In Text Questions (Text Book Page No. 74, 75,77,78,79,81,83,84,86,87,88,91,94) - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Recap (Text Book Page No. 74 \& 75) : Chapter 3 - Algebra - 8th Maths Guide Samacheer Kalvi Solutions
Question 1.
Write the number of terms in the following expressions
(i) $x+y+z-x y z$
Answer:
4 terms
(ii) $m^{2} n^{2} c^{2}$
Answer:
1 term
(iii) $\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}-\mathrm{ab} \mathrm{b}^{2} \mathrm{c}^{2}+\mathrm{a}^{2} \mathrm{bc}^{2}+3 \mathrm{abc}$
Answer:
4 terms
(iv) $8 x^{2}-4 x y+7 x y^{2}$
Answer:
3 terms

Question 2.
Identify the numerical co-efficient of each term in the following expressions.
(i) $2 x^{2}-5 x y+6 y^{2}+7 x-10 y+9$
Answer:
Numerical co efficient in $2 x^{2}$ is 2
Numerical co efficient in $-5 x y$ is $-5$
Numerical co efficient in $6 y^{2}$ is 6
Numerical co efficient in $7 x$ is 7
Numerical co efficient in $-10 \mathrm{y}$ is $-10$
Numerical co-efficient in 9 is 9
(ii) $\frac{z}{3}+\frac{2 y}{5}-x y+7$
Answer:
Numerical co efficient in $\frac{z}{3}$ is $\frac{1}{3}$
Numerical co efficient in $\frac{2 y}{5}$ is $\frac{2}{5}$
Numerical co efficient in $-\mathrm{xy}$ is $-1$
Numerical co efficient in 7 is 7

Question $3 .$
Pick out the like terms from the following:

Like Terms
The variables of the terms along with their respective exponents must be same
Examples: $\mathrm{x}^{2}, 4 \mathrm{x}^{2}$
$\begin{aligned}
&a^{2} b^{2},-5 a^{2} b^{2} \\
&2 m,-7 m
\end{aligned}$

Question $4 .$
Add: $2 \mathrm{x}, 6 \mathrm{y}, 9 \mathrm{x}-2 \mathrm{y}$
Answer:
$\begin{aligned}
&2 x+6 y+9 x-2 y \\
&=2 x+9 x+6 y-2 y \\
&=(2+9) x+(6-2) y \\
&=11 x+4 y
\end{aligned}$

Question $5 .$
$\begin{aligned}
&\text { Simplify: }\left(5 x^{3} y^{3}-3 x^{2} y^{2}+x y+7\right)+\left(2 x y+x^{3} y^{3}-5\right. \\
&\text { Answer: } \\
&\left(5 x^{3} y^{3}-3 x^{2} y^{2}+x y+7\right)+\left(2 x y+x^{3} y^{3}-5+2 x^{2} y^{2}\right) \\
&=5 x^{3} y^{3}+x^{3} y^{3}-3 x^{2} y^{2}+2 x^{2} y^{2}+x y+2 x y+7-5 \\
&=(5+1) x^{3} y^{3}+(-3+2) x^{2} y^{2}+(1+2) x y+2 \\
&=6 x^{3} y^{3}-x^{2} y^{2}+3 x y+2
\end{aligned}$

Question $6 .$
The sides of a triangle are $2 x-5 y+9,3 y+6 x-7$ and $-4 x+y+10$. Find the perimeter of the triangle.
Answer:
Perimeter of the triangle $=$ Sum of three sides
$=(2 x-5 y+9)+(3 y+6 x-7)+(-4 x+y+10)$ $=2 x-5 y+9+3 y+6 x-7-4 x+y+10$ $=2 x+6 x-4 x-5 y+3 y+y+9-7+10$ $=(2+6-4) x+(-5+3+1) y+(9-7+10)$ $=4 x-y+12$ $\therefore$ Perimeter of the triangle $=4 x-y+12$ units.

Question $7 .$
Subtract - 2mn from $6 \mathrm{mn}$.
Answer:
$6 \mathrm{mn}-(-2 \mathrm{mn})=6 \mathrm{mn}+(+2 \mathrm{mn})$
$=(6+2) m n$
$=8 \mathrm{mn}$

Question $8 .$
Subtract $6 a^{2}-5 a b+3 b^{2}$ from $4 a^{2}-3 a b+b^{2}$.
Answer:
$\begin{aligned}
&\left(4 \mathrm{a}^{2}-3 \mathrm{ab}+\mathrm{b}^{2}\right)-\left(6 \mathrm{a}^{2}-5 \mathrm{ab}+3 \mathrm{~b}^{2}\right) \\
&=\left(4 \mathrm{a}^{2}-6 \mathrm{a}^{2}\right)+(-3 \mathrm{ab}-(-5 \mathrm{ab})]+\left(\mathrm{b}^{2}-3 \mathrm{~b}^{2}\right) \\
&=(4-6) \mathrm{a}^{2}+(-3 \mathrm{ab}+(+5 \mathrm{ab})]+(1-3) \mathrm{b}^{2} \\
&=[4+(-6)] \mathrm{a}^{2}+(-3+5) \mathrm{ab}+[1+(-3)] \mathrm{b}^{2} \\
&=-2 \mathrm{a}^{2}+2 \mathrm{ab}-2 \mathrm{~b}^{2}
\end{aligned}$

Question $9 .$
The length of $a \log$ is $3 a+4 b-2$ and a piece $(2 a-b)$ is removed from it. What is the length of the remaining $\log$ ?

Answer:
Length of the $\log =3 a+4 b-2$
Length of the piece removed $=2 a-b$
Remaining length of the $\log =(3 a+4 b-2)-(2 a-b)$
$\begin{aligned}
&=(3 a-2 a)+[4 b-(-b)]-2 \\
&=(3-2) a+(4+1) b-2 \\
&=a+5 b-2
\end{aligned}$

Question $10 .$
A tin had ' $x$ ' litre oil. Another tin had $\left(3 x^{2}+6 x-5\right)$ litre of oil. The shopkeeper added ( $\left.x+7\right)$ litre more to the second tin. Later he sold $\left(x^{2}+6\right)$ litres of oil from the second tin How much oil was left in the second tin?
Answer:
Quantity of öil in the second tin $=3 x^{2}+6 x-5$ litres.
Quantity of oil added $=x+7$ litres
$\therefore$ Total quantity of oil in the second tin
$\begin{aligned}
&=\left(3 x^{2}+6 x-5\right)+(x+7) \text { litres } \\
&=3 x^{2}+(6 x+x)+(-5+7)=3 x^{4}+(6+1) x+2 \\
&=3 x^{2}+7 x+2 \text { litres }
\end{aligned}$
Quantity of oil sold $=x^{2}+6$ litres
$\therefore$ Quantity of oil left in the second tin
$\begin{aligned}
&=\left(3 x^{2}+7 x+2\right)-\left(x^{2}+6\right)=\left(3 x^{2}-x^{2}\right)+7 x+(2-6) \\
&=(3-1) x^{2}+7 x+(-4)=2 x^{2}+7 x-4
\end{aligned}$
Quantity of oil left $=2 x^{2}+7 x-4$ litres
 

Think (Text Book Page No. 77)
Question 1.
Every algebraic expression is a polynomial. Is this statement true? Why?
Answer:
No, This statement is not true. Because Polynomials contain only whole numbers as the powers of their variables. But an algebraic expression may contains fractions and negative powers on their variables. Eg. $2 y^{2}+5 y^{-1}-3$ is a an algebraic expression. But not a polynomial.

Try These (Text Book Page No. 78)
Find the product of
(i) $3 a b^{2},-2 a^{2} b^{3}$
Answer:
$\begin{aligned}
&\left(3 a b^{2}\right) \times\left(-2 a^{2} b^{3}\right)=(+) \times(-) \times(3 \times 2) \times\left(a \times a^{2}\right) \times\left(b^{2} \times b^{3}\right) \\
&=-6 a^{3} b^{5}
\end{aligned}$
(ii) $4 x y, 5 y^{2} x,\left(-x^{2}\right)$
Answer:
$\begin{aligned}
&(4 x y) \times\left(5 y^{2} x\right) \times\left(-x^{2}\right)=(+) \times(+) \times(-) \times(4 \times 5 \times 1) \times\left(x \times x \times x^{2}\right) \times\left(y \times y^{2}\right) \\
&=-20 x^{4} y^{3}
\end{aligned}$
(iii) $2 \mathrm{~m},-5 \mathrm{n},-3 \mathrm{p}$
Answer:
$\begin{aligned}
&(2 \mathrm{~m}) \times(-5 \mathrm{n}) \times(-3 \mathrm{p})=(+) \times(-) \times(-) \times(2 \times 5 \times 3) \times \mathrm{m} \times \mathrm{n} \times \mathrm{p} \\
&=+30 \mathrm{mnp} \\
&=30 \mathrm{mnp}
\end{aligned}$

Think (Text Book Page No. 79)
why $3+(4 x-7 y) \neq 12 x-21$ y?
Answer:
Addition and multiplication are different $3+(4 x-7 y)=3+4 x-7 y$
We can add only like terms.

Try These (Text Book Page No. 79)
Question $1 .$
Multiply
(i) $\left(5 x^{2}+7 x-3\right)$ by $-4 x^{2}$
Answer:
$\begin{aligned}
&\left(5 x^{2}+7 x-3\right) \times\left(-4 x^{2}\right)=5 x^{2}\left(-4 x^{2}\right)+7 x\left(-4 x^{2}\right)-3\left(-4 x^{2}\right) \\
&=-20 x^{2}-28 x^{2}+12 x^{2}
\end{aligned}$
(ii) $(10 x-7 y+5 z)$ by $6 x y z$
Answer:
$\begin{aligned}
&(10 x-7 y+5 z) \times 6 x y z=6 x y z(10 x-7 y+5 z)[\because \text { Multiplication is commutative] } \\
&=6 x y z(10 x)+6 x y z(-7 y)+6 x y z(5 z) \\
&=(6 \times 10)(x \times x \times y \times z)+(6 \times-7)+(x \times y \times y \times z)+(6 \times 5)(x \times y \times z \times z)
\end{aligned}$
(iii) $(a b+3 b c-5 c a)$ by $3 a^{2} b c$
Answer:
$\begin{aligned}
&(a b+3 b c-5 c a) \times\left(3 a^{2} b c\right)=a b\left(3 a^{2} b c\right)+3 b c\left(3 a^{2} b c\right)-5 c a\left(3 a^{2} b c\right) \\
&=3 a^{3} b^{2} c+9 a^{2} b^{2} c^{2}-15 a^{3} b c^{2}
\end{aligned}$

(iv) $\left(4 m^{2}-3 m+7\right) b y-5 m^{3}$
Answer:
$\begin{aligned}
&\left(4 m^{2}-3 m+7\right) \times\left(-5 m^{3}\right)=4 m^{2}\left(-5 m^{3}\right)-(3 m)\left(-5 m^{3}\right)+7\left(-5 m^{3}\right) \\
&=-20 m^{5}+15 m^{4}-35 m^{3}
\end{aligned}$

Try These (Text Book Page No. 81)
Multiply
(i) $(a-5)$ and $(a+4)$
Answer:
$\begin{aligned}
&(a-5)(a+4)=a(a+ \\
&=(a \times a)+(a \times 4)+ \\
&=a^{2}+4 a-5 a-20 \\
&=a^{2}-a-20
\end{aligned}$
(ii) $(a+b)$ and $(a-b)$
Answer:
$\begin{aligned}
&(a+b)(a-b)=a(a-b)+b(a-b) \\
&=(a \times a)+(a \times-b)+(b \times a)+b(-b) \\
&=a^{2}-a b+a b-b=a^{2}-b^{2}
\end{aligned}$
(iii) $\left(m^{4}+n^{4}\right)$ and $(m-n)$
Answer:
$\begin{aligned}
&\left(m^{4}+n^{4}\right)(m-n)=m^{4}(m-n)+n^{4}(m-n) \\
&\left(m^{4} \times m\right)+\left(m^{4} \times(-n)\right)+\left(n^{4} \times m\right)+\left(n^{4} \times(-n)\right) \\
&=m^{5}-m^{4} n+m n^{4}-n^{5}
\end{aligned}$
(iv) $(2 x+3)(x+4)$
Answer:
$\begin{aligned}
&(2 x+3)(x+4)=2 x(x+4)+3(x+4) \\
&=\left(2 x^{2} \times x\right)+(2 x \times 4)+(3 \times x)+(3 \times 4) \\
&=2 x^{2}+8 x+3 x+12 \\
&=2 x^{2}+11 x+12
\end{aligned}$

(v) $x-5)(3 x+7)$
Answer:
$\begin{aligned}
&(x-5)(3 x+7)=x(3 x+7)-5(3 x+7) \\
&=(x \times 3 x)+(x \times 7)+(-5 \times 3 x)+(-5 \times 7) \\
&=3 x^{2}+7 x-15 x-35 \\
&=3 x^{2}-8 x-35
\end{aligned}$
(vi) $(x-2)(6 x-3)$
Answer:
$(x-2)(6 x-3)=x(6 x-3)-2(6 x-3)$
$\begin{aligned}
&=(x \times 6 x)+(x \times(-3)-(2 \times 6 x)-(2 \times 3) \\
&=6 x^{2}-3 x-12 x+6 \\
&=6 x^{2}-15 x+6
\end{aligned}$

Think (Text Book Page No. 81)
(i) $\operatorname{In} 3 x^{2}\left(x^{4}-7 x^{3}+2\right)$ what is the highest power in the expression?
Answer:
$\begin{aligned}
&3 x^{2}\left(x^{4}-7 x^{3}+2\right)=\left(3 x^{2}\right)\left(x^{4}\right)+3 x^{2}\left(-7 x^{3}\right)+\left(3 x^{2}\right) 2 \\
&=3 x^{6}-21 x^{5}+6 x^{2}
\end{aligned}$
(ii) Is $-5 y^{2}+2 y-6=-\left(5 y^{2}+2 y-6\right)$ ? If not, correct the mistake.
Answer:
$\mathrm{No},-5 \mathrm{y}^{2}+2 \mathrm{y}-6=-\left(5 \mathrm{y}^{2}+2 \mathrm{y}-6\right)$

Think (Text Book Page No. 83)

(i) $\operatorname{In} 3 x^{2}\left(x^{4}-7 x^{3}+2\right)$ what is the highest power in the expression?
Answer:
$\begin{aligned}
&3 x^{2}\left(x^{4}-7 x^{3}+2\right)=\left(3 x^{2}\right)\left(x^{4}\right)+3 x^{2}\left(-7 x^{3}\right)+\left(3 x^{2}\right) 2 \\
&=3 x^{6}-21 x^{5}+6 x^{2}
\end{aligned}$
(ii) Is $-5 \mathrm{y}^{2}+2 \mathrm{y}-6=-\left(5 \mathrm{y}^{2}+2 \mathrm{y}-6\right)$ ? If not, correct the mistake.
Answer:
No, $-5 y^{2}+2 y-6=-\left(5 y^{2}+2 y-6\right)$

Think (Text Book Page No. 83)
Are the following correct?
(i) $\frac{x^{3}}{x^{5}}=x^{8-3}=x^{5}$
Answer:
$\frac{x^{3}}{x^{5}}=\mathrm{x}^{8-3}=\mathrm{x}^{5} \text { (or) } \frac{x^{3}}{x^{5}}=\frac{1}{x^{5-3}}=\frac{1}{x^{5}}$
$\therefore$ The given answer is wrong
(ii) $\frac{10 m^{4}}{10 m^{4}}=0$
Answer:
$\begin{aligned}
&\frac{10 m^{4}}{10 m^{4}}=\frac{10}{10} m^{4-4}=1 m^{0}=1 \\
&{\left[\because m^{0}=1\right]}
\end{aligned}$
$\therefore$ The given answer is not correct.
(iii) When a monomial is divided by itself, we will get $1 .$ ?
Answer:
When a monomial is divided by itself we get 1 .
$\frac{x}{x}=x^{1-1}=x^{0}=1$
$\therefore$ The given statement is correct.

Try These (Text Book Page No. 83)
Divide
(i) $12 x^{3} y^{2}$ by $x^{2} y$
Answer:
$=12 x^{3-2} y^{2-1}=12 x^{1} y^{1}=12 x y$
(ii) $-20 a^{5} b^{2}$ by $2 a^{3} b^{7}$
Answer:
$\frac{-20 a^{5} b^{2}}{2 a^{3} b^{7}}=\frac{-20 a^{5-3}}{2 b^{7-2}}=\frac{-10 a^{2}}{b^{5}}$
Answer:
$\frac{28 a^{4} c^{2}}{21 c a^{2}}=\frac{28}{21} a^{4-2} c^{2-1}=\frac{4}{3} a^{2} c^{1}=\frac{4}{3} a^{2} c$

(iv) $\left(3 x^{2} y\right)^{3} \sqrt{6} x^{2} y^{3}$
Answer:
$\frac{\left(3 x^{2} y\right)^{3}}{6 x^{2} y^{3}}=\frac{27 x^{6} y^{b^{3}}}{6 x^{2} y^{y}}=\frac{9}{2} x^{6-2}=\frac{9}{2} x^{4}$
(v) $64 m^{4}\left(n^{2}\right)^{3}$ by $4 m n$
Answer:
$\begin{aligned}
\frac{64 m^{4}\left(n^{2}\right)^{3}}{4 m^{2} n^{2}} &=\frac{64 m^{4} n^{6}}{4 m^{2} n^{2}}=16 m^{4-2} n^{6-2} \\
&=16 m^{2} n^{4}
\end{aligned}$
(vi) $\left(8 x^{2} y^{2}\right)^{3}$ by $\left(8 x^{2} y^{2}\right)^{2}$
Answer:
$\begin{aligned}
&\frac{\left(8 x^{2} y^{2}\right)^{3}}{\left(8 x^{2} y^{2}\right)^{2}}=\frac{512 x^{6} y^{6}}{64 x^{4} y^{4}} \\
&=8 x^{6-4} y^{6-4} \\
&=8 x^{2} y^{2}
\end{aligned}$
(vii) $81 \mathrm{p}^{2} \mathrm{q}^{4}$ by $\sqrt{81 p^{2} q^{4}}$
Answer:
$\begin{aligned}
& \frac{81 p^{2} q^{4}}{\sqrt{81 p^{2} q^{4}}}=\\
=& 9 p^{2} 1 q^{4} 2 \\
=& 9 p q^{2}
\end{aligned}$

(vii) $\left(4 x^{2} y^{3}\right)^{0}$ by $\frac{\left(x^{3}\right)^{2}}{x^{4}}$
Answer:
$\frac{\left(4 x^{2} y^{3}\right)^{0}}{\frac{\left(x^{3}\right)^{2}}{x^{6}}}=\frac{1}{\frac{x^{6}}{x^{6}}}=\frac{1}{1}=1$

Think (Text Book Page No. 84)
Are the following divisions correct?
(i) $\frac{4 y+3}{4}=y+3$
Answer:
$\frac{4 y+3}{4}=\frac{4 y}{4}+\frac{3}{4}=y+\frac{3}{4}$ is the correct answer.
$\therefore$ The given statement is not correct
(ii) $\frac{5 m^{2}+9}{9}=5 m^{2}$
Answer:
$\frac{5 m^{2}+9}{9}=\frac{5 m^{2}}{9}+\frac{9}{9}=\frac{5}{9} m^{2}+1$ is the correct answer.
$\therefore$ The given statement is not correct
(iii) $\frac{2 x^{2}+8}{4}=2 x^{2}+2$. If not, correct it.
Answer:
$\frac{2 x^{2}+8}{4}=\frac{2 x^{2}}{4}+\frac{8}{4}=\frac{1}{2} x^{2}+2$ is the correct answer.
$\therefore$ The given statement is not correct

Try These (Text Book Page No. 84)
(i) $\left(16 y^{5}-8 y^{2}\right) \div 4 y$
Answer:
$\frac{16 y^{5}-8 y^{2}}{4 y}=\frac{16 y^{5}}{4 y}-\frac{8 y^{2}}{4 y}=4 y^{5-1}-2 y^{2-1}=4 y^{4}-2 y$
(ii) $\left(\mathrm{p}^{5} \mathrm{q}^{2}+24 \mathrm{p}^{3} \mathrm{q}-128 \mathrm{q}^{3}\right) \div 6 \mathrm{q}$
Answer:

$\frac{p^{5} q^{5}+24 p^{3} q-128 q^{3}}{6 q}=\frac{p^{5} q^{2}}{6 q}+\frac{24 p^{3} q}{6 q}-\frac{128 q^{3}}{6 q}$ $=\frac{1}{6} p^{5} q^{2-1}+4 p^{3} q^{1-1}-\frac{64}{3} q^{3-1}$ $=\frac{1}{6} p^{5} q^{1}+4 p^{3} q^{0}-\frac{64}{3} q^{2}=\frac{1}{6} p^{5} q+4 p^{3}-\frac{64}{3} q^{2}$ (iii) $\left(4 m^{2} \mathrm{n}+9 \mathrm{n}^{2} \mathrm{~m}+3 \mathrm{mn}\right) \div 4 \mathrm{mn}$ Answer: $4 m^{2} n+9 n^{2} m+3 m n+4 m n=\frac{4 m^{2} n+9 n^{2} m+3 m n}{4 m n}=\frac{4 m^{2} n}{4 m n}+\frac{9 n^{2} m}{4 m n}+\frac{3 m n}{4 m n}$ $=m^{2-1} n^{1-1}+\frac{9}{4} m^{1-1} n^{2-1}+\frac{3}{4} m^{1-1} n^{1-1}$ $=m^{1} n^{0}+\frac{9}{4} m^{0} n^{1}+\frac{3}{4} m^{0} n^{0}=m+\frac{9}{4} n+\frac{3}{4}$
Answer:
$4 m^{2} n+9 n^{2} m+3 m n+4 m n=\frac{4 m^{2} n+9 n^{2} m+3 m n}{4 m n}=\frac{4 m^{2} n}{4 m n}+\frac{9 n^{2} m}{4 m n}+\frac{3 m n}{4 m n}$
$=m^{2-1} n^{1-1}+\frac{9}{4} m^{1-1} n^{2-1}+\frac{3}{4} m^{1-1} n^{1-1}$
$=m^{1} n^{0}+\frac{9}{4} m^{0} n^{1}+\frac{3}{4} m^{0} n^{0}=m+\frac{9}{4} n+\frac{3}{4}$
$\left[\because n^{0}=1\right]$
 

Try These (Text Book Page No. 86)

Expand the following
Question $1 .$
$(\mathrm{p}+2)^{2}=$
Answer:
$(\mathrm{p}+2)^{2}=\mathrm{p}^{2}+2(\mathrm{p})(2)+2^{2}$
$=\mathrm{p}^{2}+4 \mathrm{p}+4$

Question $2 .$
$(3-a)^{2}=$
Answer:
$\begin{aligned}
&(3-a)^{2}=3^{2}-2(3)(a)+a^{2} \\
&=9-6 a+a^{2}
\end{aligned}$

Question $3 .$
$$
\left(6^{2}-x^{2}\right)=
$$
Answer:
$$
\left(6^{2}-x^{2}\right)=(6+x)(6-x)
$$

Question 4 .
$(a+b)^{2}-(a-b)^{2}=$
Answer:
$\begin{aligned}
&(a+b)^{2}-(a-b)^{2}=a^{2}+2 a b+b^{2}-\left(a^{2}-2 a b+b^{2}\right) \\
&=a^{2}+2 a b+b^{2}-a^{2}+2 a b-b^{2} \\
&=(1-1) a^{2}+(2+2) a b+(+1-1) b^{2}=4 a b
\end{aligned}$

Question $5 .$
$(a+b)^{2}=(a+b) \times$
Answer:
$(a+b)^{2}=(a+b) \times(a+b)$

Question $6 .$
$(m+n)(\ldots . .)=m^{2}-n^{2}$
Answer:
$(m+n)(m-n)=m^{2}-n^{2}$

Question $7 .$
$(m+\ldots \ldots)^{2}=m^{2}+14 m+49$
Answer:
$(m+7)^{2}=m^{2}+14 m+49$

Question $8 .$
$\left(k^{2}-49\right)=(k+\ldots)(k-\ldots)$
Answer:
$\mathrm{k}^{2}-49=\mathrm{k}^{2}-7^{2}=(\mathrm{k}+7)(\mathrm{k}-7)$

Question $9 .$
$m^{2}-6 m+9=$
Answer:
$m^{2}-6 m+9=(m-3)^{2}$
 

Question $10 .$
$(m-10)(m+5)=$
Answer:
$(m-10)(m+5)=m^{2}+(-10+5) m+(-10)(5)=m^{2}-5 m-50$

Think (Text Book Page No. 87)
Which is correct? $(3 a)^{2}$ is equal to
(i) $3 \mathrm{a}^{2}$
(ii) $3^{2} a$
(iii) $6 \mathrm{a}^{2}$
(iv) $9 \mathrm{a}^{2}$
Answer:

(iv) $9 \mathrm{a}^{2}$
Hint:
$(3 a)^{2}=3^{2} a^{2}=9 a^{2}$

Try These (Text Book Page No. 88)
Expand using appropriate identities.
Question $1 .$
$(3 p+2 q)^{2}$
Answer:
$(3 p+2 q)^{2}$
Comparing $(3 p+2 q)^{2}$ with $(a+b)^{2}$, we get $a=3 p$ and $b=2 q$.
$\begin{aligned}
&(a+b)^{2}=a^{2}+2 a b+b^{2} \\
&(3 p+2 q)^{2}=(3 p)^{2}+2(3 p)(2 q)+(2 q)^{2} \\
&=9 p^{2}+12 p q+4 q^{2}
\end{aligned}$

Question $2 .$
$(105)^{2}$
Answer:
$(105)^{2}=(100+5)^{2}$
Comparing $(100+5)^{2}$ with $(a+b)^{2}$, we get $a=100$ and $b=5$.
$\begin{aligned}
&(a+b)^{2}=a^{2}+2 a b+b^{2} \\
&(100+5)^{2}=(100)^{2}+2(100)(5)+5^{2} \\
&=10000+1000+25 \\
&105^{2}=11,025
\end{aligned}$

Question $3 .$
$(2 x-5 d)^{2}$
Answer:
$(2 \mathrm{x}-5 \mathrm{~d})^{2}$
Comparing with $(a-b)^{2}$, we get
$\begin{aligned}
&a=2 x, b=5 d \\
&(a-b)^{2}=a^{2}-2 a b+b^{2} \\
&(2 x-5 d)^{2}-(2 x)^{2}-2(2 x)(5 d)+(5 d)^{2} \\
&=2^{2} x^{2}-20 x d+5 d^{2} \\
&=4 x^{2}-20 x d+25 d^{2}
\end{aligned}$

Question $3 .$
$(2 x-5 d)^{2}$
Answer:
$(2 \mathrm{x}-5 \mathrm{~d})^{2}$
Comparing with $(a-h)^{2}$, we get
$\begin{aligned}
&a=2 x, b=5 d \\
&(a-b)^{2}=a^{2}-2 a b+b^{2} \\
&(2 x-5 d)^{2}-(2 x)^{2}-2(2 x)(5 d)+(5 d)^{2} \\
&=2^{2} x^{2}-20 x d+5^{2} d^{2} \\
&=4 x^{2}-20 x d+25 d^{2}
\end{aligned}$
 

Question $4 .$
$(98)^{2}$
Answer:
$(98)^{2}=(100-2)^{2}$
Comparing $(100-2)^{2}$ with $(a-b)^{2}$ we get
$\mathrm{a}=100, \mathrm{~b}=2$
$(a-b)^{2}=a^{2}-2 a b+b^{2}$
$\begin{aligned}
&(100-2)^{2}=100^{2}-2(100)(2)+2^{2} \\
&=10000-400+4 \\
&=9600+4 \\
&=9604
\end{aligned}$

Question 5 .
$(y-5)(y+5)$
Answer:
$(y-5)(y+5)$
Comparing $(y-5)(y+5)$ with $(a-b)(a+b)$ we get $a=y ; b=5$
$\begin{aligned}
&(a-b)(a+b)=a^{2}-b^{2} \\
&(y-5)(y+5)=y^{2}-5^{2} \\
&=y^{2}-25
\end{aligned}$

Question 6 .
$(3 x)^{2}-5^{2}$
Answer:
$(3 x)^{2}-5^{2}$
Comparing $(3 x)^{2}-5^{2}$ with $a^{2}-b^{2}$ we have
$\begin{aligned}
&a=3 x ; b=5 \\
&\left(a^{2}-b^{2}\right)=(a+b)(a-b) \\
&(3 x)^{2}-5^{2}=(3 x+5)(3 x-5) \\
&=3 x(3 x-5)+5(3 x-5) \\
&=(3 x)(3 x)-(3 x)(5)+5(3 x)-5(5) \\
&=9 x^{2}-15 x+15 x-25 \\
&=9 x^{2}-25
\end{aligned}$

Question $7 .$
$\begin{aligned}
&(2 m+n)(2 m+p) \\
&\text { Answer: } \\
&(2 m+n)(2 m+p) \\
&\text { Comparing }(2 m+n)(2 m+p) \text { with }(x+a)(x+b) \text { we have } \\
&x=2 n ; a=n ; b=p \\
&(x-a)(x+b)=x^{2}+(a+b) x+a b \\
&(2 m+n)(2 m+p)=\left(2 m^{2}\right)+(n+p)(2 m)+(n)(p) \\
&=2^{2} m^{2}+n(2 m)+p(2 m)+n p \\
&=4 m^{2}+2 m n+2 m p+n p
\end{aligned}$

Question 8 .
$203 \times 197$
Answer:
$203 \times 197=(200+3)(200-3)$
Comparing $(a+b)(a-b)$ we have
$\begin{aligned}
&a=200, b=3 \\
&(a+b)(a-b)=a^{2}-b^{2} \\
&(200+3)(200-3)=200^{2}-3^{2} \\
&203 \times 197=40000-9 \\
&203 \times 197=39991
\end{aligned}$
 

Question $9 .$
Find the area of the square whose side is $(x-2)$ units.
Answer:
Side of a square $=x-2$
$\therefore$ Area $=$ Side $\times$ Side
$=(x-2)(x-2)=x(x-2)-2(x-2)$
$=x(x)+(x)(-2)+(-2)(x)+(-2)(-2)$
$=x^{2}-2 x-2 x+4$
$=x^{2}-4 x+4$ units square

Question $10 .$
Find the area of the rectangle whose length and breadth are $(y+4)$ units and $(y-3)$ units.
Answer:
Length of the rectangle $=y+4$
breadth of the rectangle $=y-3$
Area of the rectangle $=$ length $x$ breadth
$\begin{aligned}
&=(y+4)(y-3)=y^{2}+(4+(-3)) y+(4)(-3) \\
&=y^{2}+y-12
\end{aligned}$

Try These (Text Book Page No. 91)
Expand:
(i) $(x+5)^{3}$
Answer:
Comparing $(x+5)^{3}$ with $(a+b)^{3}$, we have $a=x$ and $b=4$.
$\begin{aligned}
&(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\
&(x+5)^{3}=x^{3}+3 x^{2}(5)+3(x)(5)^{2}+5^{3} \\
&=x^{3}+15 x^{2}+75 x+125
\end{aligned}$
(ii) $(y-2)^{3}$
Answer:
Comparing $(y-2)^{3}$ with $(a-b)^{3}$ we have $a=y b=z$
$\begin{aligned}
&(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \\
&(y-2)^{2}=y^{3}-3 y^{2}(2)+3 y(2)^{2}+2^{3} \\
&=y^{3}-6 y^{2}+12 y+8
\end{aligned}$
(iii) $(x+1)(x+4)(x+6)$
Answer:
Comparing $(x+1)(x+4)(x+6)$ with $(x+a)(x+b)(x+c)$ we have $\mathrm{a}=1 \mathrm{~b}=4$ and $\mathrm{c}=6$

Try These (Text Book Page No. 94)
Find the factors

Answer:

Think (Text Book Page No. 94)
$x^{2}-4(x-2)=\left(x^{2}-4\right)(x-2)$ Is this correct? If not correct it.
Answer:
$\begin{aligned}
&(3 a)^{2}=3^{2} a^{2}=9 a^{2} \\
&x^{2}-4(x-2)=x^{2}-4 x+8
\end{aligned}$