DAV Class 8 Maths Chapter 8 Worksheet 2
Polynomials Worksheet 2
1. Divide the following monomials by the given monomial.
(i) \(6x^3\text{ by } 3x^2\)
Solution
\[ \begin{align*} 6x^3 \div 3x^2 &= \frac{6x^3}{3x^2} \\[6pt] &= \frac{\cancel6^2}{\cancel3_1} \times \frac{x^3}{x^2}\\[6pt] &= 2 \times x^{3-2}\\ &= 2 x \end{align*} \]
Answer \( \color{red} 2 x\)
(ii) \(-35x^4 \text{ by } -7x^3\)
Solution
\[ \begin{align*} -35x^4 \div (-7x^3) &= \frac{-35x^4}{-7x^3} \\[6pt] &= \frac{\cancel{-35}^5}{\cancel{-7}_1}\times \frac{x^4}{x^3} \\[6pt] &= 5 \times x^{4-3} \\ &= 5x \end{align*} \]Answer \(\color{red}{5x}\)
(iii) \(-5z^2 \text{ by } \sqrt{5}\,z\)
Solution
\[ \begin{align*} -5z^2 \div (\sqrt5\,z) &= \frac{-5z^2}{\sqrt5\,z} \\[6pt] &= -\frac{5}{\sqrt5}\times \frac{z^2}{z} \\[6pt] &= -\frac{\sqrt{5} \times \cancel{\sqrt5}}{\cancel{\sqrt5}}\times z^{2-1} \\[6pt] &= -\sqrt5 \times z^{1} \\[6pt] &= -\sqrt5\,z \end{align*} \]Answer \(\color{red}{-\sqrt5\,z}\)
(iv) \(16p^4 \text{ by } -6p^2\)
Solution
\[ \begin{align*} 16p^4 \div (-6p^2) &= \frac{16p^4}{-6p^2} \\[6pt] &= -\,\frac{\cancel{16}^8}{\cancel6_3}\times p^{4-2} \\[6pt] &= -\,\frac{8}{3}\,p^{2} \end{align*} \]Answer \(\color{red}{-\dfrac{8}{3}p^{2}}\)
(v) \(4\sqrt2\,y^3 \text{ by } 3\sqrt2\,y^2\)
Solution
\[ \begin{align*} 4\sqrt2\,y^3 \div (3\sqrt2\,y^2) &= \frac{4\sqrt2\,y^3}{3\sqrt2\,y^2} \\[6pt] &= \frac{4 \cancel{\sqrt2} }{3 \cancel{\sqrt2} } \times y^{3-2} \\[6pt] &= \frac{4}{3}\,y \end{align*} \]Answer \(\color{red}{\dfrac{4}{3}y}\)
(vi) \(\dfrac{3}{4}p^2 \text{ by } \dfrac{4}{3}p^2\)
Solution
\[ \begin{align*} &= \left(\frac{3p^2}{4}\right) \div \left(\frac{4p^2}{3}\right) \\[6pt] &= \frac{3\cancel{p^2}}{4} \times \frac{3}{4\cancel{p^2}} \\[6pt] &= \frac{9}{16} \end{align*} \]Answer \(\color{red}{\dfrac{9}{16}}\)
2. Divide the following polynomials by a monomial.
(i) \(6x^4 - 24x^3 + 15x^2 + 9 \text{ by } (-3x^2)\)
Solution
\[ \begin{aligned} & = \frac{6x^4 - 24x^3 + 15x^2 + 9}{-3x^2} \\[6pt] &= \frac{6x^4}{-3x^2}\;-\;\frac{24x^3}{-3x^2}\;+\;\frac{15x^2}{-3x^2}\;+\;\frac{9}{-3x^2} \\[8pt] &= -\frac{\cancel{6}^{\,2}}{\cancel{3}_{\,1}}\,x^{4-2} \;+\;\frac{\cancel{24}^{\,8}}{\cancel{3}_{\,1}}\,x^{3-2} \;-\;\frac{\cancel{15}^{5} \ \cancel{x^2}}{\cancel{3}_{1} \ \cancel{x^2}} \;-\;\frac{\cancel9^3}{\cancel3 \ x^2} \\[8pt] &= -2x^2 \;+\; 8x \;-\; 5 \;-\; \frac{3}{x^2} \end{aligned} \]Answer \(\color{red}{-2x^2 + 8x - 5 - \dfrac{3}{x^2}}\)
(ii) \(-12x + 22x^2 - 16x^3 + 4 \text{ by } 2x\)
Solution
\[ \begin{aligned} &= \frac{-12x + 22x^2 - 16x^3 + 4}{2x} \\[6pt] &= \frac{-12x}{2x} + \frac{22x^2}{2x} - \frac{16x^3}{2x} + \frac{4}{2x} \\[8pt] &= -\frac{\cancel{12}^{\,6} \cancel x}{\cancel{2}_{\,1} \cancel x} \;+\;\frac{\cancel{22}^{\,11}}{\cancel{2}_{\,1}}\,x^{2-1} \;-\;\frac{\cancel{16}^{\,8}}{\cancel{2}_{\,1}}\,x^{3-1} \;+\;\frac{\cancel{4}^{\,2}}{\cancel{2}\,x} \\[8pt] &= -6 + 11x - 8x^2 + \frac{2}{x} \\[4pt] &= -8x^2 + 11x - 6 + \dfrac{2}{x} \end{aligned} \]Answer \(\color{red}{-8x^2 + 11x - 6 + \dfrac{2}{x}}\)
(iii) \(\dfrac{2}{3}z^4 - \dfrac{1}{3}z^2 - 1 \text{ by } \dfrac{1}{3}z\)
Solution
\[ \begin{aligned} &= \frac{\dfrac{2}{3}z^4 - \dfrac{1}{3}z^2 - 1}{\dfrac{1}{3}z} \\ \\ &= \frac{\dfrac{2}{3}z^4}{\dfrac{1}{3}z} \;-\;\frac{\cancel{\dfrac{1}{3}}z^2}{\cancel{\dfrac{1}{3}}z} \;-\;\frac{1}{\dfrac{z}{3}} \\ \\ &= \left({\frac{2}{\cancel3} \times \frac{\cancel3}{1}}\right) z^{4-1} \;-\;z^{2-1} \;-\;1 \times \frac{3}{z} \\[8pt] &= 2z^3 - z - \frac{3}{z} \end{aligned} \]Answer \(\color{red}{2z^3 - z - \dfrac{3}{z}}\)
(iv) \(2\sqrt{2}\,q^4 + 4\sqrt{2}\,q^3 + q^2 \text{ by } -2\sqrt{2}\,q^2\)
Solution
\[ \begin{aligned} &= \frac{2\sqrt{2}\,q^4 + 4\sqrt{2}\,q^3 + q^2}{-2\sqrt{2}\,q^2} \\[6pt] &= \frac{2\sqrt{2}\,q^4}{-2\sqrt{2}\,q^2} \;+\;\frac{4\sqrt{2}\,q^3}{-2\sqrt{2}\,q^2} \;+\;\frac{\cancel{q^2}}{-2\sqrt{2}\,\cancel{q^2}} \\[8pt] &= -\frac{\cancel{2\sqrt{2}}}{\cancel{2\sqrt{2}}}\,q^{4-2} \;-\;\frac{\cancel{4\sqrt{2}}^{ \ 2}}{\cancel{2\sqrt{2}}_1}\,q^{3-2} \;-\;\frac{1}{2\sqrt{2}} \\[8pt] &= -q^2 - 2q - \frac{1}{2\sqrt{2}} \end{aligned} \]Answer \(\color{red}{-q^2 - 2q - \dfrac{1}{2\sqrt{2}}}\)
3. Divide the following polynomials by a monomial using long division method. Also, write the quotient and the remainder.
(i) \(-8x^3 + 6x^4 - 4 + 12x \text{ by } 2x^2\)
Solution
\[ \begin{aligned} \color{magenta} (6x^4-8x^3+12x-4 )\div 2x^2 \end{aligned} \] \[ \begin{array}{l} \hspace{1.cm} \ 3x^2 - 4x \\ 2x^2 \enclose{longdiv}{ \ \ 6x^4-8x^3+12x-4}\\ \hspace{.7cm}\overset{\color{red} \scriptsize (-)}{+}\,6x^4 \phantom{00} \big\downarrow \\\hline \hspace{1.2cm}0 \ - 8x^3 \\\hspace{1.74cm}\overset{\color{red} \scriptsize (+)}{-}8x^3 \phantom{000} \smash{\downarrow \hspace{.031cm} \!\!\rule[1ex]{.1pt}{11ex}} \phantom{00000} \smash{\downarrow \hspace{.02cm} \!\!\rule[1ex]{.1pt}{11ex}} \\\hline \hspace{2.3cm}0 + \ 12x \ -4 \\ \hline \end{array} \] \[ \begin{aligned} \color{green} \text{Quotient} & = \color{red} 3x^2 - 4x \\ \color{green} \text{Remainder} & = \color{red} 12x -4 \end{aligned} \](ii) \(5x^{10}-9x^8-9x^5+7x \text{ by } x^5\)
Solution
\[ \begin{aligned} \color{magenta}(5x^{10}-9x^8-9x^5+7x)\div x^5 \end{aligned} \] \[ \begin{array}{l} \hspace{.8cm} 5x^5 \ - \ 9x^3 \ - \ 9 \\ x^5 \enclose{longdiv}{ \ \ 5x^{10}-9x^8-9x^5+7x}\\ \hspace{.5cm}\overset{\color{red} \scriptsize (-)}{+}5x^{10} \phantom{00} \big\downarrow \\ \hline \hspace{1.2cm}0 \ - \ 9x^8 \\ \hspace{1.65cm}\overset{\color{red} \scriptsize (+)}{-} \ 9x^8 \phantom{000} \smash{\downarrow \hspace{.03cm} \!\!\rule [1ex]{.5pt}{11ex}} \\ \hline \hspace{2.4cm}0 \ - 9x^5\\ \hspace{2.8cm}\overset{\color{red} \scriptsize (+)}{-}\,9x^5 \phantom{000} \smash{\downarrow \hspace{.028cm} \!\!\rule [1ex]{.5pt}{19ex}}\\ \hline \hspace{3.4cm}0 \ + \ 7x \end{array} \] \[ \begin{aligned} \color{green}\text{Quotient} &= \color{red}{5x^5 - 9x^3 - 9} \\ \color{green}\text{Remainder} &= \color{red}{7x} \end{aligned} \](iii) \(5z^3-6z^2+7z \text{ by } 2z\)
Solution
\[ \begin{aligned} \color{magenta}(5z^3-6z^2+7z)\div 2z \end{aligned} \] \[ \begin{array}{l} \hspace{.5cm} \ \dfrac{5}{2}z^2 \ - \ 3z \ + \ \dfrac{7}{2} \\ 2z \enclose{longdiv}{ \ \ 5z^{3}-6z^{2}+7z}\\ \hspace{.4cm}\overset{\color{red} \scriptsize (-)}{+} \ 5z^{3} \phantom{0.0} \big\downarrow\\ \hline \hspace{1.2cm}0 - 6z^{2} \\ \hspace{1.6cm} \overset{\color{red} \scriptsize (+)}{-}6z^{2} \phantom{0.0} \smash{\downarrow \hspace{.025cm} \!\!\rule[1ex]{.4pt}{11ex}}\\ \hline \hspace{2. cm}0 \ + \ 7z\\ \hspace{2.5cm}- \ 7z\\ \hline \hspace{3cm}0 \end{array} \] \[ \begin{aligned} \color{green}\text{Quotient} &= \color{red}{\dfrac{5}{2}z^{2}-3z+\dfrac{7}{2}} \\[6pt] \color{green}\text{Remainder} &= \color{red}{0} \end{aligned} \]