\displaystyle \textbf{Question 1: }\text{Divide:}
\displaystyle \text{(i) }45x^7\text{ by }-9x^4\qquad \text{(ii) }-60x^3y^2\text{ by }-15xy
\displaystyle \text{(iii) }-\frac{3}{4}x^2yz^3\text{ by } -\frac{2}{3}x^2yz
\displaystyle \text{(iv) }63a^4b^3c^6\text{ by }-14a^2b^5c^4
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{45x^7}{-9x^4} =\frac{45}{-9}x^{7-4}=-5x^3.

\displaystyle \text{(ii) }\frac{-60x^3y^2}{-15xy} =\frac{-60}{-15}x^{3-1}y^{2-1}=4x^2y.

\displaystyle \text{(iii) } \frac{-\frac{3}{4}x^2yz^3}{-\frac{2}{3}x^2yz}
\displaystyle =\left(-\frac{3}{4}\div-\frac{2}{3}\right) x^{2-2}y^{1-1}z^{3-1}=\frac{9}{8}z^2.

\displaystyle \text{(iv) }\frac{63a^4b^3c^6}{-14a^2b^5c^4}
\displaystyle =\frac{63}{-14}a^{4-2}b^{3-5}c^{6-4} =-\frac{9}{2}a^2b^{-2}c^2
\displaystyle =-\frac{9a^2c^2}{2b^2}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Divide:}
\displaystyle \text{(i) }6ax-3cx+15x\text{ by }-3x
\displaystyle \text{(ii) }10mn^2-15m^2n^2+5m^3n\text{ by }-5mn
\displaystyle \text{(iii) }14x^3y^4-7x^4y^3-28x^3y^6\text{ by }-7x^3y^2
\displaystyle \text{(iv) }18a^6b^3-30a^4b^5+6a^4b^4\text{ by }6a^2b^2
\displaystyle \text{(v) }8a^6b-16a^2b^2-6a^4b^3\text{ by }-2a^2b
\displaystyle \text{(vi) }\frac{1}{2}p^2q^3-\frac{5}{3}p^3q^2+ \frac{1}{4}p^3q^3\text{ by }-\frac{1}{4}p^2q^2
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\frac{6ax-3cx+15x}{-3x}
\displaystyle =\frac{6ax}{-3x}-\frac{3cx}{-3x} +\frac{15x}{-3x}=-2a+c-5.

\displaystyle \text{(ii) }\frac{10mn^2-15m^2n^2+5m^3n}{-5mn}
\displaystyle =\frac{10mn^2}{-5mn}-\frac{15m^2n^2}{-5mn} +\frac{5m^3n}{-5mn}
\displaystyle =-2n+3mn-m^2.

\displaystyle \text{(iii) } \frac{14x^3y^4-7x^4y^3-28x^3y^6}{-7x^3y^2}
\displaystyle =\frac{14x^3y^4}{-7x^3y^2} -\frac{7x^4y^3}{-7x^3y^2} -\frac{28x^3y^6}{-7x^3y^2}
\displaystyle =-2y^2+xy+4y^4.

\displaystyle \text{(iv) } \frac{18a^6b^3-30a^4b^5+6a^4b^4}{6a^2b^2}
\displaystyle =\frac{18a^6b^3}{6a^2b^2} -\frac{30a^4b^5}{6a^2b^2} +\frac{6a^4b^4}{6a^2b^2}
\displaystyle =3a^4b-5a^2b^3+a^2b^2.

\displaystyle \text{(v) } \frac{8a^6b-16a^2b^2-6a^4b^3}{-2a^2b}
\displaystyle =\frac{8a^6b}{-2a^2b} -\frac{16a^2b^2}{-2a^2b} -\frac{6a^4b^3}{-2a^2b}
\displaystyle =-4a^4+8b+3a^2b^2.

\displaystyle \text{(vi) } \frac{\frac{1}{2}p^2q^3-\frac{5}{3}p^3q^2+\frac{1}{4}p^3q^3} {-\frac{1}{4}p^2q^2}
\displaystyle =\frac{\frac{1}{2}p^2q^3}{-\frac{1}{4}p^2q^2} -\frac{\frac{5}{3}p^3q^2}{-\frac{1}{4}p^2q^2} +\frac{\frac{1}{4}p^3q^3}{-\frac{1}{4}p^2q^2}
\displaystyle =-2q+\frac{20}{3}p-pq.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Divide:}
\displaystyle \text{(i) }x^2+8x+15\text{ by }x+5\qquad \text{(ii) }4x^2+11x-3\text{ by }x+3
\displaystyle \text{(iii) }-2x^2+x+3\text{ by }x+1
\displaystyle \text{(iv) }x^3-9x^2+26x-24\text{ by }x-4
\displaystyle \text{(v) }x^3-3x^2y+3xy^2-y^3\text{ by }x-y
\displaystyle \text{Answer:}

\displaystyle \text{(i)}
\displaystyle \begin{array}{rll} x+5 & )\overline{x^2+8x+15}( & x+3\\ (-)&x^2+5x&\\ \hline &\hspace{1.0cm}3x+15&\\ (-)&\hspace{1.0cm}3x+15&\\ \hline &\hspace{1.75cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }x+3.

\displaystyle \text{(ii)}
\displaystyle \begin{array}{rll} x+3 & )\overline{4x^2+11x-3}( & 4x-1\\ (-)&4x^2+12x&\\ \hline &\hspace{1.0cm}-x-3&\\ (-)&\hspace{1.0cm}-x-3&\\ \hline &\hspace{1.75cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }4x-1.

\displaystyle \text{(iii)}
\displaystyle \begin{array}{rll} x+1 & )\overline{-2x^2+x+3}( & -2x+3\\ (-)&-2x^2-2x&\\ \hline &\hspace{1.0cm}3x+3&\\ (-)&\hspace{1.0cm}3x+3&\\ \hline &\hspace{1.75cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }-2x+3.

\displaystyle \text{(iv)}
\displaystyle \begin{array}{rll} x-4 & )\overline{x^3-9x^2+26x-24}( & x^2-5x+6\\ (-)&x^3-4x^2&\\ \hline &\hspace{0.5cm}-5x^2+26x-24&\\ (-)&\hspace{0.5cm}-5x^2+20x&\\ \hline &\hspace{2.0cm}6x-24&\\ (-)&\hspace{2.0cm}6x-24&\\ \hline &\hspace{2.75cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }x^2-5x+6.

\displaystyle \text{(v)}
\displaystyle \begin{array}{rll} x-y & )\overline{x^3-3x^2y+3xy^2-y^3}( & x^2-2xy+y^2\\ (-)&x^3-x^2y&\\ \hline &\hspace{0.5cm}-2x^2y+3xy^2-y^3&\\ (-)&\hspace{0.5cm}-2x^2y+2xy^2&\\ \hline &\hspace{2.0cm}xy^2-y^3&\\ (-)&\hspace{2.0cm}xy^2-y^3&\\ \hline &\hspace{2.75cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }x^2-2xy+y^2.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Divide:}
\displaystyle \text{(i) }12x^2+11x+2\text{ by }4x+1\qquad \text{(ii) }6x^2+x-15\text{ by }2x-3
\displaystyle \text{(iii) }6x^3-x^2-26x-21\text{ by }3x-7
\displaystyle \text{(iv) }12x^2+7xy-12y^2\text{ by }3x+4y
\displaystyle \text{Answer:}

\displaystyle \text{(i)}
\displaystyle \begin{array}{rll} 4x+1 & )\overline{12x^2+11x+2}( & 3x+2\\ (-)&12x^2+3x&\\ \hline &\hspace{1.0cm}8x+2&\\ (-)&\hspace{1.0cm}8x+2&\\ \hline &\hspace{1.75cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }3x+2.

\displaystyle \text{(ii)}
\displaystyle \begin{array}{rll} 2x-3 & )\overline{6x^2+x-15}( & 3x+5\\ (-)&6x^2-9x&\\ \hline &\hspace{1.0cm}10x-15&\\ (-)&\hspace{1.0cm}10x-15&\\ \hline &\hspace{1.75cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }3x+5.

\displaystyle \text{(iii)}
\displaystyle \begin{array}{rll} 3x-7 & )\overline{6x^3-x^2-26x-21}( & 2x^2+5x+3\\ (-)&6x^3-14x^2&\\ \hline &\hspace{0.5cm}13x^2-26x-21&\\ (-)&\hspace{0.5cm}13x^2-\frac{91}{3}x&\\ \hline &\hspace{2.0cm}\frac{13}{3}x-21& \end{array}
\displaystyle \therefore \text{the given division does not have } 2x^2+5x+3\text{ as its quotient.}

\displaystyle \text{(iv)}
\displaystyle \begin{array}{rll} 3x+4y & )\overline{12x^2+7xy-12y^2}( & 4x-3y\\ (-)&12x^2+16xy&\\ \hline &\hspace{1.5cm}-9xy-12y^2&\\ (-)&\hspace{1.5cm}-9xy-12y^2&\\ \hline &\hspace{2.75cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }4x-3y.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Divide:}
\displaystyle \text{(i) }x^3-2x-1\text{ by }x^2-x-1
\displaystyle \text{(ii) }x^3-6x^2+11x-6\text{ by }x^2-5x+6
\displaystyle \text{(iii) }6x^5+4x^4-3x^3-1\text{ by }3x^2-x+1
\displaystyle \text{(iv) }6x^5-28x^3+3x^2+30x-9\text{ by }2x^2-6
\displaystyle \text{Answer:}

\displaystyle \text{(i)}
\displaystyle \begin{array}{rll} x^2-x-1 & )\overline{x^3+0x^2-2x-1}( & x+1\\ (-)&x^3-x^2-x&\\ \hline &\hspace{1.0cm}x^2-x-1&\\ (-)&\hspace{1.0cm}x^2-x-1&\\ \hline &\hspace{1.75cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }x+1.

\displaystyle \text{(ii)}
\displaystyle \begin{array}{rll} x^2-5x+6 & )\overline{x^3-6x^2+11x-6}( & x-1\\ (-)&x^3-5x^2+6x&\\ \hline &\hspace{1.0cm}-x^2+5x-6&\\ (-)&\hspace{1.0cm}-x^2+5x-6&\\ \hline &\hspace{1.75cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }x-1.

\displaystyle \text{(iii)}
\displaystyle \begin{array}{rll} 3x^2-x+1 & )\overline{6x^5+4x^4-3x^3+0x^2+0x-1}( & 2x^3+2x^2-x-1\\ (-)&6x^5-2x^4+2x^3&\\ \hline &\hspace{0.5cm}6x^4-5x^3+0x^2+0x-1&\\ (-)&\hspace{0.5cm}6x^4-2x^3+2x^2&\\ \hline &\hspace{1.2cm}-3x^3-2x^2+0x-1&\\ (-)&\hspace{1.2cm}-3x^3+x^2-x&\\ \hline &\hspace{2.0cm}-3x^2+x-1&\\ (-)&\hspace{2.0cm}-3x^2+x-1&\\ \hline &\hspace{3.0cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }2x^3+2x^2-x-1.

\displaystyle \text{(iv)}
\displaystyle \begin{array}{rll} 2x^2-6 & )\overline{6x^5+0x^4-28x^3+3x^2+30x-9}( & 3x^3-5x+\frac{3}{2}\\ (-)&6x^5-18x^3&\\ \hline &\hspace{0.7cm}-10x^3+3x^2+30x-9&\\ (-)&\hspace{0.7cm}-10x^3+30x&\\ \hline &\hspace{2.5cm}3x^2-9&\\ (-)&\hspace{2.5cm}3x^2-9&\\ \hline &\hspace{3.0cm}0& \end{array}
\displaystyle \therefore \text{the quotient is } 3x^3-5x+\frac{3}{2}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Divide:}
\displaystyle \text{(i) }x^3+27\text{ by }x^2-3x+9
\displaystyle \text{(ii) }x^4-81\text{ by }x+3
\displaystyle \text{(iii) }27x^3-8\text{ by }3x-2
\displaystyle \text{(iv) }x^6-8\text{ by }x^2-2
\displaystyle \text{(v) }x^6-y^6\text{ by }x-y
\displaystyle \text{(vi) }16x^4-81y^4\text{ by }2x-3y
\displaystyle \text{Answer:}

\displaystyle \text{(i)}
\displaystyle \begin{array}{rll} x^2-3x+9 & )\overline{x^3+0x^2+0x+27}( & x+3\\ (-)&x^3-3x^2+9x&\\ \hline &\hspace{0.7cm}3x^2-9x+27&\\ (-)&\hspace{0.7cm}3x^2-9x+27&\\ \hline &\hspace{2.5cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }x+3.

\displaystyle \text{(ii)}
\displaystyle \begin{array}{rll} x+3 & )\overline{x^4+0x^3+0x^2+0x-81}( & x^3-3x^2+9x-27\\ (-)&x^4+3x^3&\\ \hline &\hspace{0.5cm}-3x^3+0x^2+0x-81&\\ (-)&\hspace{0.5cm}-3x^3-9x^2&\\ \hline &\hspace{1.6cm}9x^2+0x-81&\\ (-)&\hspace{1.6cm}9x^2+27x&\\ \hline &\hspace{2.7cm}-27x-81&\\ (-)&\hspace{2.7cm}-27x-81&\\ \hline &\hspace{3.7cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }x^3-3x^2+9x-27.

\displaystyle \text{(iii)}
\displaystyle \begin{array}{rll} 3x-2 & )\overline{27x^3+0x^2+0x-8}( & 9x^2+6x+4\\ (-)&27x^3-18x^2&\\ \hline &\hspace{1.0cm}18x^2+0x-8&\\ (-)&\hspace{1.0cm}18x^2-12x&\\ \hline &\hspace{2.3cm}12x-8&\\ (-)&\hspace{2.3cm}12x-8&\\ \hline &\hspace{3.2cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }9x^2+6x+4.

\displaystyle \text{(iv)}
\displaystyle \begin{array}{rll} x^2-2 & )\overline{x^6+0x^4+0x^2-8}( & x^4+2x^2+4\\ (-)&x^6-2x^4&\\ \hline &\hspace{0.9cm}2x^4+0x^2-8&\\ (-)&\hspace{0.9cm}2x^4-4x^2&\\ \hline &\hspace{2.0cm}4x^2-8&\\ (-)&\hspace{2.0cm}4x^2-8&\\ \hline &\hspace{2.8cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }x^4+2x^2+4.

\displaystyle \text{(v)}
\displaystyle \begin{array}{rll} x-y & )\overline{x^6-y^6}( & x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5\\ (-)&x^6-x^5y&\\ \hline &\hspace{0.7cm}x^5y-y^6&\\ (-)&\hspace{0.7cm}x^5y-x^4y^2&\\ \hline &\hspace{1.7cm}x^4y^2-y^6&\\ (-)&\hspace{1.7cm}x^4y^2-x^3y^3&\\ \hline &\hspace{2.8cm}x^3y^3-y^6&\\ (-)&\hspace{2.8cm}x^3y^3-x^2y^4&\\ \hline &\hspace{3.8cm}x^2y^4-y^6&\\ (-)&\hspace{3.8cm}x^2y^4-xy^5&\\ \hline &\hspace{4.9cm}xy^5-y^6&\\ (-)&\hspace{4.9cm}xy^5-y^6&\\ \hline &\hspace{5.7cm}0& \end{array}
\displaystyle \therefore \text{the quotient is } x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5.

\displaystyle \text{(vi)}
\displaystyle \begin{array}{rll} 2x-3y & )\overline{16x^4-81y^4}( & 8x^3+12x^2y+18xy^2+27y^3\\ (-)&16x^4-24x^3y&\\ \hline &\hspace{0.8cm}24x^3y-81y^4&\\ (-)&\hspace{0.8cm}24x^3y-36x^2y^2&\\ \hline &\hspace{2.0cm}36x^2y^2-81y^4&\\ (-)&\hspace{2.0cm}36x^2y^2-54xy^3&\\ \hline &\hspace{3.3cm}54xy^3-81y^4&\\ (-)&\hspace{3.3cm}54xy^3-81y^4&\\ \hline &\hspace{4.5cm}0& \end{array}
\displaystyle \therefore \text{the quotient is } 8x^3+12x^2y+18xy^2+27y^3.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Divide and find the quotient and remainder:}
\displaystyle \text{(i) }6a^2-31a+47\text{ by }2a-5
\displaystyle \text{(ii) }2x^4-x^3+10x^2+8x-5\text{ by }x^2-x+6
\displaystyle \text{(iii) }3t^5+7t^4-11t^3+8t^2-32t+5\text{ by}
\displaystyle t^2+3t+2
\displaystyle \text{(iv) }x^6+3x^2+10\text{ by }x^3+1
\displaystyle \text{Answer:}

\displaystyle \text{(i)}
\displaystyle \begin{array}{rll} 2a-5 & )\overline{6a^2-31a+47}( & 3a-8\\ (-)&6a^2-15a&\\ \hline &\hspace{1.0cm}-16a+47&\\ (-)&\hspace{1.0cm}-16a+40&\\ \hline &\hspace{2.65cm}7& \end{array}
\displaystyle \text{Quotient: }3a-8
\displaystyle \text{Remainder: }7

\displaystyle \text{(ii)}
\displaystyle \begin{array}{rll} x^2-x+6 & )\overline{2x^4-x^3+10x^2+8x-5}( & 2x^2+x-1\\ (-)&2x^4-2x^3+12x^2&\\ \hline &\hspace{1.5cm}x^3-2x^2+8x-5&\\ (-)&\hspace{1.5cm}x^3-x^2+6x&\\ \hline &\hspace{2.0cm}-x^2+2x-5&\\ (-)&\hspace{2.0cm}-x^2+x-6&\\ \hline &\hspace{3.0cm}x+1& \end{array}
\displaystyle \text{Quotient: }2x^2+x-1
\displaystyle \text{Remainder: }x+1

\displaystyle \text{(iii)}
\displaystyle \begin{array}{rll} t^2+3t+2 & )\overline{3t^5+7t^4-11t^3+8t^2-32t+5}( & 3t^3-2t^2-11t+45\\ (-)&3t^5+9t^4+6t^3&\\ \hline &\hspace{0.8cm}-2t^4-17t^3+8t^2-32t+5&\\ (-)&\hspace{0.8cm}-2t^4-6t^3-4t^2&\\ \hline &\hspace{1.7cm}-11t^3+12t^2-32t+5&\\ (-)&\hspace{1.7cm}-11t^3-33t^2-22t&\\ \hline &\hspace{3.0cm}45t^2-10t+5&\\ (-)&\hspace{3.0cm}45t^2+135t+90&\\ \hline &\hspace{4.5cm}-145t-85& \end{array}
\displaystyle \text{Quotient: }3t^3-2t^2-11t+45
\displaystyle \text{Remainder: }-145t-85

\displaystyle \text{(iv)}
\displaystyle \begin{array}{rll} x^3+1 & )\overline{x^6+0x^3+3x^2+10}( & x^3-1\\ (-)&x^6+x^3&\\ \hline &\hspace{0.9cm}-x^3+3x^2+10&\\ (-)&\hspace{0.9cm}-x^3-1&\\ \hline &\hspace{2.0cm}3x^2+11& \end{array}
\displaystyle \text{Quotient: }x^3-1
\displaystyle \text{Remainder: }3x^2+11
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Show by division method that } (2a^2-a+3)\text{ is a factor of}
\displaystyle (6a^5-a^4+4a^3-5a^2-a-15).
\displaystyle \text{Answer:}

\displaystyle \begin{array}{rll} 2a^2-a+3 & )\overline{6a^5-a^4+4a^3-5a^2-a-15}( & 3a^3+a^2-2a-5\\ (-)&6a^5-3a^4+9a^3&\\ \hline &\hspace{1.0cm}2a^4-5a^3-5a^2-a-15&\\ (-)&\hspace{1.0cm}2a^4-a^3+3a^2&\\ \hline &\hspace{2.0cm}-4a^3-8a^2-a-15&\\ (-)&\hspace{2.0cm}-4a^3+2a^2-6a&\\ \hline &\hspace{3.0cm}-10a^2+5a-15&\\ (-)&\hspace{3.0cm}-10a^2+5a-15&\\ \hline &\hspace{4.0cm}0& \end{array}
\displaystyle \text{Quotient: }3a^3+a^2-2a-5
\displaystyle \text{Remainder: }0
\displaystyle \therefore (2a^2-a+3)\text{ is a factor of } (6a^5-a^4+4a^3-5a^2-a-15).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{What must be subtracted from } 8x^4+14x^3-2x^2+7x-8
\displaystyle \text{so that the resulting polynomial is exactly divisible by } (4x^2+3x-2)?
\displaystyle \text{Answer:}

\displaystyle \begin{array}{rll} 4x^2+3x-2 & )\overline{8x^4+14x^3-2x^2+7x-8}( & 2x^2+2x-1\\ (-)&8x^4+6x^3-4x^2&\\ \hline &\hspace{1.0cm}8x^3+2x^2+7x-8&\\ (-)&\hspace{1.0cm}8x^3+6x^2-4x&\\ \hline &\hspace{2.0cm}-4x^2+11x-8&\\ (-)&\hspace{2.0cm}-4x^2-3x+2&\\ \hline &\hspace{3.0cm}14x-10& \end{array}
\displaystyle \text{Quotient: }2x^2+2x-1
\displaystyle \text{Remainder: }14x-10
\displaystyle \therefore \text{we must subtract }14x-10 \text{ from the given polynomial.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }(2x-3y)\text{ units, }(7x+y) \text{ units,}
\displaystyle (x+12y)\text{ units and }(3y-4x)\text{ units are the lengths of the}
\displaystyle \text{sides of a quadrilateral, find the perimeter of the quadrilateral.}
\displaystyle \text{Answer:}
\displaystyle \text{Perimeter}=(2x-3y)+(7x+y)+(x+12y)+(3y-4x)
\displaystyle =(2+7+1-4)x+(-3+1+12+3)y
\displaystyle =6x+13y\text{ units}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{There are }(3x+5y)\text{ shelves in a library}
\displaystyle \text{and on each shelf there are }(5x+3y)\text{ books. How many books}
\displaystyle \text{are there in the library?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of books}=(3x+5y)(5x+3y)
\displaystyle =15x^2+9xy+25xy+15y^2
\displaystyle =15x^2+34xy+15y^2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The length and breadth of a rectangle are } (a+5b)\text{ units}
\displaystyle \text{and }(7a-b)\text{ units respectively. The perimeter of the rectangle}
\displaystyle \text{is equal to the perimeter of a square. Find how much the area of the}
\displaystyle \text{square exceeds the area of the rectangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Perimeter of rectangle} =2(a+5b)+2(7a-b)
\displaystyle =2a+10b+14a-2b=16a+8b.
\displaystyle \text{Area of rectangle}=(a+5b)(7a-b)
\displaystyle =7a^2-ab+35ab-5b^2=7a^2+34ab-5b^2.
\displaystyle \text{Since the perimeters of the rectangle and square are equal,}
\displaystyle \text{side of square}=\frac{16a+8b}{4}=4a+2b.
\displaystyle \text{Area of square}=(4a+2b)^2
\displaystyle =16a^2+16ab+4b^2.
\displaystyle \text{Area of square}-\text{Area of rectangle}
\displaystyle =(16a^2+16ab+4b^2)-(7a^2+34ab-5b^2)
\displaystyle =9a^2-18ab+9b^2=9(a-b)^2.
\displaystyle \therefore \text{the area of the square exceeds the area of the rectangle by}
\displaystyle 9(a-b)^2\text{ square units}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If a sum of Rs. }(16x^3-46x^2+39x-9) \text{ is to be}
\displaystyle \text{divided equally among }(8x-3)\text{ persons, find the amount received}
\displaystyle \text{by each person.}
\displaystyle \text{Answer:}

\displaystyle \begin{array}{rll} 8x-3 & )\overline{16x^3-46x^2+39x-9}( & 2x^2-5x+3\\ (-)&16x^3-6x^2&\\ \hline &\hspace{1.0cm}-40x^2+39x-9&\\ (-)&\hspace{1.0cm}-40x^2+15x&\\ \hline &\hspace{2.0cm}24x-9&\\ (-)&\hspace{2.0cm}24x-9&\\ \hline &\hspace{3.0cm}0& \end{array}
\displaystyle \text{Quotient: }2x^2-5x+3
\displaystyle \text{Remainder: }0
\displaystyle \therefore \text{each person will receive Rs. }(2x^2-5x+3).
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The product of two numbers is }(x^6-y^6). \text{ If one of the}
\displaystyle \text{numbers is }(x-y),\text{ find the other number.}
\displaystyle \text{Answer:}

\displaystyle \begin{array}{rll} x-y & )\overline{x^6-y^6}( & x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5\\ (-)&x^6-x^5y&\\ \hline &\hspace{1.0cm}x^5y-y^6&\\ (-)&\hspace{1.0cm}x^5y-x^4y^2&\\ \hline &\hspace{2.0cm}x^4y^2-y^6&\\ (-)&\hspace{2.0cm}x^4y^2-x^3y^3&\\ \hline &\hspace{3.0cm}x^3y^3-y^6&\\ (-)&\hspace{3.0cm}x^3y^3-x^2y^4&\\ \hline &\hspace{4.0cm}x^2y^4-y^6&\\ (-)&\hspace{4.0cm}x^2y^4-xy^5&\\ \hline &\hspace{5.0cm}xy^5-y^6&\\ (-)&\hspace{5.0cm}xy^5-y^6&\\ \hline &\hspace{6.0cm}0& \end{array}
\displaystyle \therefore \text{the other number is } x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Divide }6x^4-13x^3+2x^2+22x-24 \text{ by the product of}
\displaystyle (2x-3)\text{ and }(x^2-2x+2).
\displaystyle \text{Answer:}

\displaystyle \text{Step 1: Divide by }(x^2-2x+2).

\displaystyle \begin{array}{rll} x^2-2x+2 & )\overline{6x^4-13x^3+2x^2+22x-24}( & 6x^2-x-12\\ (-)&6x^4-12x^3+12x^2&\\ \hline &\hspace{1.0cm}-x^3-10x^2+22x-24&\\ (-)&\hspace{1.0cm}-x^3+2x^2-2x&\\ \hline &\hspace{2.0cm}-12x^2+24x-24&\\ (-)&\hspace{2.0cm}-12x^2+24x-24&\\ \hline &\hspace{3.0cm}0& \end{array}

\displaystyle \text{Step 2: Divide }(6x^2-x-12)\text{ by }(2x-3).

\displaystyle \begin{array}{rll} 2x-3 & )\overline{6x^2-x-12}( & 3x+4\\ (-)&6x^2-9x&\\ \hline &\hspace{1.0cm}8x-12&\\ (-)&\hspace{1.0cm}8x-12&\\ \hline &\hspace{1.5cm}0& \end{array}
\displaystyle \therefore \text{the quotient is }3x+4.
\displaystyle \\

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