DAV Class 8 Maths Chapter 8 Worksheet 3
Polynomials Worksheet 3
1. Using factor method, divide the following polynomials by a binomial.
(i) \( x^2 + 3x + 2 \text{ by }x+1 \)
Solution
\[ \begin{align*} &= \frac{ x^2 \ {\color{magenta} + \ 3x} + 2}{(x+1)} \\[8pt] &= \frac{ x^2 \ {\color{magenta} + \ 2x + 1x} + 2}{(x+1)} \\[8pt] &= \frac{ x(x + 2) + 1(x + 2)}{(x+1)} \\[8pt] &= \frac{ (x + 2)\cancel{(x+1)}}{\cancel{(x+1)}} \\[8pt] &= x + 2 \end{align*} \]
Answer \( \color{red} x + 2 \)
(ii) \( x^2 - 7x - 18 \text{ by } x-9 \)
Solution
\[ \begin{aligned} &= \frac{ x^2 \ {\color{magenta} - \ 7x} - 18}{(x-9)} \\[8pt] &= \frac{x^2 \ {\color{magenta}- \ 9x+ 2x} - 18}{(x-9)} \\[6pt] &= \frac{x(x-9) + 2(x-9)}{(x-9)} \\[6pt] &= \frac{\cancel{(x-9)}(x+2)}{\cancel{(x-9)}} \\[6pt] &= \ x+2 \end{aligned} \]Answer \( \color{red}{x+2} \)
(iii) \( p^2 + 6p + 8 \text{ by } p+2 \)
Solution
\[ \begin{aligned} &= \frac{ p^2 \ {\color{magenta} + \ 6p} + 8}{(p+2)} \\[8pt] &= \frac{ p^2 \ {\color{magenta} + \ 2p + 4p} + 8}{(p+2)} \\[8pt] &= \frac{ p(p+2) + 4(p+2)}{(p+2)} \\[8pt] &= \frac{ \cancel{(p+2)}(p+4)}{\cancel{(p+2)}} \\[8pt] &= \ p+4 \end{aligned} \]Answer \( \color{red}{p+4} \)
(iv) \( p^2 - p - 42 \text{ by } p+6 \)
Solution
\[ \begin{aligned} &= \frac{ p^2 \ {\color{magenta} - \ p} - 42}{(p+6)} \\[8pt] &= \frac{ p^2 \ {\color{magenta} + \ 6p - 7p} - 42}{(p+6)} \\[8pt] &= \frac{ p(p+6) - 7(p+6)}{(p+6)} \\[8pt] &= \frac{ \cancel{(p+6)}(p-7)}{\cancel{(p+6)}} \\[8pt] &= \ p-7 \end{aligned} \]Answer \( \color{red}{p-7} \)
(v) \( y^2 + 12y + 35 \text{ by } y+7 \)
Solution
\[ \begin{aligned} &= \frac{ y^2 \ {\color{magenta} + \ 12y} + 35}{(y+7)} \\[8pt] &= \frac{ y^2 \ {\color{magenta} + \ 7y + 5y} + 35}{(y+7)} \\[8pt] &= \frac{ y(y+7) + 5(y+7)}{(y+7)} \\[8pt] &= \frac{ \cancel{(y+7)}(y+5)}{\cancel{(y+7)}} \\[8pt] &= \ y+5 \end{aligned} \]Answer \( \color{red}{y+5} \)
(vi) \( z^2 - 10z + 16 \text{ by } z-2 \)
Solution
\[ \begin{aligned} &= \frac{ z^2 \ {\color{magenta} - \ 10z} + 16}{(z-2)} \\[8pt] &= \frac{ z^2 \ {\color{magenta} - \ 2z - 8z} + 16}{(z-2)} \\[8pt] &= \frac{ z(z-2) - 8(z-2)}{(z-2)} \\[8pt] &= \frac{ \cancel{(z-2)}(z-8)}{\cancel{(z-2)}} \\[8pt] &= \ z-8 \end{aligned} \]Answer \( \color{red}{z-8} \)
(vii) \( x^4 + 3x^2 - 10 \text{ by } x^2+5 \)
Solution
\[ \begin{aligned} & \color{red} \text{Method - 1} \\[6pt] &= \frac{x^4 + 3x^2 - 10}{x^2+5} \\[6pt] \text{Put }& x^2 = y \\[4pt] &= \frac{ y^2 \ {\color{magenta} + \ 3y} - 10}{(y+5)} \\[8pt] &= \frac{ y^2 \ {\color{magenta} + \ 5y - 2y} - 10}{(y+5)} \\[8pt] &= \frac{ y(y+5) - 2(y+5)}{(y+5)} \\[8pt] &= \frac{ \cancel{(y+5)}(y-2)}{\cancel{(y+5)}} \\[8pt] &= \ y-2 \\[4pt] & \implies x^2-2 \\ \\ & \color{red} \text{Method - 2} \\[6pt] &= \frac{ x^4 \ {\color{magenta} + \ 3x^2} - 10}{(x^2 + 5)} \\[8pt] &= \frac{x^4 \ {\color{magenta} + \ 5x^2 - 2x^2} - 10}{(x^2 + 5)} \\[6pt] &= \frac{x^2(x^2 + 5) - 2(x^2 + 5)}{(x^2 + 5)} \\[6pt] &= \frac{\cancel{(x^2 + 5)}(x^2 - 2)}{\cancel{(x^2 + 5)}} \\[6pt] &= x^2 - 2\end{aligned} \]Answer \( \color{red}{x^2-2} \)
(viii) \( x^2 - 7x + 12 \text{ by } x-4 \)
Solution
\[ \begin{aligned} &= \frac{ x^2 \ {\color{magenta} - \ 7x} + 12}{(x-4)} \\[8pt] &= \frac{ x^2 \ {\color{magenta} - \ 4x - 3x} + 12}{(x-4)} \\[8pt] &= \frac{ x(x-4) - 3(x-4)}{(x-4)} \\[8pt] &= \frac{ \cancel{(x-4)}(x-3)}{\cancel{(x-4)}} \\[8pt] &= \ x-3 \end{aligned} \]Answer \( \color{red}{x-3} \)
2. Using long division method, divide the following polynomials by a binomial and check your answer.
(i) \(4p^3 - 4p^2 + 6p - \dfrac{5}{2} \text{ by } 2p - 1 \)
Solution
\[ \begin{aligned} \text{Dividend} & = \color{magenta}4p^3 - 4p^2 + 6p - \frac{5}{2} \\[6pt] \text{Divisor} & = \color{magenta} 2p - 1 \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm} \color{green} 2p^2 \ - \ 1p \ + \ \dfrac{5}{2} \\ 2p - 1 \enclose{longdiv}{ \ \ 4p^3 - 4p^2 + 6p - \dfrac{5}{2}}\\ \hspace{1.1cm}\overset{\color{red} \scriptsize (-)}{+} \ 4p^3 \overset{\color{red} \scriptsize (+)}{-} \ 2p^2 \\ \hline \hspace{1.8cm}0 \ - \ 2p^2 + 6p - \dfrac{5}{2} \\ \hspace{2.3cm}\overset{\color{red} \scriptsize (+)}{-} \ 2p^2 \overset{\color{red} \scriptsize (-)}{+} 1p \\ \hline \hspace{3cm}0 \ + 5p - \dfrac{5}{2} \\[6pt] \hspace{3.4cm}\overset{\color{red} \scriptsize (-)}{+} \ 5p \overset{\color{red} \scriptsize (+)}{-} \ \dfrac{5}{2} \\ \hline \hspace{4.5cm} 0 \\ \end{array} \] \[ \begin{aligned} \color{green}\text{Quotient} &= 2p^2 - 1p \ + \frac{5}{2} \\ \color{green}\text{Remainder} &= 0 \\ \\ &\text{Check} \\ \color{magenta} \text{Dividend} & = \color{magenta} \text{Divisor } \times \text{ Quotient + Remainder }\\[6pt] LHS & \implies \color{green} 4p^3 - 4p^2 + 6p - \frac{5}{2} \\[6pt] RHS & = \text{Divisor } \times \text{ Quotient + Remainder } \\[6pt] & = \left[(2p - 1) \times \left( 2p^2 - 1p + \frac{5}{2} \right)\right] + 0 \\[6pt] & = \left[2p \left( 2p^2 - 1p + \frac{5}{2} \right) - 1 \left( 2p^2 - 1p + \frac{5}{2} \right)\right] \\[6pt] & = \left[4p^3 - 2p^2 + 5p - 2p^2 + 1p - \frac{5}{2} \right] \\[6pt] & = 4p^3 - 2p^2 - 2p^2 + 5p + 1p - \frac{5}{2} \\[6pt] RHS & \implies \color{green} 4p^3 - 4p^2 + 6p - \frac{5}{2} \\[6pt] LHS & = RHS \\ & \text{Hence verified} \end{aligned} \](ii) \( x - 6 + 15x^2 \text{ by } 2 + 3x \)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{15x^2 + x - 6} \\[6pt] \text{Divisor} &= \color{magenta}{3x + 2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm} \color{green}{5x \ - \ 3} \\ 3x + 2 \enclose{longdiv}{\ \ 15x^2 + x - 6}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+} 15x^2 \overset{\color{red}\scriptsize(-)}{+}10x \\ \hline \hspace{1.9cm}0 - 9x - 6 \\ \hspace{2.3cm}\overset{\color{red}\scriptsize(+)}{-}9x \overset{\color{red}\scriptsize(+)}{-}6 \\ \hline \hspace{3.3cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 5x - 3 \\ \color{green}{\text{Remainder}} &= 0 \\[6pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor } \times \text{ Quotient } + \text{ Remainder}} \\[6pt] LHS & \implies \color{green} 15x^2 + x - 6 \\[6pt] RHS & = \text{Divisor } \times \text{ Quotient + Remainder } \\[6pt] & = [(3x+2)\times(5x-3)] + 0 \\[6pt] & = [3x(5x-3) +2 (5x-3)] \\[6pt] &= 15x^2 - 9x + 10x - 6 \\[6pt] RHS & \implies \color{green} 15x^2 + x - 6 \\[6pt] LHS & = RHS \\ & \text{Hence verified} \end{aligned} \](iii) \(4x^3 - 37x^2 + 52x - 15 \text{ by } 4x - 5\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{4x^3 - 37x^2 + 52x - 15} \\[6pt] \text{Divisor} &= \color{magenta}{4x - 5} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm} \color{green}{x^2 \ - \ 8x \ + \ 3} \\ 4x - 5 \enclose{longdiv}{\ \ 4x^3 - 37x^2 + 52x - 15}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+} 4x^3 \ \ \overset{\color{red}\scriptsize(+)}{-} 5x^2 \\ \hline \hspace{1.9cm}0 \ - 32x^2 + 52x - 15 \\ \hspace{2.3cm}\overset{\color{red}\scriptsize(+)}{-}32x^2 \ \ \overset{\color{red}\scriptsize(-)}{+}40x \\ \hline \hspace{3.2cm}0 \ + 12x - 15 \\ \hspace{3.6cm}\overset{\color{red}\scriptsize(-)}{+}12x \ \ \overset{\color{red}\scriptsize(+)}{-}15 \\ \hline \hspace{4.4cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= x^2 - 8x + 3 \\[6pt] \color{green}\text{Remainder}& = 0 \\[6pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor } \times \text{ Quotient } + \text{ Remainder}} \\[6pt] LHS & \implies \color{green} 4x^3 - 37x^2 + 52x - 15 \\[6pt] RHS & = \text{Divisor } \times \text{ Quotient + Remainder } \\[6pt] &= [(4x-5)(x^2-8x+3)] + 0 \\[6pt] &= [4x(x^2-8x+3) -5(x^2-8x+3)] \\[6pt] &= 4x^3 - 32x^2 + 12x - 5x^2 + 40x - 15 \\[6pt] &= 4x^3 - 32x^2 - 5x^2 + 12x + 40x - 15 \\[6pt] RHS & \implies \color{green} 4x^3 - 37x^2 + 52x - 15 \\[6pt] LHS & = RHS \\ & \text{Hence verified} \end{aligned} \](iv) \(y^3 - 8 \text{ by } y - 2\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}y^3 + 0y^2 + 0y - 8 \\[6pt] \text{Divisor} &= \color{magenta}{y - 2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm} \color{green}{y^2 \ + \ 2y \ + \ 4} \\ y - 2 \enclose{longdiv}{\ \ y^3 + 0y^2 + 0y - 8}\\ \hspace{1cm}\overset{\color{red}\scriptsize(-)}{+} y^3 \overset{\color{red}\scriptsize(+)}{-} 2y^2 \\ \hline \hspace{1.4cm}0 + 2y^2 + 0y - 8 \\ \hspace{1.8cm}\overset{\color{red}\scriptsize(-)}{+} 2y^2 \overset{\color{red}\scriptsize(+)}{-} 4y \\ \hline \hspace{2.5cm}0 + 4y - 8 \\ \hspace{2.8cm}\overset{\color{red}\scriptsize(-)}{+} \ 4y \overset{\color{red}\scriptsize(+)}{-} 8 \\ \hline \hspace{4cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= y^2 + 2y + 4 \\[6pt] \color{green}{\text{Remainder}} &= 0 \\[6pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor } \times \text{ Quotient } + \text{ Remainder}} \\[6pt] LHS &\implies \color{green}{y^3 - 8} \\[6pt] RHS &= [(y-2)(y^2+2y+4)] + 0 \\[6pt] RHS &= [y(y^2+2y+4) -2(y^2+2y+4)] \\[6pt] &= y^3 + 2y^2 + 4y -2y^2 - 4y - 8 \\[6pt] &= y^3 + \cancel{2y^2} - \cancel{2y^2} + \cancel{4y} - \cancel{4y} - 8 \\[6pt] &= y^3 - 8 \\[6pt] RHS &\implies \color{green}{y^3 - 8} \\[6pt] LHS & = RHS \\ & \text{Hence verified} \end{aligned} \](v) \(29z - 6z^2 - 28 \text{ by } 3z - 4\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{-6z^2 + 29z - 28} \\[6pt] \text{Divisor} &= \color{magenta}{3z - 4} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm} \color{green}{-2z \ + \ 7} \\ 3z - 4 \enclose{longdiv}{ \ -6z^2 + 29z - 28}\\ \hspace{1.6cm}\overset{\color{red}\scriptsize(+)}{-}\ 6z^2 \overset{\color{red}\scriptsize(-)}{+} \ 8z \\ \hline \hspace{2.4cm} 0 + 21z - 28 \\ \hspace{2.7cm}\overset{\color{red}\scriptsize(-)}{+} \ 21z \overset{\color{red}\scriptsize(+)}{-} 28 \\ \hline \hspace{3.6cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= -2z + 7 \\ \color{green}{\text{Remainder}} &= 0 \\[6pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor} \times \text{Quotient} + \text{Remainder}} \\[6pt] LHS &\implies \color{green}{-6z^2 + 29z - 28} \\[6pt] RHS &= [(3z-4)(-2z+7)] + 0 \\[6pt] &= 3z(-2z+7) - 4(-2z+7) \\[6pt] &= -6z^2 + 21z + 8z - 28 \\[6pt] &= -6z^2 + 29z - 28 \\[6pt] RHS &\implies \color{green}{-6z^2 + 29z - 28} \\[6pt] LHS & = RHS \\ & \text{Hence verified} \end{aligned} \](vi) \( p^4 + p^3 - p^2 + 1 \text{ by } p - 1 \)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{p^4 + p^3 - p^2 + 0p + 1} \\[4pt] \text{Divisor} &= \color{magenta}{p - 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.6cm}\color{green}{p^3 \ + \ 2p^2 \ + \ p \ + \ 1} \\ p-1 \enclose{longdiv}{\ \ p^4 + p^3 - p^2 + 0p + 1}\\ \hspace{.9cm}\overset{\color{red}\scriptsize(-)}{+}p^4 \ \ \overset{\color{red}\scriptsize(+)}{-}p^3 \\ \hline \hspace{1.5cm}0 \ + \ 2p^3 - p^2 + 0p + 1\\ \hspace{2.0cm}\overset{\color{red}\scriptsize(-)}{+}2p^3 \ \ \overset{\color{red}\scriptsize(+)}{-}2p^2 \\ \hline \hspace{2.9cm}0 \ + \ p^2 + 0p + 1 \\ \hspace{3.3cm}\overset{\color{red}\scriptsize(-)}{+}p^2 \ \ \overset{\color{red}\scriptsize(+)}{-}p \\ \hline \hspace{4.2cm}0 \ + \ p + 1 \\ \hspace{4.6cm}\overset{\color{red}\scriptsize(-)}{+}p \ \ \overset{\color{red}\scriptsize(+)}{-}1 \\ \hline \hspace{5.2cm} 2 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}}&=p^3+2p^2+p+1 \\ \color{green}{\text{Remainder}} & = 2 \\[6pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor} \times \text{Quotient} + \text{Remainder}} \\[6pt] LHS &\implies \color{green}p^4 + p^3 - p^2 + 1 \\[6pt] RHS &= [(p-1)(p^3+2p^2+p+1)] + 2 \\[6pt] &= [p(p^3+2p^2+p+1) - 1(p^3+2p^2+p+1)] + 2 \\[6pt] &= [p^4+2p^3+p^2 + p - p^3 - 2p^2 - p - 1] + 2 \\[6pt] &= p^4+2p^3 - p^3 + p^2 - 2p^2 + p - p - 1 + 2 \\[6pt] RHS &\implies \color{green}p^4+p^3-p^2+1 \\[6pt] LHS & = RHS \\ & \text{Hence verified} \end{aligned} \](vii) \( 12z^3 + 4z + 3z^2 + 1 \text{ by } 4z + 1 \)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{12z^3 + 3z^2 + 4z + 1} \\[6pt] \text{Divisor} &= \color{magenta}{4z + 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{ 3z^2 \ + \ 1} \\ 4z + 1 \enclose{longdiv}{\ 12z^3 + 3z^2 + 4z + 1}\\ \hspace{1.15cm}\overset{\color{red}\scriptsize(-)}{+} 12z^3 \overset{\color{red}\scriptsize(-)}{+}3z^2 \\ \hline \hspace{2.9cm}0 \ + 4z \ + \ 1 \\ \hspace{3.3cm}\overset{\color{red}\scriptsize(-)}{+} \ 4z \ \overset{\color{red}\scriptsize(-)}{+}1 \\ \hline \hspace{4.35cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 3z^2 + 1 \\ \color{green}{\text{Remainder}} &= 0 \\[6pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor} \times \text{Quotient} + \text{Remainder}} \\[6pt] LHS &\implies \color{green}{12z^3 + 3z^2 + 4z + 1} \\[6pt] RHS &= \big[(4z+1) (3z^2+1)\big] + 0 \\[6pt] &= \big[4z(3z^2+1) + 1(3z^2+1)\big] \\[6pt] &= 12z^3 + 4z + 3z^2 + 1 \\[6pt] RHS &\implies \color{green}{12z^3 + 3z^2 + 4z + 1} \\[6pt] LHS &= RHS \\ &\text{Hence verified} \end{aligned} \](viii) \( 8x^2 - 2 - 3x + 12x^3 \text{ by } 4x^2 - 1 \)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{12x^3 + 8x^2 - 3x - 2} \\[6pt] \text{Divisor} &= \color{magenta}{4x^2 - 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{ \ 3x \ + \ 2} \\ 4x^2 - 1 \enclose{longdiv}{\ \ 12x^3 + 8x^2 - 3x - 2}\\ \hspace{1.5cm}\overset{\color{red}\scriptsize(-)}{+}12x^3 \hspace{1cm} \overset{\color{red}\scriptsize(+)}{-}3x \\ \hline \hspace{2cm}0 \ + \ 8x^2 \ + \ 0 \ - 2 \\ \hspace{2.5cm}\overset{\color{red}\scriptsize(-)}{+}\ 8x^2 \hspace{1.2cm} \overset{\color{red}\scriptsize(+)}{-}2 \\ \hline \hspace{4cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 3x + 2 \\ \color{green}{\text{Remainder}} &= 0 \\[6pt] &\text{Check} \\ \color{magenta}{\text{Dividend}} &= \color{magenta}{\text{Divisor} \times \text{Quotient} + \text{Remainder}} \\[6pt] LHS &\implies \color{green}{12x^3 + 8x^2 - 3x - 2} \\[6pt] RHS &= \big[(4x^2-1) (3x+2)\big] + 0 \\[6pt] &= 4x^2(3x+2) - 1(3x+2) \\[6pt] &= 12x^3 + 8x^2 - 3x - 2 \\[6pt] RHS &\implies \color{green}{12x^3 + 8x^2 - 3x - 2} \\[6pt] LHS &= RHS \\ &\text{Hence verified} \end{aligned} \]3. Using long division method, check whether the second polynomial is the factor of the first polynomial.
(i) \(x^4 - 16 \; ; \; x + 2\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{x^4 + 0x^3 + 0x^2 + 0x - 16} \\[6pt] \text{Divisor} &= \color{magenta}{x + 2} \end{aligned} \]\[ \begin{array}{l} \hspace{1.7cm}\color{green}{x^3 \ - \ 2x^2 \ + \ 4x \ - \ 8} \\ x+2 \enclose{longdiv}{ \ x^4 + 0x^3 + 0x^2 + 0x - 16}\\ \hspace{.8cm}\overset{\color{red}\scriptsize(-)}{+} \ x^4 \overset{\color{red}\scriptsize(-)}{+}2x^3 \\ \hline \hspace{1.4cm}0 - 2x^3 + 0x^2 + 0x - 16 \\ \hspace{1.8cm}\overset{\color{red}\scriptsize(+)}{-}2x^3 \overset{\color{red}\scriptsize(+)}{-}4x^2 \\ \hline \hspace{2.6cm} 0 + 4x^2 + 0x - 16 \\ \hspace{3cm}\overset{\color{red}\scriptsize(-)}{+}4x^2 \overset{\color{red}\scriptsize(-)}{+}8x \\ \hline \hspace{3.7cm}0 - 8x - 16 \\ \hspace{4.1cm}\overset{\color{red}\scriptsize(+)}{-}8x \overset{\color{red}\scriptsize(+)}{-}16 \\ \hline \hspace{5cm}0 \end{array} \]\[ \begin{aligned} \color{green}{\text{Quotient}} &= x^3 - 2x^2 + 4x - 8 \\ \color{green}{\text{Remainder}} &= 0 \\[6pt] \therefore \ \color{red}(x+2) & \color{red} \text{ is a factor of } (x^4-16) \end{aligned} \](ii) \(z^4 - z^3 + 3z^2 - 2z + 2 \; ; \; z^2 + 2\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{z^4 - z^3 + 3z^2 - 2z + 2} \\[6pt] \text{Divisor} &= \color{magenta}{z^2 + 2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{z^2 \ - \ z \ + \ 1} \\ z^2+2 \enclose{longdiv}{\ z^4 - z^3 + 3z^2 - 2z + 2}\\ \hspace{1.1cm}\overset{\color{red}\scriptsize(-)}{+}z^4 \hspace{.8cm} \overset{\color{red}\scriptsize(-)}{+}2z^2 \\ \hline \hspace{1.6cm}0 - z^3 + z^2 - 2z + 2\\ \hspace{2cm}\overset{\color{red}\scriptsize(+)}{-}z^3 \hspace{.9cm} \overset{\color{red}\scriptsize(+)}{-}2z \\ \hline \hspace{2.4cm}0 + z^2 + 0z + 2\\ \hspace{2.8cm}\overset{\color{red}\scriptsize(-)}{+}z^2 \hspace{.9cm} \overset{\color{red}\scriptsize(-)}{+}2 \\ \hline \hspace{4cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= z^2 - z + 1 \\ \color{green}{\text{Remainder}} &= 0 \\[6pt] \therefore \ \color{red}(z^2+2) & \color{red} \text{ is a factor of } (z^4 - z^3 + 3z^2 - 2z + 2) \end{aligned} \](iii) \(x^3 - 4x^2 - 3x + 5 \; ; \; x - 3\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{x^3 - 4x^2 - 3x + 5} \\[6pt] \text{Divisor} &= \color{magenta}{x - 3} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{x^2 \ - \ x \ - \ 6} \\ x-3 \enclose{longdiv}{\ \ x^3 - 4x^2 - 3x + 5}\\ \hspace{1.05cm}\overset{\color{red}\scriptsize(-)}{+}x^3 \overset{\color{red}\scriptsize(+)}{-}3x^2 \\ \hline \hspace{1.7cm}0 - x^2 - 3x + 5\\ \hspace{2.1cm}\overset{\color{red}\scriptsize(+)}{-}x^2 \overset{\color{red}\scriptsize(-)}{+}3x \\ \hline \hspace{2.6cm}0 - 6x + 5\\ \hspace{3cm}\overset{\color{red}\scriptsize(+)}{-}6x \overset{\color{red}\scriptsize(-)}{+}18 \\ \hline \hspace{4cm}-13 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= x^2 - x - 6 \\ \color{green}{\text{Remainder}} &= -13 \\[6pt] \therefore \ \color{red}(x-3) & \color{red} \text{ is not a factor of } (x^3 - 4x^2 - 3x + 5) \end{aligned} \](iv) \(4p^3 - 12p^2 - 37p - 15 \; ; \; 2p + 1\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{4p^3 - 12p^2 - 37p - 15} \\[6pt] \text{Divisor} &= \color{magenta}{2p + 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{2p^2 \ - \ 7p \ - \ 15} \\ 2p+1 \enclose{longdiv}{ \ 4p^3 - 12p^2 - 37p - 15}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+}4p^3 \overset{\color{red}\scriptsize(-)}{+}2p^2 \\ \hline \hspace{1.7cm}0 - 14p^2 - \ 37p - 15\\ \hspace{2.25cm}\overset{\color{red}\scriptsize(+)}{-}14p^2 \overset{\color{red}\scriptsize(+)}{-}7p \\ \hline \hspace{3.05cm}0 - 30p - 15\\ \hspace{3.4cm}\overset{\color{red}\scriptsize(+)}{-}30p \overset{\color{red}\scriptsize(+)}{-}15 \\ \hline \hspace{4.4cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 2p^2 - 7p - 15 \\ \color{green}{\text{Remainder}} &= 0 \\[6pt] \therefore \ \color{red}(2p+1) & \color{red} \text{ is a factor of } (4p^3 - 12p^2 - 37p - 15) \end{aligned} \](v) \(4q^3 - 6q^2 - 4q + 3 \; ; \; 2q - 1\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{4q^3 - 6q^2 - 4q + 3} \\[6pt] \text{Divisor} &= \color{magenta}{2q - 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{2q^2 \ - \ 2q \ - \ 3} \\ 2q-1 \enclose{longdiv}{\ \ 4q^3 - 6q^2 - 4q + 3}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+}4q^3 \overset{\color{red}\scriptsize(+)}{-}2q^2 \\ \hline \hspace{1.8cm}0 - 4q^2 - 4q + 3\\ \hspace{2.25cm}\overset{\color{red}\scriptsize(+)}{-}4q^2 \overset{\color{red}\scriptsize(-)}{+}2q \\ \hline \hspace{2.9cm}0 - 6q + 3\\ \hspace{3.3cm}\overset{\color{red}\scriptsize(+)}{-}6q \overset{\color{red}\scriptsize(-)}{+}3 \\ \hline \hspace{4cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 2q^2 - 2q - 3 \\ \color{green}{\text{Remainder}} &= 0 \\[6pt] \therefore \ \color{red}(2q-1) & \color{red} \text{ is a factor of } (4q^3 - 6q^2 - 4q + 3) \end{aligned} \](vi) \(20x^2 - 32x - 16 \; ; \; 5x + 2\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{20x^2 - 32x - 16} \\[6pt] \text{Divisor} &= \color{magenta}{5x + 2} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{4x \ - \ 8} \\ 5x+2 \enclose{longdiv}{\ \ 20x^2 - 32x - 16}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+}20x^2 \overset{\color{red}\scriptsize(-)}{+}8x \\ \hline \hspace{2cm}0 - 40x - 16\\ \hspace{2.4cm}\overset{\color{red}\scriptsize(+)}{-}40x \overset{\color{red}\scriptsize(+)}{-}16 \\ \hline \hspace{3.15cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 4x - 8 \\ \color{green}{\text{Remainder}} &= 0 \\[6pt] \therefore\ \color{red}(5x+2) & \color{red} \text{ is a factor of } (20x^2 - 32x - 16) \end{aligned} \](vii) \(12a^2 - 7a - 2 \; ; \; 4a - 1\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{12a^2 - 7a - 2} \\[6pt] \text{Divisor} &= \color{magenta}{4a - 1} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{3a \ - \ 1} \\ 4a-1 \enclose{longdiv}{\ \ 12a^2 - 7a - 2}\\ \hspace{1.2cm}\overset{\color{red}\scriptsize(-)}{+}12a^2 \overset{\color{red}\scriptsize(+)}{-}3a \\ \hline \hspace{2.1cm}0 - 4a - 2\\ \hspace{2.4cm}\overset{ \color{red} \scriptsize(+)}{-} \ 4a \overset{\color{red}\scriptsize(-)}{+}1 \\ \hline \hspace{3.15cm}-3 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 3a - 1 \\ \color{green}{\text{Remainder}} &= -3 \\[6pt] \therefore\ \color{red} (4a-1) & \color{red} \text{ is not a factor of } (12a^2 - 7a - 2) \end{aligned} \](viii) \(6x^3 + 19x^2 + 13x - 3 \; ; \; 2x + 3\)
Solution
\[ \begin{aligned} \text{Dividend} &= \color{magenta}{6x^3 + 19x^2 + 13x - 3} \\[6pt] \text{Divisor} &= \color{magenta}{2x + 3} \end{aligned} \] \[ \begin{array}{l} \hspace{1.7cm}\color{green}{3x^2 \ + \ 5x \ - \ 1} \\ 2x+3 \enclose{longdiv}{\ \ 6x^3 + 19x^2 + 13x - 3}\\ \hspace{1.3cm}\overset{\color{red}\scriptsize(-)}{+} 6x^3 \overset{\color{red}\scriptsize(-)}{+}9x^2 \\ \hline \hspace{2cm}0 + 10x^2 + 13x - 3\\ \hspace{2.3cm}\overset{\color{red}\scriptsize(-)}{+} \ 10x^2 \overset{\color{red}\scriptsize(-)}{+}15x \\ \hline \hspace{3.3cm}0 - 2x - 3\\ \hspace{3.6cm}\overset{\color{red}\scriptsize(+)}{-} \ 2x \overset{\color{red}\scriptsize(+)}{-}3 \\ \hline \hspace{4.5cm}0 \end{array} \] \[ \begin{aligned} \color{green}{\text{Quotient}} &= 3x^2 + 5x - 1 \\ \color{green}{\text{Remainder}} &= 0 \\[6pt] \therefore\ \color{red}(2x+3) & \color{red} \text{ is a factor of } (6x^3 + 19x^2 + 13x - 3) \end{aligned} \]