Q1. Divide:
(i) $-70a^3$ by $14a^2$
(ii) $24x^3y^3$ by $-8y^2$
(iii) $15a^4b$ by $-5a^3b$
(iv) $-24x^4d^3$ by $-2x^2d^5$
(v) $63a^4b^5c^6$ by $-9a^2b^4c^3$
(vi) $8x -10y + 6c$ by 2
(vii) $15a^3b^4 - 10a^4b^3 - 25a^3b^6$ by $-5a^3b^2$
(viii) $-14x^6y^3 - 21x^4y^5 + 7x^5y^4$ by $7x^2y^2$
(ix) $a^2 + 7a + 12$ by $a + 4$
(x) $x^2 + 3x - 54$ by $x - 6$
(xi) $12x^2 + 7xy -12y62$ by $3x + 4y$
(xii) $x^6 - 8$ by $x^2 - 2$
(xiii) $6x^3 - 13x^2 - 13x + 30$ by $2x^2 - x - 6$
(xiv) $4a^2 + 12ab + 9b^2 - 25c^2$ by $2a + 3b + 5c$
(xv) $16 + 8x + x^6 - 8x^3 - 2x^4 + x^2$ by $x + 4 - x^3$
Show Answer
(i) $-70a^3$ by $14a^2$
$\Rightarrow \dfrac{-70a^3}{14a^2} = -5a^{3-2} = -5a$
(ii) $24x^3y^3$ by $-8y^2$
$\Rightarrow \dfrac{24x^3y^3}{-8y^2} = -3x^3y^{3-2} = -3x^3y$ Ans
(iii) $15a^4b$ by $-5a^3b$
$\Rightarrow \dfrac{15a^4b}{-5a^3b} = -3a^{4-3}b^{1-1} = -3a$ Ans
(iv) $-24x^4d^3$ by $-2x^2d^5$
$\Rightarrow \dfrac{-24x^4d^3}{-2x^2d^5} = \dfrac{12x^{4-2}}{d^{5-3}} = \dfrac{12x^2}{d^2}$ Ans
(v) $63a^4b^5c^6$ by $-9a^2b^4c^3$
$\Rightarrow \dfrac{63a^4b^5c^6}{-9a^2b^4c^3} = -7a^{4-2}b^{5-4}c^{6-3} = -7a^2bc^3$ Ans
(vi) $8x -10y + 6c$ by 2
$\Rightarrow \dfrac{8x -10y + 6c}{2}$
$\Rightarrow \dfrac{8x}{2} - \dfrac{10y}{2} + \dfrac{6c}{2}$
$\Rightarrow 4x - 5y + 3c$ Ans
(vii) $15a^3b^4 - 10a^4b^3 - 25a^3b^6$ by $-5a^3b^2$
$\Rightarrow \dfrac{15a^3b^4 - 10a^4b^3 - 25a^3b^6}{-5a^3b^2}$
$\Rightarrow \dfrac{15a^3b^4}{-5a^3b^2} - \dfrac{10a^4b^3}{-5a^3b^2} - \dfrac{25a^3b^6}{-5a^3b^2}$
$\Rightarrow -3a^{3-3}b^{4-2} - (-2a^{4-3}b^{3-2}) - (-5a^{3-3}b^{6-2})$
$\Rightarrow (-3b^2) - (-2ab) - (-5b^4)$
$\Rightarrow (-3b^2) + 2ab + 5b^4$ Ans
(viii) $-14x^6y^3 - 21x^4y^5 + 7x^5y^4$ by $7x^2y^2$
$\Rightarrow \dfrac{-14x^6y^3}{7x^2y^2} - \dfrac{21x^4y^5}{7x^2y^2} + \dfrac{7x^5y^4}{7x^2y^2}$
$\Rightarrow -2x^{6-2}y^{3-2} - 3x^{4-2}y^{5-2} + x^{5-2}y^{4-2})$
$\Rightarrow (-2x^4y) - 3x^2y^3 + x^3y^2$ Ans
(ix) $a^2 + 7a + 12$ by $a + 4$
$$ \begin{array}{ c c c c c c c c c } a + 4 & \Bigg) & a^2 & + & 7a & + & 12 & \Bigg( & a + 3\\ & & a^2 & + & 4a\\ & (-) & & (-)\\\\ \hline \\& & & & 3a & + & 12 \\& & & & 3a & + & 12\\ & & & (-) & & (-)\\\\ \hline \\& & & & 0 & & 0\\\\ \end{array} $$
$\therefore a + 3$ Ans
(x) $x^2 + 3x - 54$ by $x - 6$
$$ \begin{array}{ c c c c c c c c c } x - 6 & \Bigg) & x^2 & + & 3x & - & 54 & \Bigg( & x + 9\\ & & x^2 & + & 6x\\ & (-) & & (+)\\\\ \hline \\& & & & 9x & - & 54 \\& & & & 9x & - & 54\\ & & & (-) & & (+)\\\\ \hline \\& & & & 0 & & 0\\ \end{array} $$
$\therefore x + 9$ Ans
(xi) $12x^2 + 7xy - 12y^2$ by $3x + 4y$
$$ \begin{array}{ c c c c c c c c c } 3x + 4y & \Bigg) & 12x^2 & + & 7xy & - & 12y^2 & \Bigg( & 4x - 3y\\ & & 12x^2 & + & 16xy\\ & (-) & & (-)\\\\ \hline \\& & & & -9xy & - & 12y^2 \\& & & & -9xy & - & 12y^2\\ & & & (+) & & (+)\\\\ \hline \\& & & & 0 & & 0\\ \end{array} $$
$\therefore 4x - 3y$ Ans
(xii) $x^6 - 8$ by $x^2 - 2$
$$ \begin{array}{ c c c c c c c c c } x^2 - 2 & \Bigg) & x^6 & - & 8 & \Bigg( & x^4 + 2x^2 + 4\\ & & x^6 & + & 2x^4\\ & (-) & & (+)\\\\ \hline \\& & & & 2x^4 & - & 8 \\& & & & 2x^4 & - & 4x^2\\ & & & (-) & & (+)\\\\ \hline \\& & & & 0 & & 4x^2 & - & 8\\ \\& & & & & & 4x^2 & - & 8\\ \\& & & & & (-) & & (+) &\\ \hline \\& & & & & & 0 & & 0\\ \end{array} $$
$\therefore x^4 + 2x^2 + 4$ Ans
(xiii) $6x^3 - 13x^2 - 13x + 30$ by $2x^2 - x - 6$
$$ \begin{array}{ c c c c c c c c c c } 2x^2 - x - 6 & \Bigg) & 6x^3 & - & 13x^2 & - & 13x & + & 30 \Bigg( & 3x - 5\\ & & 6x^3 & - & 3x^2 & - & 18x\\ & (-) & & (+) & & (+)\\ \hline \\& & & & -10x^2 & + & 5x & + & 30 \\& & & & -10x^2 & + & 5x & + & 30 \\& & & (+) & & (-) & & (-)\\ \hline \\& & & & 0 & & 0 & & 0\\ \end{array} $$
$\therefore 3x - 5$ Ans
(xiv) $4a^2 + 12ab + 9b^2 - 25c^2$ by $2a + 3b + 5c$
$$ \begin{array}{ c c c c c c c c c c c c c } 2a + 3b + 5c & \Bigg) & 4a^2 & + & 12ab & + & 9b^2 & - & 25c^2 \Bigg( & 2a + 3b - 5c\\ & & 4a^2 & + & 6ab & + & 10ac\\ & (-) & & (-) & & (-)\\ \hline \\& & 0 & & 6ab & + & 9b^2 & - & 10ac & - & 25c^2 \\& & & & 6ab & + & 9b^2 & + & 15bc \\& & & (-) & & (-) & & (-)\\ \hline \\& & & & 0 & & 0 & - & 10ac & - & 15bc & - & 25c^2 \\& & & & & & & - & 10ac & - & 15bc & - & 25c^2 \\& & & & & & & (+) & & (+) & & (+) &\\ \hline \\& & & & & & & & 0 & & 0 & & 0 \end{array} $$
$\therefore 2a + 3b - 5c$ Ans
(xv) $16 + 8x + x^6 - 8x^3 - 2x^4 + x^2$ by $x + 4 - x^3$
$\Rightarrow$ Reordering divident = $x^6 - 2x^4 - 8x^3 + x^2 + 8x + 16$
$\Rightarrow$ Reordering divisor = $-x^3 + x + 4$
$$ \begin{array}{ c c c c c c c c c c c c c } -x^3 + x + 4 & \Bigg) & x^6 & - & 2x^4 & - & 8x^3 & + & 8x & 16 \Bigg( & -x^3 + x + 4\\ & & x^6 & - & x^4 & - & 4x^3\\ & (-) & & (+) & & (+)\\ \hline \\& & 0 & - & x^4 & - & 4x^3 & + & x^2 & + & 8x & + & 16 \\& & & - & x^4 & & & + & x^2 & + & 4x & & \\& & & (+) & & & & (-) & & (-) & & &\\ \hline \\& & & & 0 & - & 4x^3 & & 0 & + & 4x & + & 16 \\& & & & & - & 4x^3 & & & + & 4x & + & 16 \\& & & & & (+) & & & & (-) & & (-) &\\ \hline \\& & & & & & 0 & & & & 0 & & 0 \end{array} $$
$\therefore -x^3 + x + 4$ Ans
$\Rightarrow \dfrac{-70a^3}{14a^2} = -5a^{3-2} = -5a$
(ii) $24x^3y^3$ by $-8y^2$
$\Rightarrow \dfrac{24x^3y^3}{-8y^2} = -3x^3y^{3-2} = -3x^3y$ Ans
(iii) $15a^4b$ by $-5a^3b$
$\Rightarrow \dfrac{15a^4b}{-5a^3b} = -3a^{4-3}b^{1-1} = -3a$ Ans
(iv) $-24x^4d^3$ by $-2x^2d^5$
$\Rightarrow \dfrac{-24x^4d^3}{-2x^2d^5} = \dfrac{12x^{4-2}}{d^{5-3}} = \dfrac{12x^2}{d^2}$ Ans
(v) $63a^4b^5c^6$ by $-9a^2b^4c^3$
$\Rightarrow \dfrac{63a^4b^5c^6}{-9a^2b^4c^3} = -7a^{4-2}b^{5-4}c^{6-3} = -7a^2bc^3$ Ans
(vi) $8x -10y + 6c$ by 2
$\Rightarrow \dfrac{8x -10y + 6c}{2}$
$\Rightarrow \dfrac{8x}{2} - \dfrac{10y}{2} + \dfrac{6c}{2}$
$\Rightarrow 4x - 5y + 3c$ Ans
(vii) $15a^3b^4 - 10a^4b^3 - 25a^3b^6$ by $-5a^3b^2$
$\Rightarrow \dfrac{15a^3b^4 - 10a^4b^3 - 25a^3b^6}{-5a^3b^2}$
$\Rightarrow \dfrac{15a^3b^4}{-5a^3b^2} - \dfrac{10a^4b^3}{-5a^3b^2} - \dfrac{25a^3b^6}{-5a^3b^2}$
$\Rightarrow -3a^{3-3}b^{4-2} - (-2a^{4-3}b^{3-2}) - (-5a^{3-3}b^{6-2})$
$\Rightarrow (-3b^2) - (-2ab) - (-5b^4)$
$\Rightarrow (-3b^2) + 2ab + 5b^4$ Ans
(viii) $-14x^6y^3 - 21x^4y^5 + 7x^5y^4$ by $7x^2y^2$
$\Rightarrow \dfrac{-14x^6y^3}{7x^2y^2} - \dfrac{21x^4y^5}{7x^2y^2} + \dfrac{7x^5y^4}{7x^2y^2}$
$\Rightarrow -2x^{6-2}y^{3-2} - 3x^{4-2}y^{5-2} + x^{5-2}y^{4-2})$
$\Rightarrow (-2x^4y) - 3x^2y^3 + x^3y^2$ Ans
(ix) $a^2 + 7a + 12$ by $a + 4$
$$ \begin{array}{ c c c c c c c c c } a + 4 & \Bigg) & a^2 & + & 7a & + & 12 & \Bigg( & a + 3\\ & & a^2 & + & 4a\\ & (-) & & (-)\\\\ \hline \\& & & & 3a & + & 12 \\& & & & 3a & + & 12\\ & & & (-) & & (-)\\\\ \hline \\& & & & 0 & & 0\\\\ \end{array} $$
$\therefore a + 3$ Ans
(x) $x^2 + 3x - 54$ by $x - 6$
$$ \begin{array}{ c c c c c c c c c } x - 6 & \Bigg) & x^2 & + & 3x & - & 54 & \Bigg( & x + 9\\ & & x^2 & + & 6x\\ & (-) & & (+)\\\\ \hline \\& & & & 9x & - & 54 \\& & & & 9x & - & 54\\ & & & (-) & & (+)\\\\ \hline \\& & & & 0 & & 0\\ \end{array} $$
$\therefore x + 9$ Ans
(xi) $12x^2 + 7xy - 12y^2$ by $3x + 4y$
$$ \begin{array}{ c c c c c c c c c } 3x + 4y & \Bigg) & 12x^2 & + & 7xy & - & 12y^2 & \Bigg( & 4x - 3y\\ & & 12x^2 & + & 16xy\\ & (-) & & (-)\\\\ \hline \\& & & & -9xy & - & 12y^2 \\& & & & -9xy & - & 12y^2\\ & & & (+) & & (+)\\\\ \hline \\& & & & 0 & & 0\\ \end{array} $$
$\therefore 4x - 3y$ Ans
(xii) $x^6 - 8$ by $x^2 - 2$
$$ \begin{array}{ c c c c c c c c c } x^2 - 2 & \Bigg) & x^6 & - & 8 & \Bigg( & x^4 + 2x^2 + 4\\ & & x^6 & + & 2x^4\\ & (-) & & (+)\\\\ \hline \\& & & & 2x^4 & - & 8 \\& & & & 2x^4 & - & 4x^2\\ & & & (-) & & (+)\\\\ \hline \\& & & & 0 & & 4x^2 & - & 8\\ \\& & & & & & 4x^2 & - & 8\\ \\& & & & & (-) & & (+) &\\ \hline \\& & & & & & 0 & & 0\\ \end{array} $$
$\therefore x^4 + 2x^2 + 4$ Ans
(xiii) $6x^3 - 13x^2 - 13x + 30$ by $2x^2 - x - 6$
$$ \begin{array}{ c c c c c c c c c c } 2x^2 - x - 6 & \Bigg) & 6x^3 & - & 13x^2 & - & 13x & + & 30 \Bigg( & 3x - 5\\ & & 6x^3 & - & 3x^2 & - & 18x\\ & (-) & & (+) & & (+)\\ \hline \\& & & & -10x^2 & + & 5x & + & 30 \\& & & & -10x^2 & + & 5x & + & 30 \\& & & (+) & & (-) & & (-)\\ \hline \\& & & & 0 & & 0 & & 0\\ \end{array} $$
$\therefore 3x - 5$ Ans
(xiv) $4a^2 + 12ab + 9b^2 - 25c^2$ by $2a + 3b + 5c$
$$ \begin{array}{ c c c c c c c c c c c c c } 2a + 3b + 5c & \Bigg) & 4a^2 & + & 12ab & + & 9b^2 & - & 25c^2 \Bigg( & 2a + 3b - 5c\\ & & 4a^2 & + & 6ab & + & 10ac\\ & (-) & & (-) & & (-)\\ \hline \\& & 0 & & 6ab & + & 9b^2 & - & 10ac & - & 25c^2 \\& & & & 6ab & + & 9b^2 & + & 15bc \\& & & (-) & & (-) & & (-)\\ \hline \\& & & & 0 & & 0 & - & 10ac & - & 15bc & - & 25c^2 \\& & & & & & & - & 10ac & - & 15bc & - & 25c^2 \\& & & & & & & (+) & & (+) & & (+) &\\ \hline \\& & & & & & & & 0 & & 0 & & 0 \end{array} $$
$\therefore 2a + 3b - 5c$ Ans
(xv) $16 + 8x + x^6 - 8x^3 - 2x^4 + x^2$ by $x + 4 - x^3$
$\Rightarrow$ Reordering divident = $x^6 - 2x^4 - 8x^3 + x^2 + 8x + 16$
$\Rightarrow$ Reordering divisor = $-x^3 + x + 4$
$$ \begin{array}{ c c c c c c c c c c c c c } -x^3 + x + 4 & \Bigg) & x^6 & - & 2x^4 & - & 8x^3 & + & 8x & 16 \Bigg( & -x^3 + x + 4\\ & & x^6 & - & x^4 & - & 4x^3\\ & (-) & & (+) & & (+)\\ \hline \\& & 0 & - & x^4 & - & 4x^3 & + & x^2 & + & 8x & + & 16 \\& & & - & x^4 & & & + & x^2 & + & 4x & & \\& & & (+) & & & & (-) & & (-) & & &\\ \hline \\& & & & 0 & - & 4x^3 & & 0 & + & 4x & + & 16 \\& & & & & - & 4x^3 & & & + & 4x & + & 16 \\& & & & & (+) & & & & (-) & & (-) &\\ \hline \\& & & & & & 0 & & & & 0 & & 0 \end{array} $$
$\therefore -x^3 + x + 4$ Ans
Q2. Find the quotient and remainder:
(i) $a^3 - 5a^2 + 8a + 15$ by $a + 1$
(ii) $3x^4 + 6x^3 - 6x^2 + 2x - 7$ by $x - 3$
(iii) $6x^2 + x - 15$ by $3x + 5$
Show Answer
(i) $a^3 - 5a^2 + 8a + 15$ by $a + 1$
$$ \begin{array}{ c c c c c c c c c c } a + 1 & \Bigg) & a^3 & - & 5a^2 & + & 8a & + & 15 \Bigg( & a^2 - 6a + 14\\ & & a^3 & + & 1a^2\\ & (-) & & (-)\\ \hline \\& & 0 & - & 6a^2 & + & 8a & + & 15 \\& & & - & 6a^2 & - & 6a \\& & & (+) & & (+)\\ \hline \\& & & & 0 & + & 14a & + & 15 \\& & & & & + & 14a & + & 14 \\& & & & & (-) & & (-)\\ \hline \\& & & & & & 0 & & 1 \end{array} $$
$\therefore \text{ Quotient } = a^2 - 6a + 14$
$\text{ And, remainder } = 1$ Ans
(ii) $3x^4 + 6x^3 - 6x^2 + 2x - 7$ by $x - 3$
$$ \begin{array}{ c c c c c c c c c c c c c } x - 3 & \Bigg) & 3x^4 & + & 6x^3 & - & 6x^2 & + & 2x & - & 7 \Bigg( & 3x^3 + 15x^2 + 39x + 119\\ & & 3x^4 & - & 9x^3\\ & (-) & & (+))\\ \hline \\& & 0 & + & 15x^3 & - & 6x^2 & + & 2x & - & 7 \\& & & + & 15x^3 & - & 45x^2 \\& & & (-) & & (+)\\ \hline \\& & & & & + & 39x^2 & + & 2x & - & 7 \\& & & & & + & 39x^2 & - & 117x \\& & & & & (-) & & (+)\\ \hline \\& & & & & & 0 & + & 119x & - & 7 \\& & & & & & & + & 119x & - & 357 \\& & & & & & & (-) & & (+)\\ \hline \\& & & & & & & & 0 & & 350 \end{array} $$
$\therefore \text{ Quotient } = 3x^3 + 15x^2 + 39x + 119$
$\text{ And, remainder } = 350$ Ans
(iii) $6x^2 + x - 15$ by $3x + 5$
$$ \begin{array}{ c c c c c c c c c } 3x + 5 & \Bigg) & 6x^2 & + & x & - & 15 & \Bigg( & 2x - 3\\ & & 6x^2 & + & 10x\\ & (-) & & (-)\\ \hline \\& & 0 & - & 9x & - & 15 \\& & & - & 9x & - & 15 \\& & & (+) & & (+)\\ \hline \\& & & & 0 & & 0\\ \end{array} $$
$\therefore \text{ Quotient } = 2x - 3$
$\text{ And, remainder } = 0$ Ans
$$ \begin{array}{ c c c c c c c c c c } a + 1 & \Bigg) & a^3 & - & 5a^2 & + & 8a & + & 15 \Bigg( & a^2 - 6a + 14\\ & & a^3 & + & 1a^2\\ & (-) & & (-)\\ \hline \\& & 0 & - & 6a^2 & + & 8a & + & 15 \\& & & - & 6a^2 & - & 6a \\& & & (+) & & (+)\\ \hline \\& & & & 0 & + & 14a & + & 15 \\& & & & & + & 14a & + & 14 \\& & & & & (-) & & (-)\\ \hline \\& & & & & & 0 & & 1 \end{array} $$
$\therefore \text{ Quotient } = a^2 - 6a + 14$
$\text{ And, remainder } = 1$ Ans
(ii) $3x^4 + 6x^3 - 6x^2 + 2x - 7$ by $x - 3$
$$ \begin{array}{ c c c c c c c c c c c c c } x - 3 & \Bigg) & 3x^4 & + & 6x^3 & - & 6x^2 & + & 2x & - & 7 \Bigg( & 3x^3 + 15x^2 + 39x + 119\\ & & 3x^4 & - & 9x^3\\ & (-) & & (+))\\ \hline \\& & 0 & + & 15x^3 & - & 6x^2 & + & 2x & - & 7 \\& & & + & 15x^3 & - & 45x^2 \\& & & (-) & & (+)\\ \hline \\& & & & & + & 39x^2 & + & 2x & - & 7 \\& & & & & + & 39x^2 & - & 117x \\& & & & & (-) & & (+)\\ \hline \\& & & & & & 0 & + & 119x & - & 7 \\& & & & & & & + & 119x & - & 357 \\& & & & & & & (-) & & (+)\\ \hline \\& & & & & & & & 0 & & 350 \end{array} $$
$\therefore \text{ Quotient } = 3x^3 + 15x^2 + 39x + 119$
$\text{ And, remainder } = 350$ Ans
(iii) $6x^2 + x - 15$ by $3x + 5$
$$ \begin{array}{ c c c c c c c c c } 3x + 5 & \Bigg) & 6x^2 & + & x & - & 15 & \Bigg( & 2x - 3\\ & & 6x^2 & + & 10x\\ & (-) & & (-)\\ \hline \\& & 0 & - & 9x & - & 15 \\& & & - & 9x & - & 15 \\& & & (+) & & (+)\\ \hline \\& & & & 0 & & 0\\ \end{array} $$
$\therefore \text{ Quotient } = 2x - 3$
$\text{ And, remainder } = 0$ Ans
Q3. The area of a rectangle is $x^3 - 8x^2 + 7$ and one of its sides is $x - 1$. Find length of adjacent side.
Show Answer
$\Rightarrow$ As per given conditon, we have the following equation:
$\Rightarrow (x - 1)(\text{Required number}) = x^3 - 8x^2 + 7$
$\Rightarrow$ Required number = $\dfrac{x^3 - 8x^2 + 7}{x - 1}$
$$ \begin{array}{ c c c c c c c c c } x - 1 & \Bigg) & x^3 & - & 8x^2 & + & 7 & \Bigg( & x^2 - 7x - 7\\ & & x^3 & - & x^2\\ & (-) & & (+)\\ \hline \\& & 0 & - & 7x^2 & + & 7 \\& & & - & 7x^2 & + & 7x \\& & & (+) & & (-)\\ \hline \\& & & & 0 & - & 7x & + & 7 \\& & & & & - & 7x & + & 7 \\& & & & & (+) & & (-)\\ \hline \\& & & & & & 0 & & 0 \end{array} $$
$\therefore \text{ Required Number } = x^2 - 7x - 7$ Ans
$\Rightarrow (x - 1)(\text{Required number}) = x^3 - 8x^2 + 7$
$\Rightarrow$ Required number = $\dfrac{x^3 - 8x^2 + 7}{x - 1}$
$$ \begin{array}{ c c c c c c c c c } x - 1 & \Bigg) & x^3 & - & 8x^2 & + & 7 & \Bigg( & x^2 - 7x - 7\\ & & x^3 & - & x^2\\ & (-) & & (+)\\ \hline \\& & 0 & - & 7x^2 & + & 7 \\& & & - & 7x^2 & + & 7x \\& & & (+) & & (-)\\ \hline \\& & & & 0 & - & 7x & + & 7 \\& & & & & - & 7x & + & 7 \\& & & & & (+) & & (-)\\ \hline \\& & & & & & 0 & & 0 \end{array} $$
$\therefore \text{ Required Number } = x^2 - 7x - 7$ Ans
Q4. The product of two numbers is $16x^4 - 1$. If one number is $2x -1$. Find the other number.
Show Answer
$\Rightarrow$ As per given conditon, we have the following equation:
$\Rightarrow (2x -1)(\text{Required number}) = 16x^4 - 1$
$\Rightarrow$ Required number = $\dfrac{16x^4 - 1}{2x - 1}$
$$ \begin{array}{ c c c c c c c c c c c } 2x - 1 & \Bigg) & 16x^4 & - & 1 & \Bigg( & 8x^3 + 4x^2 + 2x + 1\\ & & 16x^4 & - & 8x^3\\ & (-) & & (+)\\ \hline \\& & 0 & + & 8x^3 & - & 1 \\& & & + & 8x^3 & - & 4x^2 \\& & & (-) & & (+)\\ \hline \\& & & & 0 & + & 4x^2 & - & 1 \\& & & & & + & 4x^2 & - & 2x \\& & & & & (-) & & (+)\\ \hline \\& & & & & & 0 & + & 2x & - & 1 \\& & & & & & & + & 2x & - & 1 \\ & & & & & & & (-) & & (+)\\ \hline \\& & & & & & & & 0 & & 0 \end{array} $$
$\therefore \text{ Required Number } = 8x^3 + 4x^2 + 2x + 1$ Ans
$\Rightarrow (2x -1)(\text{Required number}) = 16x^4 - 1$
$\Rightarrow$ Required number = $\dfrac{16x^4 - 1}{2x - 1}$
$$ \begin{array}{ c c c c c c c c c c c } 2x - 1 & \Bigg) & 16x^4 & - & 1 & \Bigg( & 8x^3 + 4x^2 + 2x + 1\\ & & 16x^4 & - & 8x^3\\ & (-) & & (+)\\ \hline \\& & 0 & + & 8x^3 & - & 1 \\& & & + & 8x^3 & - & 4x^2 \\& & & (-) & & (+)\\ \hline \\& & & & 0 & + & 4x^2 & - & 1 \\& & & & & + & 4x^2 & - & 2x \\& & & & & (-) & & (+)\\ \hline \\& & & & & & 0 & + & 2x & - & 1 \\& & & & & & & + & 2x & - & 1 \\ & & & & & & & (-) & & (+)\\ \hline \\& & & & & & & & 0 & & 0 \end{array} $$
$\therefore \text{ Required Number } = 8x^3 + 4x^2 + 2x + 1$ Ans
Q5. Divide $x^6 - y^6$ by the product of $x^2 + xy + y^2$ and $x - y$.
Show Answer
$\Rightarrow (x - y)(x^2 + xy + y^2)$
$\Rightarrow x(x^2 + xy + y^2) - y(x^2 + xy + y^2)$
$\Rightarrow x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3$
$\Rightarrow x^3 + (x^2y - x^2y) + (xy^2 - xy^2) + y^3$
$\Rightarrow x^3 - y^3$
So, we have = $\dfrac{x^6 - y^6}{x^3 - y^3}$
$$ \begin{array}{ c c c c c c c c c } x^3 - y^3 & \Bigg) & x^6 & - & y^6 & \Bigg( & x^3 + y^3\\ & & x^6 & - & x^3y^3\\ & (-) & & (+)\\ \hline \\& & 0 & + & x^3y^3 & - & y^6 \\& & & + & x^3y^3 & - & y^6 \\& & & (-) & & (+)\\ \hline \\& & & & 0 & & 0 \end{array} $$
$\therefore x^3 + y^3$ Ans
$\Rightarrow x(x^2 + xy + y^2) - y(x^2 + xy + y^2)$
$\Rightarrow x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3$
$\Rightarrow x^3 + (x^2y - x^2y) + (xy^2 - xy^2) + y^3$
$\Rightarrow x^3 - y^3$
So, we have = $\dfrac{x^6 - y^6}{x^3 - y^3}$
$$ \begin{array}{ c c c c c c c c c } x^3 - y^3 & \Bigg) & x^6 & - & y^6 & \Bigg( & x^3 + y^3\\ & & x^6 & - & x^3y^3\\ & (-) & & (+)\\ \hline \\& & 0 & + & x^3y^3 & - & y^6 \\& & & + & x^3y^3 & - & y^6 \\& & & (-) & & (+)\\ \hline \\& & & & 0 & & 0 \end{array} $$
$\therefore x^3 + y^3$ Ans