \( \color{red}{x^2-7x+42} \)
(b) If \((x-3)\) is a factor of \((2x^2+x+k)\), find \(k\).
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{2x^2+x+k}\\ \text{Divisor} &= \color{magenta}{x-3} \end{aligned} \] \[ \begin{array}{l} \hspace{1.6cm}\color{green}{2x \ + \ 7} \\ x-3\ \enclose{longdiv}{\ 2x^2 + x + k}\\ \hspace{1.05cm}\overset{\color{red}\scriptsize(-)}{+}2x^2 \overset{\color{red}\scriptsize(+)}{-}6x \\ \hline \hspace{1.5cm}0 + 7x + k\\ \hspace{1.8cm}\overset{\color{red}\scriptsize(-)}{+}7x \overset{\color{red}\scriptsize(+)}{-}21 \\ \hline \hspace{2.5cm}0+k+21 \end{array} \] \[ \begin{aligned} \text{Remainder} & = k + 21 \\ (x-3) & \text{ is a factor of } (2x^2+x+k) \\ \\ \text{Remainder} & = 0 \\ k+21& =0 \\ k&=-21 \end{aligned} \]Answer \( \color{red}{k=-21} \)
(c) Can \((x-1)\) be the remainder on division of a polynomial \(p(x)\) by \((x+3)\)? Justify.
Answer We see that degree of both \( (x + 3) \) and \((x - 1)\) is the same.
We know that degree of remainder must be less than that of divisor.
So, \((x - 1)\) can’t be the remainder on division of a polynomial \(p(x)\) by \((x + 3)\)
(d) Find the remainder when \((x^3 + x - 4x^2 + 6)\) is divided by \((x+2)\). Conclude whether \((x+2)\) is a factor.
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{x^3-4x^2+x+6}\\ \text{Divisor} &= \color{magenta}{x+2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.6cm}\color{green}{x^2 \ - \ 6x \ + \ 13} \\ x+2\ \enclose{longdiv}{\ x^3 - 4x^2 + x + 6}\\ \hspace{1.05cm}\overset{\color{red}\scriptsize(-)}{+}x^3 \overset{\color{red}\scriptsize(-)}{+}2x^2 \\ \hline \hspace{1.5cm}0 - 6x^2 + x + 6\\ \hspace{1.7cm}\overset{\color{red}\scriptsize(+)}{-}6x^2 \overset{\color{red}\scriptsize(+)}{-}12x \\ \hline \hspace{2.30cm}0 + 13x + 6\\ \hspace{2.6cm}\overset{\color{red}\scriptsize(-)}{+}13x \overset{\color{red}\scriptsize(-)}{+}26 \\ \hline \hspace{3.8cm}-20 \end{array} \] \[ \begin{aligned} \color{green}{\text{Remainder}} =-20 \\ (x+2)\ \text{is not a factor} \end{aligned} \]Answer Remainder \( \color{red}{-20} \ \); \((x+2)\) is not a factor.
(e) If \((8y+8y^2+7)\) is divided by \((1-y)\), find the remainder.
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{8y^2+8y+7}\\ \text{Divisor} &= \color{magenta}{-y+1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.6cm}\color{green}{-8y \ - \ 16} \\ -y+1\ \enclose{longdiv}{\ 8y^2 + 8y + 7}\\ \hspace{1.3cm}\overset{\color{red}\scriptsize(-)}{+}8y^2 \overset{\color{red}\scriptsize(+)}{-}8y \\ \hline \hspace{1.9cm}0 + 16y + 7\\ \hspace{2.1cm}\overset{\color{red}\scriptsize(-)}{+}16y \overset{\color{red}\scriptsize(+)}{-}16 \\ \hline \hspace{3.5cm}23 \end{array} \] \[ \color{green}{\text{Remainder}}=\boxed{23} \]Answer \( Remainder = \color{red}{23} \)
2. Divide:
(i) \( -39x^4 \text{ by } \sqrt{13}\,x^2 \)
Solution
\[ \begin{aligned} & = \frac{-39x^4}{\sqrt{13} \ x^2} \\[8pt] &= \frac{-39 \ {\color{green} \times \sqrt{13}} }{\sqrt{13} \ {\color{green} \times \sqrt{13}}} \ x^{4-2} \\[8pt] &= \frac{-\cancel{39}^3\sqrt{13}}{\cancel{13}_1} \ x^2 \\[8pt] &= -3\sqrt{13}\,x^2 \end{aligned} \]Answer \( \color{red}{-3\sqrt{13}\,x^2} \)
(ii) \( \sqrt{125}\,y^2 \text{ by } 5y^2 \)
Solution
\[ \begin{aligned} & = \frac{\sqrt{125}\,y^2}{5y^2} \\[8pt] &= \frac{\sqrt{5 \times 5\times 5} \ \cancel{y^2}}{5 \ \cancel{y^2}} \\[8pt] &= \frac{\cancel5\sqrt{5}}{\cancel5} \\[8pt] &= \sqrt{5} \end{aligned} \]Answer \( \color{red}{\sqrt{5}} \)
(iii) \( 49p^3 \text{ by } -7\sqrt{7}\,p^2 \)
Solution
\[ \begin{aligned} & = \frac{49p^3}{-7\sqrt{7}\,p^2} \\[8pt] &= \frac{-49 \ {\color{green} \times \sqrt{7}} }{7\sqrt{7} \ {\color{green} \times \sqrt{7}}}\,p^{3-2} \\[8pt] &= \frac{-49 \sqrt{7}}{7 \times 7}\,p \\[8pt] &= \frac{-\cancel{49} \sqrt{7}}{\cancel{49}} p \\[8pt] &= -\sqrt{7}\,p \end{aligned} \]Answer \( \color{red}{-\sqrt{7}\,p} \)
(iv) \( 14z^6 \text{ by } \dfrac{14}{3}\,z^3 \)
Solution
\[ \begin{aligned} & = \frac{14z^6}{\dfrac{14}{3}z^3} \\[8pt] & = \frac{\cancel{14} \times 3}{\cancel{14}} z^{6-3} \\[8pt] &= 3z^3 \end{aligned} \]Answer \( \color{red}{3z^3} \)
(v) \( 6x^3 - 4x^2 + 8x \text{ by } \dfrac{2}{3}x \)
Solution
\[ \begin{aligned} & = \frac{6x^3 - 4x^2 + 8x}{\dfrac{2}{3}x} \\[8pt] &= \frac{6x^3}{\dfrac{2}{3}x} - \frac{4x^2}{\dfrac{2}{3}x} + \frac{8 \cancel x}{\dfrac{2}{3}\cancel x} \\[8pt] &= \left(\frac{\cancel 6^{3} \times3}{\cancel2}\right)x^{3-2} \ - \left(\frac{ \cancel4^2 \times 3}{\cancel2}\right)x^{2-1} \ + \left(\frac{\cancel8^4 \times 3}{\cancel2}\right) \\[8pt] &= 9x^2 - 6x + 12 \end{aligned} \]Answer \( \color{red}{9x^2 - 6x + 12} \)
(vi) \( y^2 - 5y + 1 \text{ by } -\dfrac{1}{3}y \)
Solution
\[ \begin{aligned} & = \frac{y^2 - 5y + 1}{-\dfrac{1}{3}y} \\[8pt] & = \frac{y^2}{-\dfrac{y}{3}} - \frac{5y}{-\dfrac{y}{3}} + \frac{1}{-\dfrac{y}{3}} \\[8pt] & = \frac{-3y^{2} }{y} + \frac{15y}{y} - \frac{3}{y} \\[8pt] &= -3y + 15 - \frac{3}{y} \end{aligned} \]Answer \( \color{red}{-3y + 15 - \dfrac{3}{y}} \)
(vii) \( 18a^3 - 12a^2 + 9a \text{ by } 3a \)
Solution
\[ \begin{aligned} & = \frac{18a^3 - 12a^2 + 9a}{3a} \\[8pt] &= \frac{18a^3}{3a} - \frac{12a^2}{3a} + \frac{9a}{3a} \\[8pt] &= \frac{\cancel{18}^6}{\cancel3_1}a^{3-1} - \frac{\cancel{12}^4}{\cancel3_1}a^{2-1} + \frac{\cancel{9a}^3}{\cancel{3a}_1} \\[8pt] &= 6a^2 - 4a + 3 \end{aligned} \]Answer \( \color{red}{6a^2 - 4a + 3} \)
3. Divide and write down the quotient and remainder. Also check your answer.
(i) \(4x^2 + 7x - 11 \text{ by } 2x - 4\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{4x^2 + 7x - 11} \\ \text{Divisor} &= \color{magenta}{2x - 4} \end{aligned} \] \[ \begin{array}{l} \hspace{1.55cm}\color{green}{2x \ + \ \dfrac{15}{2}} \\ 2x - 4 \enclose{longdiv}{\ \ 4x^2 + 7x - 11}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+} 4x^2 \overset{\color{red}\scriptsize(+)}{-} 8x \\ \hline \hspace{1.8cm}0 + 15x - 11 \\ \hspace{2.25cm}\overset{\color{red}\scriptsize(-)}{+} 15x \overset{\color{red}\scriptsize(+)}{-} 30 \\ \hline \hspace{3.7cm} 19 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 2x + \dfrac{15}{2} \\ \color{green}{\text{Remainder}} &= 19 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green}4x^2 + 7x - 11 \\[6pt]RHS&= (2x-4) \left(2x+\frac{15}{2}\right) + 19 \\[6pt] &= 2x \left(2x+\frac{15}{2}\right) -4\left(2x+\frac{15}{2}\right) + 19 \\[6pt] &= 4x^2+15x-8x-30 + 19 \\[6pt] RHS &\implies \color{green} 4x^2 + 7x - 11 \\[6pt] LHS & = RHS \\ & \text{Hence verified} \end{aligned} \](ii) \(125 - 225x + 135x^2 - 27x^3 \text{ by } 5 - 3x\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{-27x^3 + 135x^2 - 225x + 125} \\ \text{Divisor} &= \color{magenta}{-3x + 5} \end{aligned} \] \[ \begin{array}{l} \hspace{2.1cm}\color{green}{9x^2 \ - \ 30x \ + \ 25} \\ -3x + 5 \enclose{longdiv}{\ -27x^3 + 135x^2 - 225x + 125}\\ \hspace{2cm}\overset{\color{red}\scriptsize(+)}{-}27x^3 \overset{\color{red}\scriptsize(-)}{+}45x^2 \\ \hline \hspace{3cm}0 + 90x^2 - 225x + 125 \\ \hspace{3.35cm}\overset{\color{red}\scriptsize(-)}{+} 90x^2 \overset{\color{red}\scriptsize(+)}{-}150x \\ \hline \hspace{4.3cm}0 - 75x + 125 \\ \hspace{4.7cm}\overset{\color{red}\scriptsize(+)}{-} 75x \overset{\color{red}\scriptsize(-)}{+}125 \\ \hline \hspace{6cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 9x^2 - 30x + 25 \\ \color{green}{\text{Remainder}} &= 0 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient} + \text{Remainder}} \\[6pt] LHS &\implies \color{green} -27x^3 + 135x^2 - 225x + 125 \\[6pt] RHS&= (-3x+5)\,(9x^2 - 30x + 25) \\[6pt] &= - 3x(9x^2-30x+25) + 5(9x^2-30x+25) \\[6pt] &= - 27x^3 + 90x^2 - 75x + 45x^2 - 150x + 125 \\[6pt] RHS &\implies \color{green} -27x^3 + 135x^2 - 225x + 125 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \](iii) \(y^3 + 5y^2 + 12y + 9 \text{ by } y + 2\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{y^3 + 5y^2 + 12y + 9} \\ \text{Divisor} &= \color{magenta}{y + 2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.55cm}\color{green}{y^2 \ + \ 3y \ + \ 6} \\ y + 2 \enclose{longdiv}{\ \ y^3 + 5y^2 + 12y + 9}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+} y^3 \overset{\color{red}\scriptsize(-)}{+} 2y^2 \\ \hline \hspace{1.6cm}0 + 3y^2 + 12y + 9\\ \hspace{2cm}\overset{\color{red}\scriptsize(-)}{+} 3y^2 \overset{\color{red}\scriptsize(-)}{+} 6y \\ \hline \hspace{2.7cm}0 + 6y + 9 \\ \hspace{3cm}\overset{\color{red}\scriptsize(-)}{+} 6y \overset{\color{red}\scriptsize(+)}{-} 12 \\ \hline \hspace{4.1cm}-3 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= y^2 + 3y + 6 \\ \color{green}{\text{Remainder}} &= -3 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green}y^3 + 5y^2 + 12y + 9 \\[6pt] RHS&= (y+2)(y^2+3y+6) + (-3) \\[6pt] &= \big[y(y^2+3y+6) + 2(y^2+3y+6)\big] - 3 \\[6pt] &= y^3+3y^2+6y + 2y^2+6y+12 - 3 \\[6pt] RHS &\implies \color{green} y^3 + 5y^2 + 12y + 9 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \](iv) \(8p^3 - 729 - 108p^2 + 486p \text{ by } 2p - 9\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{8p^3 - 108p^2 + 486p - 729} \\ \text{Divisor} &= \color{magenta}{2p - 9} \end{aligned} \] \[ \begin{array}{l} \hspace{1.55cm}\color{green}{4p^2 \ - \ 36p \ + \ 81} \\ 2p - 9 \enclose{longdiv}{\ \ 8p^3 - 108p^2 + 486p - 729}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+} 8p^3 \overset{\color{red}\scriptsize(+)}{-} 36p^2 \\ \hline \hspace{1.9cm}0 - 72p^2 + 486p - 729 \\ \hspace{2.25cm}\overset{\color{red}\scriptsize(+)}{-} 72p^2 \overset{\color{red}\scriptsize(-)}{+} 324p \\ \hline \hspace{3.35cm}0 + 162p - 729 \\ \hspace{3.7cm}\overset{\color{red}\scriptsize(-)}{+} 162p \overset{\color{red}\scriptsize(+)}{-} 729 \\ \hline \hspace{4.7cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 4p^2 - 36p + 81 \\ \color{green}{\text{Remainder}} &= 0 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient} +\text{Remainder}} \\[6pt] LHS &\implies \color{green}8p^3 - 108p^2 + 486p - 729 \\[6pt] RHS&= (2p-9)(4p^2-36p+81) \\[6pt] &= 2p(4p^2-36p+81) - 9(4p^2-36p+81) \\[6pt] &= 8p^3 - 72p^2 + 162p - 36p^2 + 324p - 729 \\[6pt] RHS &\implies \color{green} 8p^3 - 108p^2 + 486p - 729 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \](v) \(q^4 + 3q^2 - 4 \text{ by } q^2 - 1\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{q^4 + 0q^3 + 3q^2 + 0q - 4} \\ \text{Divisor} &= \color{magenta}{q^2 - 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.55cm}\color{green}{q^2 \ + \ 4} \\ q^2 - 1 \enclose{longdiv}{\ \ q^4 + 0q^3 + 3q^2 + 0q - 4}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+} q^4 \hspace{1.2cm} \overset{\color{red}\scriptsize(+)}{-} q^2 \\ \hline \hspace{2.6cm}0 + 4q^2 + 0q - 4 \\ \hspace{3.cm}\overset{\color{red}\scriptsize(-)}{+} 4q^2 \hspace{.9cm} \overset{\color{red}\scriptsize(+)}{-} 4 \\ \hline \hspace{4.35cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= q^2 + 4 \\ \color{green}{\text{Remainder}} &= 0 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient} +\text{Remainder}} \\[6pt] LHS &\implies \color{green}q^4 + 3q^2 - 4 \\[6pt] RHS&= (q^2-1)(q^2+4) \\[6pt] &= q^2(q^2+4) - 1(q^2+4) \\[6pt] &= q^4 + 4q^2 - q^2 - 4 \\[6pt] RHS &\implies \color{green} q^4 + 3q^2 - 4 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \](vi) \(p^4 + \dfrac{63}{4}p^2 - 5 \text{ by } p^2 - \dfrac{1}{4}\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{p^4 + 0p^3 + \frac{63}{4}p^2 + 0p - 5} \\ \text{Divisor} &= \color{magenta}{p^2 - \frac{1}{4}} \end{aligned} \] \[ \begin{array}{l} \hspace{1.55cm}\color{green}{p^2 \ + \ 16} \\ p^2 - \dfrac{1}{4} \enclose{longdiv}{\ p^4 + 0p^3 + \dfrac{63}{4}p^2 + 0p - 5}\\ \hspace{1.3cm}\overset{\color{red}\scriptsize(-)}{+} p^4 \hspace{1.2cm} \overset{\color{red}\scriptsize(+)}{-} \dfrac{1}{4}p^2 \\ \hline \hspace{2.9cm}0 + 16p^2 + 0p - 5 \\ \hspace{3.3cm}\overset{\color{red}\scriptsize(-)}{+} 16p^2 \hspace{.9cm} \overset{\color{red}\scriptsize(+)}{-} 4 \\ \hline \hspace{5.5cm}-1 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= p^2 + 16 \\ \color{green}{\text{Remainder}} &= -1 \\[4pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor}\times\text{Quotient}+\text{Remainder}} \\[6pt] LHS &\implies \color{green}p^4 + \frac{63}{4}p^2 - 5 \\[6pt] RHS&= \left(p^2-\frac14\right)(p^2+16) \ +\ (-1) \\[6pt] &= \left[p^2(p^2+16) - \frac14(p^2+16)\right] - 1 \\[6pt] &= p^4 + 16p^2 - \frac14 p^2 - 4 - 1 \\[6pt] RHS &\implies \color{green} p^4 + \frac{63}{4}p^2 - 5 \\[6pt] LHS &= RHS \\ & \text{Hence verified} \end{aligned} \]4. Find whether or not, the first polynomial is a factor of the second polynomial.
(i) \(3x+2 \ , \ 3x^4+5x^3-x^2+13x+10 \)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{3x^4 + 5x^3 - x^2 + 13x + 10} \\ \text{Divisor} &= \color{magenta}{3x + 2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.45cm}\color{green}{x^3 \ + \ x^2 \ - \ x \ + \ 5} \\ 3x + 2 \enclose{longdiv}{\ 3x^4 + 5x^3 - x^2 + 13x + 10}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+} 3x^4 \overset{\color{red}\scriptsize(-)}{+} 2x^3 \\ \hline \hspace{1.9cm}0 + 3x^3 - x^2 + 13x + 10\\ \hspace{2.3cm}\overset{\color{red}\scriptsize(-)}{+} 3x^3 \overset{\color{red}\scriptsize(-)}{+} 2x^2 \\ \hline \hspace{3cm}0 - 3x^2 + 13x + 10\\ \hspace{3.4cm}\overset{\color{red}\scriptsize(+)}{-} 3x^2 \overset{\color{red}\scriptsize(+)}{-} 2x \\ \hline \hspace{4.1cm}0 + 15x + 10\\ \hspace{4.4cm}\overset{\color{red}\scriptsize(-)}{+} 15x \overset{\color{red}\scriptsize(-)}{+} 10 \\ \hline \hspace{5.6cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= x^3 + x^2 - x + 5 \\ \color{green}{\text{Remainder}} &= 0 \end{aligned} \]Answer Since the remainder is \( \color{red} 0 \) , \( \color{red}{(3x+2)}\) is a \( \color{red} factor \) of \(\color{red}{3x^4+5x^3-x^2+13x+10}\)
(ii) \(x^2+1 \ , \ x^4-3x^3-4x^2+3x+2\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{x^4 - 3x^3 - 4x^2 + 3x + 2} \\ \text{Divisor} &= \color{magenta}{x^2 + 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.55cm}\color{green}{x^2 \ - \ 3x \ - \ 5} \\ x^2 + 1 \enclose{longdiv}{\ x^4 - 3x^3 - 4x^2 + 3x + 2}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+} x^4 \hspace{1.1cm} \overset{\color{red}\scriptsize(-)}{+} x^2 \\ \hline \hspace{1.7cm}0 - 3x^3 - 5x^2 + 3x + 2\\ \hspace{2.05cm}\overset{\color{red}\scriptsize(+)}{-} 3x^3 \hspace{1.2cm} \overset{\color{red}\scriptsize(+)}{-} 3x \\ \hline \hspace{2.8cm}0 - 5x^2 + 6x + 2\\ \hspace{3.15cm}\overset{\color{red}\scriptsize(+)}{-} 5x^2 \hspace{1cm} \overset{\color{red}\scriptsize(+)}{-} 5 \\ \hline \hspace{4.6cm} 6x + 7 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= x^2 - 3x - 5 \\ \color{green}{\text{Remainder}} &= 6x + 7 \\[6pt] \end{aligned} \]Answer Since the remainder is \(\color{red}{(6x+7)}\), \(\color{red}{(x^2+1)}\) is \( \color{red} not \) a factor of \(\color{red}{x^4-3x^3-4x^2+3x+2}\)