Exercise 3.5 - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E x ~} 3.5$
Question 1.
Subtract: $-2(x y)^{2}\left(y^{3}+7 x^{2} y+5\right)$ from $5 y^{2}\left(x^{2} y^{3}-2 x^{4} y+10 x^{2}\right)$
Answer:
$\begin{aligned}
&5 y^{2}\left(x^{2} y^{3}-2 x^{4} y+10 x^{2}\right)-\left[(-2)(x y)^{2}\left(y^{3}+7 x^{2} y+5\right)\right] \\
&=\left[5 y^{2}\left(x^{2} y^{3}\right)-5 y^{2}\left(2 x^{4} y\right)+5 y^{2}\left(10 x^{2}\right)\right]-\left[(-2) x^{2} y^{2}\left(y^{3}+7 x^{2} y+5\right)\right] \\
&=\left(5 y^{5} x^{2}-10 x^{4} y^{3}+50 x^{2} y^{2}\right)-\left[\left(-2 x^{2} y^{2}\right)\left(y^{3}\right)+\left(-2 x^{2} y^{2}\right)\left(7 x^{2} y\right)+\left(-2 x^{2} y^{2}\right)(5)\right] \\
&=5 x^{2} y^{5}-10 x^{4} y^{3}+50 x^{2} y^{2}-\left[-2 x^{2} y^{5}-14 x^{4} y^{3}-10 x^{2} y^{2}\right] \\
&=5 x^{2} y^{5}-10 x^{4} y^{3}+50 x^{2} y^{2}+2 x^{2} y^{5}+14 x^{4} y^{3}+10 x^{2} y^{2} \\
&=(5+2) x^{2} y^{5}+(-10+14) x^{4} y^{3}+(+50+10) x^{2} y^{2} \\
&=7 x^{2} y^{5}+4 x^{4} y^{3}+60 x^{2} y^{2}
\end{aligned}$
 

Question $2 .$
Multiply $\left(4 x^{2}+9\right)$ and $(3 x-2)$.
Answer:
$\begin{aligned}
&\left(4 x^{2}+9\right)(3 x-2)=4 x^{2}(3 x-2)+9(3 x-2) \\
&=\left(4 x^{2}\right)(3 x)-\left(4 x^{2}\right)(2)+9(3 x)-9(2) \\
&=\left(4 \times 3 \times x \times x^{2}\right)-\left(4 \times 2 \times x^{2}\right)+(9 \times 3 \times x)-18 \\
&=12 x^{3}-8 x^{2}+27 x-18\left(4 x^{3}+9\right)(3 x-2) \\
&=12 x^{3}-8 x^{2}+27 x-18
\end{aligned}$

Question $3 .$
Find the simple interest on Rs. $5 \mathrm{a}^{2} \mathrm{~b}^{2}$ for $4 \mathrm{ab}$ years at $7 \mathrm{~b} \%$ per annum.

Answer:
$\begin{aligned}
\text { Simple interest } &=\frac{\text { Principal } \times \text { years } \times \text { rate of interest }}{100} \\
&=\frac{5 a^{2} b^{2} \times 4 a b \times 7 b}{100}=\frac{(5 \times 4 \times 7)\left(a^{2} \times b^{2} \times a \times b \times b\right)}{100} \\
&=\frac{140}{100}\left(a^{2} b^{4}\right)=\frac{14}{10} a^{3} b^{4} \\
\text { Simple Interest } &=\frac{7}{5} a^{3} b^{4}
\end{aligned}$

Question $4 .$
The cost of a note book is Rs. 10ab. If Babu has Rs. (5a $\left.\mathrm{a}^{2}+20 \mathrm{ab}{ }^{2}+40 \mathrm{ab}\right)$. Then how many note books can he buy?
Answer:
For ₹ $10 \mathrm{ab}$ the number of note books can buy $=1$.
For $₹\left(5 a^{2} b+20 a b^{2}+40 a b\right)$ the number of note books can buy $=\frac{\text { Total amount }}{\text { cost of I note book }}$
$\begin{aligned}
&=\frac{5 a^{2} b+20 a b^{2}+40 a b}{10 a b}=\frac{5 a^{2} b}{10 a b}+\frac{20 a b^{2}}{10 a b}+\frac{40 a b}{10 a b} \\
&=\quad \frac{1}{2} a^{2-1} b^{1-1}+2 a^{1-1} b^{2-1}+4 a^{1-1} b^{1-1}=\frac{1}{2} a^{1} b^{0}+2 a^{0} b^{1}+4 a^{0} b^{0}
\end{aligned}$
Number of note book he can buy $=\frac{1}{2} a+2 b+4$
 

Question $5 .$
Factorise: $\left(7 y^{2}-19 y-6\right)$
Answer:
$7 y^{2}-19 y-6$ is of the form $a x^{2}+b x+c$ where $a=7 ; b=-19 ; c=-6$

The product a $\times \mathrm{c}=7 \times-6=-42$ sum $b=-19$

The middle term $-19$ y can be written as $-21 y+2 y$
$\begin{aligned}
&7 y^{2}-19 y-6=7 y^{2}-21 y+2 y-6 \\
&=7 y(y-3)+2(y-3) \\
&=(y-3)(7 y+2) \\
&7 y^{2}-19 y-6=(y-3)(7 y+2)
\end{aligned}$

Question $6 .$
A contractor uses the expression $4 x^{2}+11 x+6$ to determine the amount of wire to order when wiring a house. If the expression comes from multiplying the number of rooms times the number of outlets and he knows the number of rooms to be $(x+2)$, find the number of outlets in terms of ' $x$ '.
[Hint : factorise $4 x^{2}+11 x+6$ ]
Answer:
Given Number of rooms $=x+2$
Number of rooms $\times$ Number of outlets = amount of wire.
$(x+2) \times$ Number of outlets $=4 x^{2}+11 x+6$
Number of outlets $=\frac{4 x^{2}+11 x+6}{x+2} \ldots$ (1)
Now factorising $4 \mathrm{x}^{2}+11 \mathrm{x}+6$ which is of the form $a \mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}$ with $\mathrm{a}=4 \mathrm{~b}=11 \mathrm{c}=6$.
The product $a \times c=4 \times 6=24$
sum $b=11$

The middle term $11 x$ can be written as $8 x+3 x$
$\therefore 4 x^{2}+11 x+6=4 x^{2}+8 x+3 x+6$ $=4 x(x+2)+3(x+2)$ $4 x^{2}+11 x+6=(x+2)(4 x+3)$ Now from $(1)$ the number of outlets $=\frac{4 x^{2}+11 x+6}{x+2}=\frac{(x+2)(4 x+3)}{(x+2)}=4 x+3$
$\therefore$ Number of outlets $=4 x+3$

Question $7 .$
A mason uses the expression $x^{2}+6 x+8$ to represent the area of the floor of a room. If the decides that the length of the room will be represented by $(x+4)$, what will the width of the room be in terms of $x$ ?
Answer:
Given length of the room $=x+4$.
Area of the room $=x^{2}+6 x+8$
Length $\times$ breadth $=x^{2}+6 x+8$
breadth $=\frac{x^{2}+6 s+8}{x+4} \ldots . .$ (1)
Factorizing $x^{2}+6 x+8$, it is in the form of $a x^{2}+b x+c$
Where $a=1 b=6 \mathrm{c}=8$.
The product $a \times c=1 \times 8=8$
sum $=b=6$

The middle term $6 x$ can be written as $2 x+4 x$
$\therefore x^{2}+6 x+8=x^{2}+2 x+4 x+8$ $=x(x+2)+4(x+2)$ $x^{2}+6 x+8=(x+2)(x+4)$ Now from $(1)$ breadth $=\frac{x^{2}+6 x+8}{x+4}=\frac{(x+2)(x+4)}{(x+4)}=x+2$
breadth $=\frac{x^{2}+6 x+8}{x+4}=\frac{(x+2)(x+4)}{(x+4)}=x+2$
$\therefore$ Width of the room $=\mathrm{x}+2$
 

Question 8.
Find the missing term: $y^{2}+(-) x+56=(y+7)(y+-)$
Answer:
We have $(x+a)(x+b)=x^{2}+(a+b) x+a b$
$\begin{aligned}
&56=7 \times 8 \\
&\therefore y^{2}+(7+8) x+56=(y+7)(y+8)
\end{aligned}$

Question $9 .$
Factorise: $16 p^{4}-1$
Answer:
$\begin{aligned}
&16 p^{4}-1=2^{4} p^{4}-1=\left(2^{2}\right)^{2}\left(p^{2}\right)^{2}-1^{2} \\
&=\left(2^{2} p^{2}\right)^{2}-1^{2}
\end{aligned}$
Comparing with $a^{2}-b^{2}(a+b)(a-b)$ where $a=2^{2} p^{2}$ and $b=1$
$\begin{aligned}
&\therefore\left(2^{2} p^{2}\right)^{2}-1^{2}=\left(2^{2} p^{2}+1\right)\left(2^{2} p^{2}-1\right) \\
&=\left(4 p^{2}+1\right)\left(4 p^{2}-1\right) \\
&\therefore 16 p^{4}-1=\left(4 p^{2}+1\right)\left(4 p^{2}-1\right)=\left(4 p^{2}+1\right)\left(2^{2} p^{2}-1^{2}\right) \\
&=\left(4 p^{2}+1\right)\left[(2 p)^{2}-1^{2}\right]=\left(4 p^{2}+1\right)(2 p+1)(2 p-1)\left[\because u \operatorname{sing} a^{2}-b^{2}=(a+b)(a-b)\right] \\
&\therefore 16 p^{4}-1=\left(4 p^{2}+1\right)(2 p+1)(2 p-1)
\end{aligned}$

Question $10 .$
Factorise : $3 x^{3}-45 x^{2} y+225 x y^{2}-375 y^{3}$

Answer:
$\begin{aligned}
&=3 x^{3}-45 x^{2} y+225 x y^{2}-375 y^{3} \\
&=3\left(x^{3}-15 x^{2} y+75 x y^{2}-125 y^{3}\right) \\
&=3\left(x^{3}-3 x^{2}(5 y)+3 x(5 y)^{2}-(5 y)^{3}\right) \\
&=3(x-5 y)^{3}
\end{aligned}$