Exercise 3.7 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E x} 3.7$
Question $1 .$
Find the quotient and remainder of the following.
(i) $\left(4 x^{3}+6 x^{2}-23 x+18\right) \div(x+3)$
(ii) $\left(8 y^{3}-16 y^{2}+16 y-15\right) \div(2 y-1)$
(iii) $\left(8 \mathrm{x}^{3}-1\right) \div(2 \mathrm{x}-1)$
(iv) $\left(-18 z+14 z^{2}+24 z^{3}+18\right) \div(3 z+4)$
Solution:
(i) $\left(4 x^{3}+6 x^{2}-23 x+18\right) \div(x+3)$

Question $2 .$
The area of rectangle is $x^{2}+7 x+12$. If its breadth is $(x+3)$, then find its length.
Solution:
Area of a rectangle $=x^{2}+7 x+12$
Its breadth $=x+3$
Area $=$ breadth $\times$ length
$\begin{aligned}
&x^{2}+7 x+12=(x+3) \times \text { length } \\
&\therefore \text { Length }=\left(x^{2}+7 x+12\right) \div(x+3)
\end{aligned}$

Quotient $=x+4$; Remainder $=0$;
$\therefore$ Length $=\mathrm{x}+4$
 

Question 3.
The base of a parallelogram is $(5 x+4)$. Find its height, if the area is $25 x^{2}-16$.
Solution:
The base of a parallelogram is $(5 x+4)$
$\begin{aligned}
&\text { Area }=25 \mathrm{x}^{2}-16 \\
&\text { Area }=\mathrm{b} \times \mathrm{h}=25 \mathrm{x}^{2}-16 \\
&\text { base }=5 \mathrm{x}+4
\end{aligned}$
$\text { height }=\frac{25 x^{2}-16}{\text { base }}=\frac{25 x^{2}-16}{5 x+4}$

 Question $4 .$
The sum of $(x+5)$ observations is $\left(x^{3}+125\right)$. Find the mean of the observations. Solution:
The sum of $(x+5)$ observations is $\left(x^{3}+125\right)$
Mean $=\frac{\text { Sum of the observations }}{\text { No. of observations }}=\frac{\sum x}{n}$
$=\frac{x^{3}+125}{x+5}=\frac{x^{3}+0 x^{2}+0 x+125}{x+5}$

Quotient $=x^{2}-5 x+25$
Remainder $=0$
$\therefore$ Mean of the observations $=\mathrm{x}^{2}-5 \mathrm{x}+25$
 

 Question 5.
Find the quotient and remainder for the following using synthetic division:
(i) $\left(x^{3}+x^{2}-7 x-3\right) \div(x-3)$
(ii) $\left(\mathrm{x}^{3}+2 \mathrm{x}^{2}-\mathrm{x}-4\right) \div(\mathrm{x}+2)$
(iii) $\left(3 x^{3}-2 x^{2}+7 x-5\right) \div(x+3)$
(iv) $\left(8 x^{4}-2 x^{2}+6 x+5\right) \div(4 x+1)$
Solution:
(i) $\left(x^{3}+x^{2}-7 x-3\right) \div(x-3)$
Let $p(x)=x^{3}+x^{2}-7 x-3$
$q(x)=x-3$ To find the zero of $x-3$
$\mathrm{p}(\mathrm{x})$ in standard form ((i.e.) descending order)
$x^{3}+x^{2}-7 x-3$

Quotient is $x^{2}+4 x+5$
Remainder is 12
(ii) $\left(x^{3}+2 x^{2}-x-4\right)+(x+2)$
$p(x)=x^{3}+2 x^{2}-x-4$
Co-efficients are $12-1-4$
To find zero of $x+2$, put $x+2=0 ; x=-2$

$\therefore$ Quotient is $x^{2}-1$
Remainder is $-2$
(iii) $\left(3 x^{3}-2 x^{2}+7 x-5\right) \div(x+3)$
To find zero of the divisor $(x+3)$, put $x+3=0 ; \therefore x=-3$
Dividend in Standard form $3 \mathrm{x}^{3}-2 \mathrm{x}^{2}+7 \mathrm{x}-5$
Co-efficients are $3-27-5$
Synthetic Division

Quotient is $3 x^{2}-11 x+40$
Remainder is $-125$
(iv) $\left(8 x^{4}-2 x^{2}+6 x+5\right) \div(4 x+1)$
To find zero of the divisor $4 x+1$, put $4 x+1=0 ; 4 x=-1 ; x=-\frac{1}{4}$
Dividend in Standard form $8 x^{4}+0 x^{3}-2 x^{2}+6 x+5$
Co-efficients are 8 0-2 65
Synthetic Division

$\begin{aligned}
8 x^{4}+0 x^{3}-2 x^{2}+6 x+5 &=\left(x+\frac{1}{4}\right)\left(8 x^{3}-2 x^{2}-\frac{3 x}{2}+\frac{51}{8}\right)+\frac{109}{32} \\
&=\frac{(4 x+1)}{4} \times 4\left(2 x^{3}-\frac{x^{2}}{2}-\frac{3 x}{8}+\frac{51}{32}\right)+\frac{109}{32} \\
&=(4 x+1)\left(2 x^{3}-\frac{x^{2}}{2}-\frac{3 x}{8}+\frac{51}{32}\right)+\frac{109}{32} \\
\end{aligned}$
$\therefore$ Quotient is $=2 x^{3}-\frac{x^{2}}{2}-\frac{3 x}{8}+\frac{51}{32}$
Remainder is $\frac{109}{32}$

Question $6 .$
If the quotient obtained on dividing $\left(8 x^{4}-2 x^{2}+6 x-7\right)$ by $(2 x+1)$ is $\left(4 x^{3}+p x^{2}-q x+3\right)$, then find $p$, $\mathrm{q}$ and also the remainder.
Solution:
Let $p(x)=8 x^{4}-2 x^{2}+6 x-7$
Standard form $=8 x^{4}+0 x^{3}-2 x^{2}+6 x-7$

Quotient $4 \mathrm{x}^{3}-2 \mathrm{x}^{2}+3$ is compared with the given quotient $4 \mathrm{x}^{3}+\mathrm{px}{ }^{2}-\mathrm{qx}+3$
Co-efficients of $x^{2}$ is $p=-2$
Co-efficients of $x$ is $q=0$
Remainder is $-10$

 Question $7 .$
If the quotient obtained on dividing $3 \mathrm{x}^{3}+11 \mathrm{x}^{2}+34 \mathrm{x}+106$ by $\mathrm{x}-3$ is $3 \mathrm{x}^{2}+\mathrm{ax}+\mathrm{b}$, then find $a, b$ and also the remainder.
Solution:
Let $p(x)=3 x^{3}+11 x^{2}+34 x+106$
$\mathrm{p}(\mathrm{x})$ in standard form
Co-efficients are 31134106
$q(x)=x-3$, its zero $x=3$
Synthetic division

Quotient is $3 x^{2}+20 x+94$, it is compared with the given quotient $3 x^{2}+a x+b$
Co-efficient of $\mathrm{x}$ is $\mathrm{a}=20$
Constant term is $\mathrm{b}=94$
Remainder $r=388$