Question 13 Marks
Divide
(i) $24 x y^2 z^3$ by $6 y z^2$
(ii) $63 a^2 b^4 c^6$ by $7 a^2 b^2 c^3$
Answer(i) $24 x y^2 z^3$ by $6 y z^2 \Rightarrow \frac{24 x y^2 z^3}{6 y z^2}$
$\begin{array}{l}\text { Now, } 24 x y^2 z^3=2 \times 2 \times 2 \times 3 \times x \times y \times y \times z \times z \times z \\ \text { and } \quad 6 y z^2=2 \times 3 \times y \times z \times z \\ \therefore \frac{24 x y^2 z^3}{6 y z^2}=\frac{2 \times 2 \times 2 \times 3 \times x \times y \times y \times z \times z \times z}{2 \times 3 \times y \times z \times z}\end{array}$
$=2 \times 2 \times x \times y \times z=4 x y z$
(ii) $63 a^2 b^4 c^6$ by $7 a^2 b^2 c^3 \Rightarrow \frac{63 a^2 b^4 c^6}{7 a^2 b^2 c^3}$
Now, $63 a^2 b^4 c^6=3 \times 3 \times 7 \times a \times a \times b \times b \times b \times b$
$
\times c \times c \times c \times c \times c \times c
$
and $7 a^2 b^2 c^3=7 \times a \times a \times b \times b \times c \times c \times c$
$\therefore \frac{63 a^2 b^4 c^6}{7 a^2 b^2 c^3}$
$\begin{aligned} & 3 \times 3 \times 7 \times a \times a \times b \times b \times b \times b \times c \\ = & \frac{\times c \times c \times c \times c \times c}{7 \times a \times a \times b \times b \times c \times c \times c} \\ = & 3 \times 3 \times b \times b \times c \times c \times c \\ = & 9 b^2 c^3\end{aligned}$
View full question & answer→Question 23 Marks
Factorise
(i) $12 x+36$
(ii) $22 y-33 z$
(iii) $14 p q+35 p q r$
Answer
(i) We have, $12 x=2 \times 2 \times 3 \times x$
$
36=2 \times 2 \times 3 \times 3
$
Both the terms have 2,2 and 3 as common factors.
Therefore,
$\begin{aligned} 12 x+36 & =(2 \times 2 \times 3 \times x)+(2 \times 2 \times 3 \times 3) \\ & =2 \times 2 \times 3 \times[(x+3)]=12(x+3)\end{aligned}$
which is the required factor form.
(ii) We have, $22 y-33 z$
$
\begin{array}{l}
22 y=2 \times 11 \times y \\
33 z=3 \times 11 \times z
\end{array}
$
Both the terms have 11 as common factor.
Therefore, $22 y-33 z=(2 \times 11 \times y)-(3 \times 11 \times z)$
$
=11 \times(2 y-3 z)=11(2 y-3 z)
$
which is the required factor form.
(iii) We have, $14 p q+35 p q r$
$
\begin{aligned}
14 p q & =2 \times 7 \times p \times q \\
35 p q r & =5 \times 7 \times p \times q \times r
\end{aligned}
$
Both the two terms have 7, $p$ and $q$ as common factors.
Therefore, $14 p q+35 p q r=(2 \times 7 \times p \times q)$
$
\begin{array}{l}
\quad-(5 \times 7 \times p \times q \times r) \\
=7 \times p \times q[(2)-(5 \times r)] \\
=7 p q(2-5 r)
\end{array}
$
which is the required factor form.
View full question & answer→Question 33 Marks
Factorise the expressions and divide them as directed.
(i) $\left(y^2+7 y+10\right)÷(y+5)$
(ii) $\left(m^2-14 m-32\right)÷(m+2)$
(iii) $\left(5 p^2-25 p+20\right)÷(p-1)$
(iv) $4 y z\left(z^2+6 z-16\right)÷2 y(z+8)$
(v) $5 p q\left(p^2-q^2\right)÷2 p(p+q)$
(vi) $12 x y\left(9 x^2-16 y^2\right) ÷ 4 x y(3 x+4 y)$
(vii) $39 y^3\left(50 y^2-98\right)÷26 y^2(5 y+7)$
Answer(i) Here, $y^2+7 y+10=y^2+5 y+2 y+10$
$
\begin{array}{l}
{\left[\begin{array}{l}
\because a b=10 \text { and } a+b=7 \\
\therefore a=5, b=2
\end{array}\right]} \\
=y(y+5)+2(y+5)=(y+5)(y+2)
\end{array}
$
Now,
$
\begin{aligned}
\left(y^2+7 y+10\right)÷(y+5) & =\frac{y^2+7 y+10}{(y+5)} \\
& =\frac{(y+5)(y+2)}{(y+5)} \\
& =(y+2)
\end{aligned}
$
(ii) Here, $m^2-14 m-32=m^2+(-16+2) m-32$
$
\left[\begin{array}{l}
\because a b=-32 \text { and } a+b=-14 \\
\therefore a=-16, b=2
\end{array}\right]
$
$
\begin{array}{l}
=m^2-16 m+2 m-32 \\
=m(m-16)+2(m-16) \\
=(m-16)(m+2)
\end{array}
$
Now, $\left(m^2-14 m-32\right)÷(m+2)$
$
\begin{array}{l}
=\frac{\left(m^2-14 m-32\right)}{(m+2)} \\
=\frac{(m-16)(m+2)}{(m+2)}=(m-16)
\end{array}
$
(iii)
$
\begin{array}{l}
\text { Here, } 5 p^2-25 p+20=5\left(p^2-5 p+4\right) \\
=5\left[p^2+(-4-1) p+4\right] \\
=5\left(p^2-4 p-p+4\right) \\
{\left[\begin{array}{l}
\because a b=4 \text { and } a+b=-5 \\
\therefore a=-4, b=-1
\end{array}\right]} \\
=5[p(p-4)-1(p-4)] \\
=5(p-4)(p-1) \\
\text { Now, }\left(5 p^2-25 p+20\right)÷(p-1) \\
=\frac{\left(5 p^2-25 p+20\right)}{(p-1)} \\
=\frac{5(p-4)(p-1)}{(p-1)}=5(p-4)
\end{array}
$
Part (iv) to (vii) Do same as above
Ans. (iv) $2 z(z-2)$
(v) $\frac{5}{2} q(p-q)$
(vi) $3(3 x-4 y)$
(vii) $3 y(5 y-7)$
View full question & answer→Question 43 Marks
Divide as directed.
(i) $5(2 x+1)(3 x+5)÷(2 x+1)$
(ii) $26 x y(x+5)(y-4)÷13 x(y-4)$
(iii) $52 p q r(p+q)(q+r)(r+p)÷104 p q(q+r)(r+p)$
(iv) $20(y+4)\left(y^2+5 y+3\right)÷5(y+4)$
(v) $x(x+1)(x+2)(x+3)÷x(x+1)$
Answer(i)
$
\text { } \begin{aligned}
5(2 x+1) & (3 x+5) ÷ (2 x+1) \\
= & \frac{5(2 x+1)(3 x+5)}{(2 x+1)} \\
= & 5(3 x+5)
\end{aligned}
$
(ii)
$
\begin{aligned}
26 x y(x+5) & (y-4) ÷ 13 x(y-4) \\
& =\frac{26 x y(x+5)(y-4)}{13 x(y-4)}=\frac{26 y(x+5)}{13} \\
= & \frac{2 \times 13 y(x+5)}{13}=2 y(x+5)
\end{aligned}
$
(iii)
$
\begin{array}{l}
52 p q r(p+q)(q+r)(r+p) \div 104 p q(q+r)(r+p) \\
=\frac{52 p q r(p+q)(q+r)(r+p)}{104 p q(q+r)(r+p)} \\
=\frac{52 r(p+q)}{104}=\frac{52 r(p+q)}{2 \times 52}=\frac{1}{2} r(p+q)
\end{array}
$
(iv)
$
\begin{aligned}
20(y+4) & \left(y^2+5 y+3\right) ÷ 5(y+4) \\
& =\frac{20(y+4)\left(y^2+5 y+3\right)}{5(y+4)} \\
& =\frac{4 \times 5\left(y^2+5 y+3\right)}{5} \\
& =4\left(y^2+5 y+3\right)
\end{aligned}
$
(v)
$
\begin{array}{l}
x(x+1)(x+2)(x+3)÷x(x+1) \\
=\frac{x(x+1)(x+2)(x+3)}{x(x+1)} \\
=(x+2)(x+3)
\end{array}
$
View full question & answer→Question 53 Marks
Work out the following divisions
(i) $(10 x-25) \div 5$
(ii) $(10 x-25) \div(2 x-5)$
(iii) $10 y(6 y+21) \div 5(2 y+7)$
(iv) $9 x^2 y^2(3 z-24) \div 27 x y(z-8)$
(v) $96 a b c(3 a-12)(5 b-30) \div 144(a-4)(b-6)$
Answer(i) $(10 x-25) \div 5=\frac{(10 x-25)}{5}=\frac{10 x}{5}-\frac{25}{5}=2 x-.5$
(ii) $(10 x-25) \div(2 x-5)=\frac{(10 x-25)}{(2 x-5)}=\frac{5(2 x-5)}{(2 x-5)}$
[taking 5 common in numerator] $=5$
(iii)
$
\begin{array}{l}
10 y(6 y+21) \div 5(2 y+7) \\
= \frac{10 y(6 y+21)}{5(2 y+7)}=\frac{2 \times 5 \times y \times 3(2 y+7)}{5(2 y+7)}
\end{array}
$
[taking 3 common in numerator]
$
=2 \times y \times 3=6 y
$
(iv)
$
\begin{aligned}
9 x^2 y^2(3 z-24) & \div 27 x y(z-8)=\frac{9 x^2 y^2(3 z-24)}{27 x y(z-8)} \\
& =\frac{3 \times 3 \times x \times x \times y \times y \times 3(z-8)}{3 \times 3 \times 3 \times x \times y(z-8)}=x y
\end{aligned}
$
(v)
$\begin{aligned} 96 a b c & (3-12)(5 b-30) ÷ 144(a-4)(b-6) \\ & =\frac{96 a b c(3 a-12)(5 b-30)}{144(a-4)(b-6)} \\ & =\frac{96 a b c \times 3(a-4) \times 5(b-6)}{144(a-4)(b-6)} \\ & =\frac{96 a b c \times 3 \times 5}{144}=\frac{96 a b c \times 5}{48} \\ & =2 a b c \times 5=10 a b c\end{aligned}$
View full question & answer→Question 63 Marks
Divide the given polynomial by the gliven monomial.
(i) $\left(5 x^2-6 x\right) ÷ 3 x$
(ii) $\left(3 y^8-4 y^6+5 y^4\right) \div y^4$
(iii) $8\left(x^3 y^2 z^2+x^2 y^3 z^2+x^2 y^2 z^3\right) ÷ 4 x^2 y^2 z^2$
(iv) $\left(x^3+2 x^2+3 x\right) ÷ 2 x$
(v) $\left(p^3 q^6-p^6 q^3\right) \div p^3 q^3$
Answer(i)
$
\begin{aligned}
\left(5 x^2-6 x\right)+3 x & =\frac{5 x^2-6 x}{3 x}=\frac{5 x^2}{3 x}-\frac{6 x}{3 x} \\
& =\frac{5}{3} x-2=\frac{5 x-6}{3}=\frac{1}{3}(5 x-6)
\end{aligned}
$
(ii)
$
\begin{aligned}
\left(3 y^8-4 y^6\right. & \left.+5 y^4\right)+y^4 \\
& =\frac{3 y^8-4 y^6+5 y^4}{y^4}=\frac{3 y^8}{y^4}-\frac{4 y^6}{y^4}+\frac{5 y^4}{y^4} \\
& =3 y^4-4 y^2+5
\end{aligned}
$
(iii)
$
\begin{array}{l}
8\left(x^3 y^2 z^2+x^2 y^3 z^2+x^2 y^2 z^3\right) \div 4 x^2 y^2 z^2 \\
=\frac{8\left(x^3 y^2 z^2+x^2 y^3 z^2+x^2 y^2 z^3\right)}{4 x^2 y^2 z^2} \\
=\frac{2\left(x^3 y^2 z^2+x^2 y^3 z^2+x^2 y^2 z^3\right)}{x^2 y^2 z^2} \\
=2\left[\frac{x^3 y^2 z^2}{x^2 y^2 z^2}+\frac{x^2 y^3 z^2}{x^2 y^2 z^2}+\frac{x^2 y^2 z^3}{x^2 y^2 z^2}\right] \\
=2(x+y+z)
\end{array}
$
(iv)
$
\begin{aligned}
\left(x^3+2 x^2+3 x\right) & \div 2 x \\
& =\frac{x^3+2 x^2+3 x}{2 x} \\
& =\frac{x^3}{2 x}+\frac{2 x^2}{2 x}+\frac{3 x}{2 x} \\
& =\frac{x^2}{2}+\frac{2 x}{2}+\frac{3}{2} \\
& =\frac{x^2+2 x+3}{2}=\frac{1}{2}\left(x^2+2 x+3\right)
\end{aligned}
$
(v)
$
\begin{aligned}
\left(p^3 q^6-p^6 q^3\right) & \div p^3 q^3 \\
& =\frac{\left(p^3 q^6-p^6 q^3\right)}{p^3 q^3}=\frac{p^3 q^6}{p^3 q^3}-\frac{p^6 q^3}{p^3 q^3} \\
& =q^3-p^3
\end{aligned}
$
View full question & answer→Question 73 Marks
Carry out the following divisions.
(i) $28 x^4+56 x$
(ii) $-36 y^3+9 y^2$
(iii) $66 p q^2 r^3+11 q r^2$
(iv) $34 x^3 y^3 z^3+51 x y^2 z^3$
(v) $12 a^8 b^8+\left(-6 a^6 b^4\right)$
Answer(i)
$\begin{array}{l}\text { } \begin{array}{l} 28 x^4+56 x \\ \text { Here, } 28 x^4=2 \times 2 \times 7 \times x \times x \times x \times x \\ \text { and } \quad 56 x=2 \times 2 \times 2 \times 7 \times x \\ \therefore \quad \frac{28 x^4}{56 x}=\frac{2 \times 2 \times 7 \times x \times x \times x \times x}{2 \times 2 \times 2 \times 7 \times x} \\ =\frac{x \times x \times x}{2}=\frac{1}{2} x^3\end{array}\end{array}$
(ii)
$\begin{array}{l}-36 y^3+9 y^2 \\ \text { Here, }-36 y^3=(-1) \times 2 \times 2 \times 3 \times 3 \times y \times y \times y \\ \text { and } \quad 9 y^2=3 \times 3 \times y \times y \\ \therefore \quad \frac{-36 y^3}{9 y^2}=\frac{(-1) \times 2 \times 2 \times 3 \times 3 \times y \times y \times y}{3 \times 3 \times y \times y} \\ =(-1) \times 2 \times 2 \times y=-4 y\end{array}$
(iii)
$
\begin{array}{l}
66 p q^2 r^3+11 q r^2 \\
\text { Here, } 66 p q^2 r^3=2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r \\
\text { and } \quad 11 q r^2=11 \times q \times r \times r \\
\begin{aligned}
\therefore \frac{66 p q^2 r^3}{11 q r^2} & =\frac{2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r}{11 \times q \times r \times r} \\
& =2 \times 3 \times p \times q \times r=6 p q r
\end{aligned}
\end{array}
$
(iv) Do same as part (iii) Ans. $\frac{2}{3} x^2 y$
(v) Do same as part (iii) Ans. $-2 a^2 b^4$
View full question & answer→Question 83 Marks
Factorise the following expressions.
(i) $p^2+6 p+8$ $\qquad$ (ii) $q^2-10 q+21$ $\qquad$ (iii) $p^2+6 p-16$
Answer(i) $p^2+6 p+8$
Here, $a b=8$ and $a+b=6$
On solving these equations, we get $a=4$ and $b=2$
Now,
$
\begin{aligned}
p^2+6 p+8 & =p^2+(4+2) p+8 \\
& =p^2+4 p+2 p+8 \\
& =p(p+4)+2(p+4) \\
& =(p+4)(p+2)
\end{aligned}
$
Hence, the required factors of given expression are $(p+4)$ and $(p+2)$.
Alternate method
$
\begin{aligned}
p^2+6 p+8 & =p^2+6 p+9-1 \quad \quad[\because 8=9-1] \\
& =\left(p^2+6 p+9\right)-1 \\
& =\left[(p)^2+2(p)(3)+(3)^2\right]-(1)^2 \\
& =(p+3)^2-(1)^2 \\
& {\left[\because a^2+2 a b+b^2=(a+b)^2\right] } \\
& =(p+3+1)(p+3-1)
\end{aligned}
$
[using Identity III]
$
=(p+4)(p+2)
$
(ii) $q^2-10 q+21$
Here, $a b=21$ and $a+b=-10$
Possible values of $a$ and $b$ are 7,3 or $-7,-3$.
$
\begin{array}{l}
\text { But } \quad 7+3=10 \neq-10 \qquad[not possible]\\
\therefore \quad a=-7 \text { and } b=-3 \\
\text { Now, } q^2-10 q+21=q^2+(-7-3) q+21 \\
=q^2-7 q-3 q+21 \\
=q(q-7)-3(q-7) \\
=(q-7)(q-3)
\end{array}
$
Hence, the required factors of given expression are
$(a-7)$ and $(a-3)$.
(iii) Do same as part (i) Ans. $(p+8)(p-2)$
View full question & answer→Question 93 Marks
Factorise.
(i) $a^4-b^4$
(ii) $p^4-81$
(iii) $x^4-(y+z)^4$
(iv) $x^4-(x-z)^4$
(v) $a^4-2 a^2 b^2+b^4$
Answer(i) $a^4-b^4=\left(a^2\right)^2-\left(b^2\right)^2$
On comparing with $a^2-b^2$, we get
$
\begin{aligned}
a & =a^2 \text { and } b=b^2 \\
\therefore \quad a^4-b^4 & =\left(a^2+b^2\right)\left(a^2-b^2\right) \\
& =\left(a^2+b^2\right)(a+b)(a-b)
\end{aligned}
$
[using Identity III]
and $a^2+b^2$ cannot be factorised further
(ii) $p^4-81=\left(p^2\right)^2-(9)^2$
On comparing with $a^2-b^2$, we get $a=p^2$ and $b=9$
$
\begin{aligned}
\therefore p^4-81= & \left(p^2+9\right)\left(p^2-9\right) \\
& {\left[\because\left(a^2-b^2\right)=(a+b)(a-b)\right] } \\
= & \left(p^2+9\right)\left[(p)^2-(3)^2\right] \\
= & \left(p^2+9\right)(p+3)(p-3) \text { [using Identity III] }
\end{aligned}
$
and $\left(p^2+9\right)$ cannot be factorised further
(iii) $x^4-(y+z)^4=\left(x^2\right)^2-\left[(y+z)^2\right]^2$
On comparing with $a^2-b^2$, we get $a = x^2$ and $b=(y+z)^2$
$\therefore x^4$ $
-(y+z)^4=\left[x^2+(y+z)^2\right]\left[x^2-(y+z)^2\right]
$
[using Identity III]
$
=\left[x^2+(y+z)^2\right]+[(x+y+z) { x-(y+z) }]
$
[again, using Identity III in second term]
$
=\left[x^2+(y+z)^2\right][(x+y+z)(x-y-z)]
$
(iv) $x^4-(x-z)^4=\left(x^2\right)^2-\left\{(x-z)^2\right\}^2$
On comparing with $a^2-b^2$, we get
$
\begin{array}{l}
a=x^2 \text { and } b=(x-z)^2 \\
\therefore x^4-(x-z)^4=\left[x^2+(x-z)^2\right]\left[x^2-(x-z)^2\right]
\end{array}
$
[using Identity III]
$
=\left[x^2+(x-z)^2\right][(x+x-z)(x-(x-z))]
$
[again, using Identity III in second term]
$
\begin{array}{l}
=\left[x^2+(x-z)^2\right][(2 x-z)(x-x+z)] \\
=\left[x^2+(x-z)^2\right][z(2 x-z)] \\
=\left[x^2+\left(x^2-2 x z+z^2\right)\right][z(2 x-z)]
\end{array}
$
[using Identity II]
$
\begin{array}{l}
=\left[2 x^2+z^2-2 x z\right][z(2 x-z)] \\
=\left(2 x^2+z^2-2 x z\right) z(2 x-z) \\
=z\left(2 x^2+z^2-2 x z\right)(2 x-z)
\end{array}
$
(v) $a^4-2 a^2 b^2+b^4=\left(a^2\right)^2-2\left(a^2\right)(b)^2+\left(b^2\right)^2$
On comparing with $a^2-2 a b+b^2$, we get $a=a^2$ and $b=b^2$
$\therefore\left(a^2\right)^2-2\left(a^2\right)\left(b^2\right)+\left(b^2\right)^2=\left(a^2-b^2\right)^2$ $\qquad$ [using Identity II]
$=[(a+b)(a-b)]^2\qquad$ [using Identity III]
$=(a+b)^2(a-b)^2 \quad\left[\because(a b)^2=a^2 b^2\right]$
$=(a+b)(a+b)(a-b)(a-b)$
View full question & answer→Question 103 Marks
Factorise the expressions.
(i) $a x^2+b x$
(ii) $7 p^2+21 q^2$
(iii) $2 x^3+2 x y^2+2 x z^2$
(iv) $a m^2+b m^2+b n^2+a n^2$
(v) $(l m+l)+m+1$
(vi) $y(y+z)+9(y+z)$
(vii) $5 y^2-20 y-8 z+2 y z$
(viii) $10 a b+4 a+5 b+2$
(ix) $6 x y-4 y+6-9 x$
Answer(i) $a x^2+b x$
Here, $a x^2=a \times x \times x$ and $b x=b \times x$
Both the terms have $x$ as common factor.
$
\therefore a x^2+b x=a \times x \times x+b \times x=x(a x+b)
$
[taking $x$ common]
Hence the factors of given ecpression are $x$ and $ax+b$.
$7p^2$ + $2lq^2$
Here, $7 p^2=7 \times p \times p$ and $21 q^2=7 \times 3 \times q \times q$
Both the terms have 7 as common factor.
$
\begin{aligned}
\therefore 7 p^2+21 q^2 & =7 \times p \times p+3 \times 7 \times q \times q \\
& =7\left(p^2+3 q^2\right) \quad[\text { taking } 7 \text { common }]
\end{aligned}
$
Hence, the factors of given expression are 7 and $\left(p^2+3 q^2\right)$.
(iii) $2 x^3+2 x y^2+2 x z^2$
Here, $2 x^3=2 \times x \times x \times x, 2 x y^2=2 \times x \times y \times y$ and $2 x z^2=2 \times x \times z \times z$
The above given three terms have 2 and $x$ as common factors.
$
\therefore 2 x^3+2 x y^2+2 x z^2=2 \times x(x \times x+y \times y+z \times z)
$
[taking 2 and $x$ common]
$
=2 x\left(x^2+y^2+z^2\right)
$
Hence, the factors of given expression are $2 x$ and
$\left(x^2+y^2+z^2\right)$.
(iv) $a m^2+b m^2+b n^2+a n^2=m^2(a+b)+n^2(b+a)$
[here, $m^2$ is common in first two terms and $n^2$ is common in last two terms]
$\begin{array}{l}=m^2(a+b)+n^2(a+b) \quad[\because a+b=b+a] \\ =(a+b)\left(m^2+n^2\right)\end{array}$
Hence, the factors of given expression are $(a+b)$ and $\left(m^2+n^2\right)$.
Part (v) to (ix) Do same as above
Ans. (v) $(l+1)(m+1)$ $\qquad$ (vi) $(y+9)(y+z)$
(vii) $(5 y+2 z)(y-4)$ $\qquad$ (viii) $(2 a+1)(5 b+2)$
(ix) $(2 y-3)(3 x-2)$
View full question & answer→Question 113 Marks
Factorise.
(i) $4 p^2-9 q^2$
(ii) $63 a^2-112 b^2$
(iii) $49 x^2-36$
(iv) $16 x^5-144 x^3$
(v) $(l+m)^2-(l-m)^2$
(vi) $9 x^2 y^2-16$
(vii) $\left(x^2-2 x y+y^2\right)-z^2$
(viii) $25 a^2-4 b^2+28 b c-49 c^2$
Answer(i) $4 p^2-9 q^2=(2 p)^2-(3 q)^2$
This expression is of the form $a^2-b^2$
On comparing, we get
$
\begin{array}{c}
a=2 p \text { and } b=3 q \\
\therefore \quad 4 p^2-9 q^2=(2 p+3 q)(2 p-3 q)
\end{array}
$
[using Identity III]
Hence, $4 p^2-9 q^2=(2 p+3 q)(2 p-3 q)$
(ii) $63 a^2-112 b^2=7 \times 9 a^2-7 \times 16 b^2=7\left[(3 a)^2-(4 b)^2\right]$
This expression is of the form $a^2-b^2$.
On comparing, we get
$
\begin{array}{c}
a=3 \text { and } b=4 b \\
\therefore \quad 7\left[(3 a)^2-(4 b)^2\right]=7(3 a+4 b)(3 a-4 b)
\end{array}
$
[using Identity III]
Hence, $63 a^2-112 b^2=7(3 a+4 b)(3 a-4 b)$
(iii) $49 x^2-36=(7 x)^2-(6)^2$
This expression is of the form $a^2-b^2$.
On comparing, we get $a=7 x$ and $b=6$
$\therefore \quad 49 x^2-36=(7 x+6)(7 x-6) \quad$ [using Identity III]
(iv)
$\begin{array}{l} 16 x^5-144 x^3=2 \times 2 \times 2 \times 2 \times x \times x \times x \times x \times x \\ -2 \times 2 \times 2 \times 2 \times 3 \times 3 \times x \times x \times x \\ = 2 \times 2 \times 2 \times 2 \times x \times x \times x(x \times x-3 \times 3) \\ =16 x^3\left(x^2-9\right)\end{array}$
Here, the expression $\left(x^2-9\right)$ is of the form $a^2-b^2$.
On comparing, we get $a=x$ and $b=3$
$\therefore \quad\left(x^2-9\right)=(x+3)(x-3) \quad$ [using Identity III]
Hence, $\quad 16 x^5-144 x^3=16 x^3\left(x^2-9\right)$
$=16 x^3(x+3)(x-3)$.
Part (v) to (viii) Do same as above.
Ans. (v) 4 lm
(vi) $(3 x y-4)(3 x y+4)$
(vii) $(x-y-z)(x-y+z)$
(viii) $(5 a-2 b+7 c)(5 a+2 b-7 c)$
View full question & answer→Question 123 Marks
Factorise the following expressions.
(i) $a^2+8 a+16$
(ii) $p^2-10 p+25$
(iii) $25 m^2+30 m+9$
(iv) $49 y^2+84 y z+36 z^2$
(v) $4 x^2-8 x+4$
(vi) $121 b^2-88 b c+16 c^2$
(vii) $(I+m)^2-4 / m$ [Hint Expand $(I+m)^2$ first]
(viii) $a^4+2 a^2 b^2+b^4$
AnswerGiven,
(i) $a^2+8 a+16=(a)^2+2(a)(4)+(4)^2=(a)^2+8 a+(4)^2$
Here, first and third terms are perfect squares.
This expression is of the form $a^2+2 a b+b^2$.
On comparing, we get $a=a$ and $b=4$
Since, $a^2+2 a b+b^2=(a+b)^2$
$
\therefore \quad a^2+8 a+16=(a+4)^2 \text { or }(a+4)(a+4)
$
(ii) $p^2-10 p+25=(p)^2-10 p+(5)^2=p^2-2(5) p+5^2$
This expression is of the form $a^2-2 a b+b^2$.
On comparing, we get $a=p$ and $b=5$
Since, $\quad a^2-2 a b+b^2=(a-b)^2$
$
\therefore p^2-10 p+25=(p-5)^2 \text { or }(p-5)(p-5)
$
(iii)
$
\begin{aligned}
25 m^2+30 m+9 & =(5 m)^2+30 m+(3)^2 \\
& =(5 m)^2+2(5) m(3)+(3)^2
\end{aligned}
$
This expression is of the form $a^2+2 a b+b^2$.
On comparing, we get $a=5 m$ and $b=3$
Since, $a^2+2 a b+b^2=(a+b)^2$
$
\therefore \quad 25 m^2+30 m+9=(5 m+3)^2 \text { or }(5 m+3)(5 m+3)
$
Part (iv) to (viii) Do same as above
Ans. (iv) $(7 y+6 z)^2$ $\qquad$ (v) $4(x-1)^2$
(vi) $(11 b-4 c)^2$ $\qquad$ (vii) $(l-m)^2$
(ix) $\left(a^2+b^2\right)^2$
View full question & answer→Question 133 Marks
Factorise
(i) $x^2+x y+8 x+8 y$
(ii) $15 x y-6 x+5 y-2$
(iii) $a x+b x-a y-b y$
(iv) $15 p q+15+9 q+25 p$
(v) $z-7+7 x y-x y z$
Answer(i)
$
\begin{aligned}
x^2+ & x y+8 x+8 y=\left(x^2+x y\right)+(8 x+8 y) \\
& =(x \times x+x \times y)+(8 \times x+8 \times y)
\end{aligned}
$
[here, in first term $x$ is common and in second term 8 is common]
$
=x(x+y)+8(x+y)=(x+y)(x+8)
$
Hence, the factors of given expression are $(x+y)$ and $(x+8)$.
(ii)
$
\begin{aligned}
15 x y- & 6 x+5 y-2=(15 x y-6 x)+(5 y-2) \\
& =(3 \times 5 \times x \times y-3 \times 2 \times x)+(1 \times 5 \times y-1 \times 2)
\end{aligned}
$
[here, in first term $3 x$ is common and in second term 1 is common]
$
\begin{array}{l}
=3 x(5 y-2)+1(5 y-2) \\
=(5 y-2)(3 x+1)
\end{array}
$
Hence, the factors of given expression are $(5 y-2)$ and $(3 x+1)$.
(iii)
$
\begin{array}{c}
a x+b x-a y-b y=(a x+b x)-(a y+b y) \\
=(a \times x+b \times x)-(a \times y+b \times y)
\end{array}
$
[here, in first term $x$ is common and in second term $y$ is common]
$
=x(a+b)-y(a+b)=(a+b)(x-y)
$
Hence, the factors of given expression are $(a+b)$ and $(x-y)$.
(iv) Do same as part (ii) Ans. $(3 q+5)(5 p+3)$.
(v) Do same as part (ii) Ans. $(1-x y)(z-7)$.
View full question & answer→Question 143 Marks
Factorise the following expressions.
(i) $7 x-42$
(ii) $6 p-12 q$
(iii) $7 a^2+14 a$
(iv) $-16 z+20 z^3$
(v) $201^2 m+30$ a $1 m$
(vi) $5 x^2 y-15 x y^2$
(vii) $10 a^2-15 b^2+20 c^2$
(viii) $-4 a^2+4 a b-4 c a$
(ix) $x^2 y z+x y^2 z+x y z^2$
(x) $a x^2 y+b x y^2+c x y z$
Answer(i) We have, $7 x=7 \times x$ and $42=2 \times 3 \times 7$
Both the terms have 7 as common factor.
Therefore,
$
\begin{aligned}
7 x-42 & =7 \times x-2 \times 3 \times 7 \\
& =7(x-2 \times 3)[\text { taking } 7 \text { common }] \\
& =7(x-6)
\end{aligned}
$
which is the required factor form.
(ii) We have,
$
\begin{aligned}
6 p & =2 \times 3 \times p \\
and \qquad12 q & =2 \times 2 \times 3 \times q
\end{aligned}
$
Both the terms have 2 and 3 as common factors.
Therefore,
$
\begin{aligned}
6 p-12 q= & 2 \times 3 \times p-2 \times 2 \times 3 \times q \\
= & 2 \times 3(p-2 \times q) \\
& \quad \text { [taking } 2 \text { and } 3 \text { common] } \\
= & 6(p-2 q)
\end{aligned}
$
which is the required factor form.
(iii) We have, $7 a^2=7 \times a \times a$ and $14 a=2 \times 7 \times a$
Both the terms have 7 and $a$ as common factors.
Therefore,
$
\begin{aligned}
7 a^2+14 a= & 7 \times a \times a+2 \times 7 \times a \\
& =7 \times a(a+2) \\
& \quad[\text { taking } 7 \text { and } a \text { common }] \\
& =7 a(a+2)
\end{aligned}
$
which is the required factor form.
(iv) We have, $\quad-16 z=(-1) \times 2 \times 2 \times 2 \times 2 \times z$
and $\quad 20 z^3=2 \times 2 \times 5 \times z \times z \times z$
Both the terms have 2,2 and $z$ as common factors.
Therefore, $-16 z+20 z^3=(-1) \times 2 \times 2 \times 2 \times 2 \times z$
$\begin{array}{r}
+2 \times 2 \times 5 \times z \times z \times z \\
=2 \times 2 \times z[(-1) \times 2 \times 2+5 \times z \times z] \\
\quad \text { taking } 2,2 \text { and } z \text { common] } \\
=4 z\left(-4+5 z^2\right)
\end{array}$
which is the required factor form.
Part (v) to (x) Do sams as above
Ans. (v) $10 lm(2 l+3 a)$ $\qquad$ (vi) $5 x y(x-3 y)$
(vii) $5\left(2 a^2-3 b^2+4 c^2\right)$ $\qquad$ (viii) $4 a(-a+b-c)$
(ix) $x y z(x+y+z)$ $\qquad$ (x) $x y(a x+b y+c z)$
View full question & answer→Question 153 Marks
Find the common factors of the given terms.
(i) $12 x, 36$
(ii) $2 y, 22 x y$
(iii) $14 p q, 28 p^2 q^2$
(iv) $2 x, 3 x^2, 4$
(v) $6 a b c, 24 a b^2, 12 a^2 b$
(vi) $16 x^3,-4 x^2, 32 x$
(vii) $10pq, 20qr, 30 rp$
(viii) $3 x^2 y^3, 10 x^3 y^2, 6 x^2 y^2 z$
Answer(i) We have, $12 x =2 \times 2 \times 3 \times x$
and $\quad 36=2 \times 2 \times 3 \times 3$
Both the terms have 2,2 and 3 as common factors
So, the common factor of given terms $=2 \times 2 \times 3=12$
(ii) We have, $\quad 2 y=2 \times y$
and $\quad 22 x y=2 \times 11 \times x \times y$
Both the terms have 2 and $y$ as common factors.
So, the common factor of given terms $=2 \times y=2 y$
(iii) We have, $14 p q=2 \times 7 \times p \times q$
and $\quad 28 p^2 q^2=2 \times 2 \times 7 \times p \times p \times q \times q$
Both the terms have 2,7,p and $q$ as common factors.
So, the common factor of given terms
$
=2 \times 7 \times p \times q=14 p q
$
Part (iv) to (viii) Do same as above
Ans. (iv) 1 $\qquad$ (v) $6 a b$ $\qquad$ (vi) $4 x$ $\qquad$ (vii) $10 \quad$ (viii) $x^2 y^2$
View full question & answer→Question 163 Marks
Verify that
$(11 p q+4 q)^2-(11 p q-4 q)^2=176 p q^2$
Answer
$\begin{aligned} \text { LHS }= & (11 p q+4 q)^2-(11 p q-4 q)^2 \\ & =(11 p q+4 q+11 p q-4 q) \times(11 p q+4 q-11 p q+4 q) \\ & \text { [using Identity III] }\end{aligned}$
where $a=11 p q+4 q$ and $b=11 p q-4 q]$
$
\begin{array}{l}
=(22 p q)(8 q) \\
=176 p q^2=\text { RHS } \qquad Hence proved
\end{array}
$
View full question & answer→Question 173 Marks
Factorise $x^4-256$.
AnswerWe have, $x^4-256$
$
\begin{array}{lr}
=\left(x^2 \times x^2\right)-(16 \times 16)=\left[\left(x^2\right)^2-(16)^2\right] \\
=\left(x^2+16\right)\left(x^2-16\right) & \text { [using Identity III] } \\
=\left(x^2+16\right)\left(x^2-4 \times 4\right) & \\
=\left(x^2+16\right)\left[(x)^2-(4)^2\right] & \\
=\left(x^2+16\right)(x+4)(x-4) & \text { [using Identity III] }
\end{array}
$
View full question & answer→Question 183 Marks
"When a monomial is divided by a monomial, the quotient is a monomial."
Siya disagrees with the statement as $x^2$ divided by $x^2$, is 1. Is Siya correct? Justify your answer.
AnswerNo, Siya is not correct.
The given statement i.e. when a monomial is divided by a monomial; the quotient is a monomial, is correct.
Justification $x^2$ and $x^2$ both are monomials.
$
\begin{array}{l}
x^2=x \times x \\
x^2=x \times x
\end{array}
$
Now, on dividing $x^2$ by $x^2$, we get
$
x^2 ÷ x^2=\frac{x \times x}{x \times x}=1
$
which is also a monomial.
View full question & answer→Question 193 Marks
Satpal solved an algebraic equation as shown below.
Algebraic equation: $(s+4)^2=36+s^2$
Step I $s^2+16+8 s=36+s^2$
Step II $8 s=52$
Step III $s=52 \times 8=132$
Did Satpal solve the equation correctly? Explain your answer.
AnswerGiven algebraic equation $(s+4)^2=36+s^2$
Step I Using Identity I on LHS, we get
$
s^2+16+8 s=36+s^2
$
Step II $8 s=36-16=20$
which shows that given Step II does not match with our Step II.
This shows that mistake of adding 16 instead of subtraction was done by Satpal in Step II.
Hence, Step II solved by Satpal was wrong.
View full question & answer→Question 203 Marks
Verify the following expression
$
\left(\frac{3 p}{7}+\frac{7}{6 p}\right)^2-\left(\frac{3}{7} p-\frac{7}{6 p}\right)^2=2
$
Answer
$
\begin{aligned}
\text { LHS }= & \left(\frac{3 p}{7}+\frac{7}{6 p}\right)^2-\left(\frac{3 p}{7}-\frac{7}{6 p}\right)^2 \\
= & {\left[\left(\frac{3 p}{7}\right)^2+\left(\frac{7}{6 p}\right)^2+2 \times\left(\frac{3 p}{7}\right) \times\left(\frac{7}{6 p}\right)\right] } \\
& -\left[\left(\frac{3 p}{7}\right)^2+\left(\frac{7}{6 p}\right)^2-2 \times\left(\frac{3 p}{7}\right) \times\left(\frac{7}{6 p}\right)\right]
\end{aligned}
$
[using Identities I in I term and Identity II in II term,
$
\text { where } \left.a=\frac{3 p}{7}, b=\frac{7}{6 p}\right]
$
$
\begin{array}{l}
=\frac{9 p^2}{49}+\frac{49}{36 p^2}+1-\left[\frac{9 p^2}{49}+\frac{49}{36 p^2}-1\right] \\
=\frac{9 p^2}{49}+\frac{49}{36 p^2}+1-\frac{9 p^2}{49}-\frac{49}{36 p^2}+1=2 \\
\text { RHS }=2 \\
\text { LHS }=\text { RHS } \qquad \text {Hence proved.}
\end{array}
$
View full question & answer→Question 213 Marks
Factorise $x^2+\frac{1}{x^2}+2-3 x-\frac{3}{x}$.
Answer
$\begin{array}{l}\text { We have, } x^2+\frac{1}{x^2}+2-3 x-\frac{3}{x} \\ =(x)^2+\left(\frac{1}{x}\right)^2+2(x) \times\left(\frac{1}{x}\right)-3 x-\frac{3}{x} \\ \quad\left[\text { since, } 1=x \times \frac{1}{x}\right]\end{array}$
$\begin{array}{l}=\left(x+\frac{1}{x}\right)^2-3\left(x+\frac{1}{x}\right) \\ \quad\left[\text { using Identity I, where } a=x, b=\frac{1}{x}\right] \\ =\left(x+\frac{1}{x}\right)\left[\left(x+\frac{1}{x}\right)-3\right]=\left(x+\frac{1}{x}\right)\left(x+\frac{1}{x}-3\right)\end{array}$
View full question & answer→Question 223 Marks
Factorise $x^2+19 x-20$ in the form $(x-a)(x+b)$.
Answer
$\begin{aligned} x^2+19 x-20 =x^2+20 x-x-20 \\ =x(x+20)-1(x+20)=(x+20)(x-1)\end{aligned}$
View full question & answer→Question 233 Marks
Divide $21(y+3)\left(y^2-16\right)$ by $7\left(y^2-y-12\right)$.
AnswerWe have, $21(y+3)\left(y^2-16\right)+7\left(y^2-y-12\right)$
$
\begin{aligned}
\text { Factorising } & 21(y+3)\left(y^2-16\right) \\
= & 21(y+3)\left[(y)^2-4 \times 4\right] \\
= & 21(y+3)\left[(y)^2-(4)^2\right]
\end{aligned}
$
$
\begin{array}{l}
=21(y+3)(y+4)(y-4) \\
=3 \times 7(y+3)(y+4)(y-4)
\end{array}
$
Now, factorising $7\left(y^2-y-12\right)$
$
\text { Since, } \begin{array}{ll}
& 4 \times(-3)=-12 \\
& 4+(-3)=1 \text { or } 4-3=1
\end{array}
$
On putting these values in given expression, we get
$
\begin{array}{l}
7\left(y^2-y-12\right)=7\left[y^2-(4-3) y-12\right] \\
=7\left(y^2-4 y+3 y-12\right) \\
=7[(y(y-4)+3(y-4)] \\
=7[(y-4)(y+3)]=7(y-4)(y+3) \\
\therefore \frac{21(y+3)\left(y^2-16\right)}{7\left(y^2-y-12\right)} \\
=\frac{3 \times 7(y+3)(y+4)(y-4)}{7(y-4)(y+3)}=3(y+4)
\end{array}
$
View full question & answer→Question 243 Marks
Factorise $\frac{1}{36} a^2 b^2-\frac{16}{49} b^2 c^2$.
AnswerWe have,
$
\begin{array}{l}
\frac{1}{36} a^2 b^2-\frac{16}{49} b^2 c^2 \\
=b^2\left(\frac{a^2}{36}-\frac{16 c^2}{49}\right)=b^2\left(\frac{a^2}{6 \times 6}-\frac{4 \times 4}{7 \times 7} c^2\right) . \\
=b^2\left[\left(\frac{a}{6}\right)^2-\left(\frac{4 c}{7}\right)^2\right] \\
=b^2\left(\frac{a}{6}+\frac{4 c}{7}\right)\left(\frac{a}{6}-\frac{4 c}{7}\right) \quad \text { [using Identity III] }
\end{array}
$
View full question & answer→