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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Exercise 3.1
Question $1 .$
Classify the following polynomials based on number of terms.
(i) $x^{3}-x^{2}$
(ii) $5 \mathrm{x}$
(iii) $4 x^{4}+2 x^{3}+1$
(iv) $4 \cdot x^{3}$
(v) $x+2$
(vi) $3 x^{2}$
(vii) $y^{4}+1$
(viii) $\mathrm{y}^{20}+\mathrm{y}^{18}+\mathrm{y}^{2}$
(ix) 6
(x) $2 u^{3}+u^{2}+3$
(xi) $u^{23}-u^{4}$
(xii) $y$
Solution:
$5 x, 3 x^{2}, 4 x^{3}, y$ and 6 are monomials because they have only one term.
$x^{3}-x^{2}, x+2, y^{4}+1$ and $u^{23}-u^{4}$ are binomials as they contain only two terms.
$4 x^{4}+2 x^{3}+1, y^{20}+y^{18}+y^{2}$ and $2 u^{3}+u^{2}+3$ are trinomials as they contain only three terms.
 

Question $2 .$
Classify the following polynomials based on their degree.
Solution:
$\mathrm{p}(\mathrm{x})=3, \mathrm{p}(\mathrm{x})=-7, \mathrm{p}(\mathrm{x})=\frac{3}{2}$ are constant polynomials
$\mathrm{p}(\mathrm{x})=\mathrm{x}+3, \mathrm{p}(\mathrm{x})=4 \mathrm{x}, \mathrm{p}(\mathrm{x})=\sqrt{3} \mathrm{x}+1$ are linear polynomials, since the highest degree of the variable $x$ is one.
$p(x)=5 x^{2}-3 x+2, p(y)=\frac{5}{2} y^{2}+1, p(x)=3 x^{2}$ are quadratic polynomials, since the highest degree of the variable is two.
$\mathrm{p}(\mathrm{x})=2 \mathrm{x}^{3}-\mathrm{x}^{2}+4 \mathrm{x}+1, \mathrm{p}(\mathrm{x})=\mathrm{x}^{3}+1, \mathrm{p}(\mathrm{y})=\mathrm{y}^{3}+3 \mathrm{y}$ are cubic polynomials, since the highest degree of the variable is three.
 

Question $3 .$
Find the product of given polynomials $p(x)=3 x^{3}+2 x-x^{2}+8$ and $q(x)=7 x+2$.

Solution:

$\begin{aligned}
&(7 x+2)\left(3 x^{3}+2 x-x^{2}+8\right)=7 x\left(3 x^{3}+2 x-x^{2}+8\right)+2 x \\
&\left(3 x^{3}+2 x-x^{2}+8\right)=21 x^{4}+14 x^{2}-7 x^{3}+56 x+6 x^{3}+4 x-2 x^{2}+16=21 x^{4}-x^{3}+12 x^{4}+60 x+16
\end{aligned}$

Question 4 .
Let $P(x)=4 x^{4}-3 x+2 x^{3}+5$ and $q(x)=x^{2}+2 x+4$ find $p(x)-q(x)$.

Solution:

Exercise $3.2$
Question 1 .
If $p(x)=5 x^{3}-3 x^{2}+7 x-9$, find
(i) $\mathrm{p}(-1)$
(ii) $\mathrm{p}(2)$.
Solution:
Given that $\mathrm{p}(\mathrm{x})=5 \mathrm{x}^{3}-3 \mathrm{x}^{2}+7 \mathrm{x}-9$
(i) $p(-1)=5(-1)^{3}-3(-1)^{2}+7(-1)-9=-5-3-7-9$ = $-24$
(ii) $\mathrm{p}(2)=5(2)^{3}-3(2)^{3}+7(2)-9=40-12+14-9$ $\therefore \mathrm{p}(2)=33$
 

Question 2.
Find the zeros of the following polynomials.
(i) $\mathrm{p}(\mathrm{x})=2 \mathrm{x}-3$
(ii) $\mathrm{p}(\mathrm{x})=\mathrm{x}-2$

Solution:
(i) Given that $p(x)=2 x-3=2\left(x-\frac{3}{2}\right)$.
we have $p\left(\frac{3}{2}\right)=2\left(\frac{3}{2}-\frac{3}{2}\right)=2(0)=0$
Hence $\frac{3}{2}$ is the zero of $p(x)$.
(ii) Given that $p(x)=x-2$. Now,
Hence 2 is the zero of $p(x)$.
 

Exercise $3.3$
Question 1.
Find the remainder using remainder theorem, when
(i) $4 x^{3}-5 x^{2}+6 x-2$ is divided by $x-1$.
(ii) $x^{3}-7 x^{2}-x+6$ is divided by $x+2$.
Solution:
(i) Let $p(x)=4 x^{3}-5 x^{2}+6 x-2$. The zero of $(x-1)$ is 1 .
When $p(x)$ is divided by $(x-1)$ the remainder is $p(1)$. Now,
$p(1)=4(1)^{3}-5(1)^{2}+6(1)-2=4-5+6-2=3$
$\therefore$ The remainder is 3 .
(ii) Let $p(x)=x^{3}-7 x^{2}-x+6$. The zero of $x+2$ is $-2$.
When $p(x)$ is divided by $x+2$, the remainder is $p(-2)$. Now,
$\mathrm{p}(-2)=(-2)^{3}-7(-2)^{2}-(-2)+6=-8-7(4)+2+6$
$=-8-28+2+6=-28$.
$\therefore$ The remainder is $-28$.
 

Question $2 .$
Find the value of a if $2 x^{3}-6 x^{2}+5 a x-9$ leaves the remainder 13 when it is divided by $x-2$.
Solution:
Let $p(x)=2 x^{3}-6 x^{2}+5 a x-9$
When $p(x)$ is divided by $(x-2)$ the remainder is $p(2)$.
Given that $p(2)=13$

$\begin{aligned}
&\Rightarrow 2(2)^{3}-6(2)^{2}+5 a(2)-9=13 \\
&\Rightarrow 2(8)-6(4)+10 a-9=13 \\
&\Rightarrow 16-24+10 a-9=13 \\
&\Rightarrow 10 a-17=13 \\
&\Rightarrow 10 a=30 \\
&\Rightarrow \therefore a=3
\end{aligned}$

Question $3 .$
If the polynomials $f(x)=a x^{3}+4 x^{2}+3 x-4$ and $g(x)=x^{3}-4 x+$ a leave the same remainder when divided by $x-3$. Find the value of a. Also find the remainder.
Solution:
Let $f(x)=a x^{3}+4 x^{2}+3 x-4$ and $g(x)=x^{3}-4 x+a$, When $f(x)$ is divided by $(x-3)$, the remainder is $f(3)$.
Now $f(3)=a(3)^{3} 4(3)^{2}+3(3)-4=27 a+36+9-4$
$f(3)=27 a+41 \ldots$ (1)
When $g(x)$ is divided by $(x-3)$, the remainder is $g(3)$.
Now $\mathrm{g}(3)=33-4(3)+\mathrm{a}=27-12+\mathrm{a}=15+\mathrm{a} \ldots(2)$
Since the remainders are same, $(1)=(2)$
Given that, $\mathrm{f}(3)=\mathrm{g}(3)$
That is $27 a+41=15+a$
$\begin{aligned}
&27 a-a=15-41 \\
&26 a=-26
\end{aligned}$
$a=\frac{-26}{26}=-1$
Sustituting $a=-1$, in $f(3)$, we get
$\mathrm{f}(3)=27(-1)+41=-27+41$ $\mathrm{f}(3)=14$ $\therefore$ The remainder is $14 .$
 

Question $4 .$
Show that $x+4$ is a factor of $x^{3}+6 x^{2}-7 x-60$.
Solution:
Let $p(x)=x^{3}+6 x^{2}-7 x-60$
By factor theorem $(x+4)$ is a factor of $p(x)$, if $p(-4)=0$
$p(-4)=(-4)^{3}+6(-4)^{2}-7(-4)-60=-64+96+28-60=0$
Therefore, $(x+4)$ is a factor of $x^{3}+6 x^{2}+-7 x-60$
 

Question $5 .$
In $(5 x+4)$ a factor of $5 x^{3}+14 x^{2}-32 x-32$

Solution:
$\text { Let } p(x)=5 x^{3}+14 x^{2}-32 x-32$
By factor theorem, $5 x+4$ is a factor, if $p\left(\frac{-4}{5}\right)=0$
$\begin{aligned}
p\left(\frac{-4}{5}\right) &=5\left(\frac{-4}{5}\right)^{3}+14\left(\frac{-4}{5}\right)^{2}-\left(32\left(\frac{-4}{5}\right)-32\right.\\
&=5\left(\frac{-64}{125}\right)+14\left(\frac{16}{25}\right)+32\left(\frac{4}{5}\right)-32 \\
&=\frac{-64}{25}+\frac{224}{25}+\frac{128}{5}-32=\frac{-65}{25}+\frac{224}{25}+\frac{640}{25}-\frac{800}{25} \\
&=\frac{-65+224+640-800}{25}=0 \\
p\left(\frac{-4}{5}\right) &=0
\end{aligned}$
Therefore, $5 x+4$ is a factor of $5 x^{3}+14 x^{2}-32 x-32$
 

Question 6 .
Find the value of $k$, if $(x-3)$ is a factor of polynomial $x^{3}-9 x^{2}+26 x+k$.
Solution:
Let $p(x)=x^{3}-9 x^{2}+26 x+k$
By factor theorem, $(x-3)$ is a factor of $p(x)$, if $p(3)=0$
$\begin{aligned}
&\mathrm{p}(3)=0 \\
&3^{3}-9(3)^{2}+26(3)+\mathrm{k}=0 \\
&27-81+78+\mathrm{k}=0 \\
&\mathrm{k}=-24
\end{aligned}$
To find the zero of $x-3$ :
Put $x-3=0$
we get $x=3$

Question $7 .$
Show that $(x-3)$ is a factor of $x^{3}+9 x^{2}-x-105$
Solution:
Let $p(x)=x^{3}+9 x^{2}-x-105$
By factor theorem, $x-3$ is a factor of $p(x)$, if $p(3)=0$
$\begin{aligned}
&\mathrm{p}(3)=3^{3}+9(3)^{3}-3-105 \\
&=27+81-3-105 \\
&=108-108 \\
&\mathrm{p}(3)=0
\end{aligned}$
To find the zero of $x-3$ :
Put $x-3=0$
we get $x=3$
Therefore, $x-3$ is a factor of $x^{3}+9 x^{2}-x-105$.
 

Question 8 .
Show that $(x+2)$ is a factor of $x^{3}-4 x^{2}-2 x+20$.
Solution:
Let $p(x)=x^{3}-4 x^{2}-2 x+20$
By factor theorem,
$(x+2)$ is factor of $\mathrm{p}(\mathrm{x})$, if $\mathrm{p}(-2)=0$
$\begin{aligned}
&\mathrm{p}(-2)=(-2)^{3}-4(-2)^{2}-2(-2)+20=-8-4(4)+4+20 \\
&P(-2)=0
\end{aligned}$

Therefore, $(x+2)$ is a factor of $x^{3}-4 x^{2}-2 x+20$
 

Exercise $3.4$
Question 1.
Expand the following using identities :
(i) $(7 x+2 y)^{2}$
(ii) $(4 \mathrm{~m}-3 \mathrm{~m})^{2}$
(iii) $(4 a+3 b)(4 a-3 b)$
(iv) $(k+2)(k-3)$
Solution:
(i) $(7 x+2 y)^{2}=(7 x)^{2}+2(7 x)(2 y)+(2 y)^{2}=49 x^{2}+28 x y+4 y^{2}$
(ii) $(4 \mathrm{~m}-3 \mathrm{~m})^{2}=(4 \mathrm{~m})^{2}-2(4 \mathrm{~m})(3 \mathrm{~m})+(3 \mathrm{~m})^{2}=16 \mathrm{~m}^{2}-24 \mathrm{mn}+9 \mathrm{n}^{2}$
(iii) $(4 a+3 b)(4 a-3 b)$ [We have $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$
Put $[\mathrm{a}=4 \mathrm{a}, \mathrm{b}=3 \mathrm{~b}$ ]
$(4 a+3 b)(4 a-3 b)=(4 a)^{2}-(3 b)^{2}=16 a^{2}-9 b^{2}$
(iv) $(k+2)(k-3)\left[\right.$ We have $\left.(x+a)(x-b)=x^{2}+(a-b) x-a b\right]$
Put $[\mathrm{x}=\mathrm{k}, \mathrm{a}=2, \mathrm{~b}=3]$
$(\mathrm{k}+2)(\mathrm{k}-3)=\mathrm{k}^{2}+(2-3) \mathrm{x}-2 \times 3=\mathrm{k}^{2}-\mathrm{x}-6$

Question 2.
Expand: $(a+b-c)^{2}$
Solution:
Replacing 'c' by '-c' in the expansion of
$\begin{aligned}
&(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \\
&(a+b+(-c))^{2}=a^{2}+b^{2}+(-c)^{2}+2 a b+2 b(-c)+2(-c) a \\
&=a^{2}+b^{2}+c^{2}+2 a b-2 b c-2 c a
\end{aligned}$

Question $3 .$
Expand: $(x+2 y+3 z)^{2}$
Solution:
We know that,

$\begin{aligned}
&(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \\
&\text { Substituting, a }=x, b=2 y \text { and } c=3 z \\
&(x+2 y+3 z)^{2}=x^{2}+(2 y)^{2}+(3 z)^{2}+2(x)(2 y)+2(2 y)(3 z)+2(3 z)(x) \\
&=x^{2}+4 y^{2}+9 z^{2}+4 x y+12 y^{2}+6 z x
\end{aligned}$

Question 4 .
Find the area of square whose side length is $\mathrm{m}+\mathrm{n}-\mathrm{q}$.
Solution:
Area of square $=$ side $\times$ side $=(m+n-q)^{2}$
We know that,
$\begin{aligned}
&(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \\
&{[m+n+(-q)]^{2}=m^{2}+n^{2}+(-q)^{2}+2 m n+2 n(-q)+2(-q) m} \\
&=m^{2}+n^{2}+q^{2}+2 m n-2 n q-2 q m
\end{aligned}$
Therefore, Area of square $=\mathrm{m}^{2}+\mathrm{n}^{2}+\mathrm{q}^{2}+2 \mathrm{mn}-2 \mathrm{nq}-2 \mathrm{qm}=\left[\mathrm{m}^{2}+\mathrm{n}^{2}+\mathrm{q}^{2}+2 \mathrm{mn}-2 \mathrm{nq}-2 \mathrm{qm}\right] \mathrm{sq}$. units.
 

EXERCISE $3.5$
Question 1.
Factorise the following
(i) $25 \mathrm{~m}^{-2}-16 \mathrm{n}^{2}$
(ii) $x^{4}-9 x^{2}$
Solution:
(i) $25 \mathrm{~m}^{2}-16 \mathrm{n}^{2}=(5 \mathrm{~m})^{2}-(4 \mathrm{n})^{2}$
$=(5 \mathrm{~m}-4 \mathrm{n})(5 \mathrm{~m}+4 \mathrm{n})\left[\because \mathrm{a}^{2}-\mathrm{b}^{2}=(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})\right]$
(ii) $x^{4}-9 x^{2}=x^{2}\left(x^{2}-9\right)=x^{2}\left(x^{2}-3^{2}\right)=x^{2}(x-3)(x+3)$

Question $2 .$
Factorise the following.
(i) $64 \mathrm{~m}^{3}+27 \mathrm{n}^{3}$
(ii) $729 \mathrm{x}^{3}-343 \mathrm{y}^{3}$
Solution:

$\begin{aligned}
&\text { (i) } 64 \mathrm{~m}^{3}+27 \mathrm{n}^{3}=(4 \mathrm{~m})^{3}+(3 \mathrm{n})^{3} \\
&=(4 \mathrm{~m}+3 \mathrm{n})\left((4 \mathrm{~m})^{2}-(4 \mathrm{~m})(3 \mathrm{n})+(3 \mathrm{n})^{2}\right)\left[\because \mathrm{a}^{2}+\mathrm{b}^{2}=(\mathrm{a}+\mathrm{b})\left(\mathrm{a}^{2}-\mathrm{ab}+\mathrm{b}^{2}\right)\right] \\
&=(4 \mathrm{~m}+3 \mathrm{n})\left(16 \mathrm{~m}^{2}-12 \mathrm{mn}+9 \mathrm{n}^{2}\right) \\
&\text { (ii) } 729 \mathrm{x}^{3}-343 \mathrm{y}^{3}=(9 \mathrm{x})^{3}-(7 \mathrm{y})^{3}=(9 \mathrm{x}-7 \mathrm{y})\left((9 \mathrm{x})^{2}+(9 \mathrm{x})(7 \mathrm{y})+(7 \mathrm{y})^{2}\right. \\
&=(9 \mathrm{x}-7 \mathrm{y})\left(81 \mathrm{x}^{2}+63 \mathrm{xy}+49 \mathrm{y}^{2}\right) \\
&{\left[\because \mathrm{a}^{3}-\mathrm{b}^{3}=(\mathrm{a}-\mathrm{b})\left(\mathrm{a}^{2}+\mathrm{ab}+\mathrm{b}^{2}\right)\right]}
\end{aligned}$

Question $3 .$
Factorise $81 x^{2}+90 x+25$
Solution:
$\begin{aligned}
&81 x^{2}+90 x+25=(9 x)^{2}+2(9 x)(5)+5^{2} \\
&=(9 x+5)^{2}\left[\because a^{2}+2 a b+b^{2}=(a+b)^{2}\right]
\end{aligned}$

Question 4 .
Find $a^{3}+b^{2}$ if $a+b=6, a b=5$.
Solution:
Given, $a+b=6, a b=5$
$a^{2}+b^{2}=(a+b)^{2}-3 a b(a+b)=(6)^{3}-3(5)(6)=126$

Question $5 .$
Factorise $(a-b)^{2}+7(a-b)+10$
Solution:
Let $a-b=p$, we get $p^{2}+7 p+10$,

$\mathrm{p}^{2}+7 \mathrm{p}+10=\mathrm{p}^{2}+5 \mathrm{p}+2 \mathrm{p}+10$ $=\mathrm{p}(\mathrm{p}+5)+2(\mathrm{p}+5)=(\mathrm{p}+5)(\mathrm{p}+2)$ Put $\mathrm{p}=\mathrm{a}-\mathrm{b}$ we get, $(\mathrm{a}-\mathrm{b})^{2}+7(\mathrm{a}-\mathrm{b})+10=(\mathrm{a}-\mathrm{b}+5)(\mathrm{a}-\mathrm{b}+2)$

Question $6 .$

Factorise $6 \mathrm{x}^{2}+17 \mathrm{x}+12$

Solution:

$ \begin{aligned} 6 x^{2}+17 x+12=&=6 x^{2}+9 x+8 x+12 \\ &=(2 x+3)(3 x+4) \end{aligned} $

Exercise 3.6

Question 1.
Factorise $x^{2}+4 x-96$
Solution:

$\begin{aligned}
x^{2}+4 x-96 &=x^{2}+12 x-8 x-96 \\
&=x(x+12)-8(x+12) \\
&=(x+12)(x-8)
\end{aligned}$

Question $2 .$
Factorise $\mathrm{x}^{2}+7 \mathrm{xy}-12 \mathrm{y}^{2}$
Solution:
$\begin{aligned}
x^{2}+7 x y+12 y^{2} &=x^{2}+4 y x+3 y x+12 y^{2} \\
&=x(x+4 y)+3 y(x+4 y) \\
&=(x+4 y)(x+3 y)
\end{aligned}$

Question $3 .$
Find the quotient and remainder when $5 x^{3}-9 x^{2}+10 x+2$ is divided by $x+2$ using synthetic division.

Solution:

$\begin{aligned}
p(x) &=5 x^{3}-9 x^{2}+10 x+2 \\
d(x) &=x+2 \\
\text { Standard form of } p(x) &=5 x^{2}-9 x^{2}+10 x+2 \text { and } \\
d(x) &=x+2
\end{aligned}$

$5 x^{3}-9 x^{2}+10 x+2=(x+2)\left(5 x^{2}-19 x+48\right)-94$
Hence the quotient is $5 x^{2}-19 x+48$ and remainder is $-94$.

Question $4 .$
Find the quotient and remainder when $-7+3 \mathrm{x}-2 \mathrm{x}^{2}+5 \mathrm{x}^{3}$ is divided by $-1+4 \mathrm{x}$ using synthetic division.
Solution:

$\begin{aligned} p(x) &=-7+3 x-2 x^{2}+5 x^{3} \\ d(x) &=-1+4 x \\ \text { Standard form of } p(x) &=5 x^{3}-2 x^{2}+3 x-7 \\ d(x) &=4 x-1 \end{aligned}$

$\begin{aligned}
5 x^{3}-2 x^{2}+3 x-7 &=\left(x-\frac{1}{4}\right)\left(5 x^{2}-\frac{3}{4} x+\frac{45}{16}\right)-\frac{403}{64} \\
&=\left(\frac{4 x-1}{4}\right) 4\left(\frac{5}{4} x^{2}-\frac{3}{16} x+\frac{45}{64}\right)-\frac{403}{64}=(4 x-1)\left(\frac{5}{4} x^{2}-\frac{3}{16} x+\frac{45}{64}\right)-\frac{403}{64}
\end{aligned}$
Hence the quotient is $\frac{5}{4} x^{2}-\frac{3}{16} x+\frac{45}{64}$ and remainder is $\frac{-40{ }^{2}}{64}$

Question $5 .$
If the quotient on dividing $5 x^{4}+4 x^{3}+2 x+1$ by $x+3$ is $5 x^{3}+9 x^{2}+b x-97$ then find the values of $a, b$ and also remainder.
Solution:

$p(x)=5 x^{4}+4 x^{3}+2 x+1$
Standard form of $p(x)=5 x^{4}+4 x^{3}+0 x^{2}+2 x+1$

Quotient $5 \mathrm{x}^{3}+11 \mathrm{x}^{2}+33 \mathrm{x}-97$ is compared with given quotient $5 \mathrm{x}^{3}+\mathrm{ax}^{2}+\mathrm{bx}-97$ co-efficient of $x^{2}$ is $-11=a$ and co-efficient of $x$ is $33=b$.
Therefore $\mathrm{a}=-11, \mathrm{~b}=33$ and remainder $=292$.
 

Exercise $3.7$
Question $1 .$
Find the quotient and the remainder when $10-4 x+3 x^{2}$ is divided by $x-2$.
Solution:
Let us first write the terms of each polynomial in descending order (or ascending order). Thus the given problem becomes $\left(3 x^{2}-4 x+10\right) \div(x-2)$

$\therefore$ Quotient $=3 x+2$ and Remainder $=14$, i.e $3 x^{2}-4 x+10=(x-2)(3 x+2)+14$ and is in the form Dividend $=($ Divisor $\times$ Quotient $)+$ Remainder

Question $2 .$
If $8 x^{3}-14 x^{2}-19 x-8$ is divided by $4 x+3$ then find the quotient and the remainder.
Solution:

$\therefore$ Quotient $=2 x^{2}-5 x-1$, Remainder $=-5$
 

Exercise $3.8$
Question 1.
Factorise $2 \mathrm{x}^{3}-\mathrm{x}^{2}-12 \mathrm{x}-9$ into linear factors.
Solution:
Let $p(x)=2 x^{3}-x^{2}-12 x-9$
Sum of the co-efficients $=2-1-12-9=-20 \neq 0$
Hence $x-1$ is not a factor
Sum of co-efficients of even powers with constant $=-1-9=-10$
Sum of co-efficients of odd powers $=2-12=-10$
Hence $x+1$ is a factor of $x$.
Now we use synthetic division to find the other factors.

Then $\mathrm{p}(\mathrm{x})=(\mathrm{x}+1)\left(2 \mathrm{x}^{2}-3 \mathrm{x}-9\right)$
Now $2 \mathrm{x}^{2}-3 \mathrm{x}-9=2 \mathrm{x}^{2}-6 \mathrm{x}+3 \mathrm{x}-9=2 \mathrm{x}(\mathrm{x}-3)+3(\mathrm{x}-3)$ $=(x-3)(2 x+3)$
Hence $2 x^{3}-x^{2}-12 x-9=(x+1)(x-3)(2 x+3)$
 

Question $2 .$
Factorize $x^{3}+3 x^{2}-13 x-15$
Solution:
Let $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}+3 \mathrm{x}^{2}-13 \mathrm{x}-15$
Sum of all the co-efficients $=1+3-13-15=-24+0$
Hence $x-1$ is not a factor
Sum of co-efficients of even powers with constant $=3-15=-12$
Sum of coefficients of odd powers $=1-13=-12$
Hence $x+1$ is a factor of $p(x)$
Now use synthetic division to find the other factors.

Hence $\mathrm{p}(\mathrm{x})=(\mathrm{x}+1)\left(\mathrm{x}^{2}+2 \mathrm{x}-15\right)$
Now $x^{3}+3 x^{2}-13 x-15=(x+1)\left(x^{2}+2 x-15\right)$
Now $x^{2}+2 x-15=x^{2}+5 x-3 x-15=x(x+5)-3(x+5)=(x+5)(x-3)$
Hence $x^{3}+3 x^{2}-13 x-15=(x+1)(x+5)(x-3)$
 

Question $3 .$
Factorize $\mathrm{x}^{3}+13 \mathrm{x}^{2}+32 \mathrm{x}+20$ into linear factors.
Solution:
Let, $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}+13 \mathrm{x}^{2}+32 \mathrm{x}+20$
Sum of all the coefficients $=1+13+32+20=66 \neq 0$
Hence, $(x-1)$ is not a factor.
Sum of coefficients of even powers with constant $=13+20=33$
Sum of coefficients of odd powers $=1+32=33$
Hence, $(x+1)$ is a factor of $p(x)$
Now we use synthetic division to find the other factors

Exercise $3.9$
Question $1 .$
Find GCD of $25 x^{3} y^{2} z, 45 x^{2} y^{4} z^{3} b$
Solution:
$\begin{aligned}
25 x^{3} y^{2} z &=5 \times 5 x^{3} y^{2} z=5 \times 5 \times x^{2} \times x \times y^{2} \times z \\
45 x^{3} y^{4} z^{3} &=5 \times 3 \times 3 \times x^{2} y^{4} z^{3}=\underline{5} \times 9 \times x^{2} \times y^{2} \times y^{2} \times z \times z^{2} \\
\text { Therefore GCD } &=5 x^{2} y^{2} z
\end{aligned}$

Question $2 .$
Find the GCD of $\left(y^{3}-1\right)$ and $(y-1)$.
Solution:
$\begin{aligned}
&y^{3}-1=(y-1)\left(y^{3}+y+1\right) \\
&y-1=y-1
\end{aligned}$
Therefore, $\mathrm{GCD}=y-1$
 

Question $3 .$
Find the GCD of $3 x^{2}-48$ and $x^{2}-7 x+12$
Solution:
$3 x^{2}-48=3\left(x^{2}-16\right)=3\left(x^{2}-4^{3}\right)=3(x+4)(x-4)$ $x^{2}-7 x+12=x^{2}-3 x-4 x+12=x(x-3)-4(x-3)=(x-3)(x-4)$ Therefore, GCD $=x-4$

Question $4 .$
Find the GCD of $a^{x}, a^{x+y}, a^{x+y+z}$.
Solution:
$\begin{aligned}
a^{x} &=\underline{a}^{x} & \\
a^{x+y} &=\underline{a}^{x} \cdot a^{y} \\
a^{x+y+z} &=\underline{a}^{x} \cdot a^{y} \cdot a^{2} \quad \therefore \mathrm{GCD}=a^{x}
\end{aligned}$

Exercise $3.10$
Question $5 .$
Solve graphically. $x-y=3 ; 2 x-y=11$
Solution:

Exercise $3.11$
Question $1 .$
Solve using the method of substitution. $5 x-y=5,3 x+y=11$
Solution:
$\begin{aligned}
&5 x-y=5 \\
&3 x+y=11
\end{aligned}$
(1) $\Rightarrow \quad y=5 x-5$
(2) $\Rightarrow$
$\begin{aligned}
3 x+5 x-5 &=11 \\
8 x &=11+5 \\
8 x &=16 \\
x &=\frac{16}{8}=2
\end{aligned}$
(1) $\Rightarrow$
$5(2)-y=5 \Rightarrow 10-5=y \Rightarrow y=5$
$\therefore$ Solution is $x=2, y=5$

Exercise $3.12$
Question 2.
Solve by the method of elimination. $2 x+y=10 ; 5 x-y=11$

Solution:

Exercise $3.13$
Question $3 .$
Solve $5 x-2 y=10 ; 3 x+y=17$ by the method of cross multiplication.
Solution:
$\begin{aligned}
&5 x-2 y-10=0 \\
&3 x+y-17=0
\end{aligned}$

$\begin{aligned} \frac{x}{34+10} &=\frac{y}{-30+85}=\frac{1}{5+6} \\ \frac{x}{44} &=\frac{y}{55}=\frac{1}{11} \\ \frac{x}{44} &=\frac{1}{11} \Rightarrow x=\frac{44}{11}=4 \\ \frac{y}{55} &=\frac{1}{11} \Rightarrow y=\frac{55}{11}=5 \quad \therefore \text { Solution is } x=4, y=5 \end{aligned}$

Exercise $3.14$
Question 1.
The age of Arjun is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
Solution:
Let the age of the father be $x$
Let the sum of the age of his sons be $y$
At present $x=2 y \Rightarrow x-2 y=0$.... (1)
After 20 years $x+20=y+40$
$x-y=40-20$

Exercise $3.15$
Multiple Choice Questions :
Question 1.
The polynomial $3 \mathrm{x}-2$ is a
(1) linear polynomial
(2) quadratic polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is linear if its degree is 1
Solution:
(1) linear polynomial
 

Question $2 .$
The polynomial $4 x^{2}+2 x-2$ is a
(1) linear polynomial
(2) quadratic polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is quadratic of its highest power of $x$ is 2
Solution:
(2) quadratic polynomial]

Question $3 .$
The zero of the polynomial $2 x-5$ is
(1) $\frac{5}{2}$
(2) $-\frac{5}{2}$
(3) $\frac{2}{5}$
(4) $-\frac{2}{5}$
Hint:
Zero is given by $2 x-5=0 \Rightarrow x=\frac{5}{2}$
Solution:
(1) $\frac{5}{2}$
 

Question $4 .$
The root of the polynomial equation $3 x-1=0$ is
(1) $x=-\frac{1}{3}$
(2) $x=\frac{1}{3}$
(3) $x=1$
(4) $x=3$
Hint:
$x=\frac{1}{3}$ gives $3\left(\frac{1}{3}\right)-1=1-1=0$
Solution:
(2) $x=\frac{1}{3}$

Question $5 .$
Zero of $(7+4 x)$ is
(1) $\frac{4}{7}$
(2) $\frac{-7}{4}$
(3) 7
(4) 4

Solution:
(2) $\frac{-7}{4}$
 

Question $6 .$
Which of the following has as a factor?
(1) $x^{2}+2 x$
(2) $(x-1)^{2}$
(3) $(x+1)^{2}$
(4) $\left(x^{2}-2^{2}\right)$
Solution:
(1) $x^{2}+2 x$
 

Question $7 .$
If $x-2$ is a factor of $q(x)$, then the remainder is
(1) $q(-2)$
(2) $x-2$
(3) 0
(4) -2
Solution:
(3) 0
 

Question 8.
$(a-b)\left(a^{2}+a b+b^{2}\right)=$
Solution:
(1) $a^{3}+b^{3}+c^{3}-3 a b c$
(2) $a^{2}-b^{2}$
(3) $a^{3}+b^{3}$
(4) $a^{3}-b^{3}$
Solution:
(4) $a^{3}-b^{3}$

Question $9 .$
The polynomial whose factors are $(x+2)(x+3)$ is
(1) $x^{2}+5 x+6$
(2) $x^{2}-4$
(3) $x^{2}-9$
(4) $x^{2}+6 x+5$
Solution:

(1) $x^{2}+5 x+6$
 

Question $10 .$
$(-a-b-c)^{2}$ is equal to
(1) $(a-b+c)^{2}$
(2) $(a+b-c)^{2}$
(3) $(-a+b+c)^{2}$
(4) $(a+b+c)^{2}$
Solution:
(4) $(a+b+c)^{2}$