# RD Sharma Class 9 Factorisation of Polynomials Exercise 6.4 Solutions

On this page you will find Maths RD Sharma Class 9 Factorisation of Polynomials Exercise 6.4 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 6 for class 9 deals with the topic of triangles and it is one of the most important chapters.

## Download RD Sharma Class 9 Factorisation of Polynomials Exercise 6.4 Solutions in PDF

Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 6 Factorisation of Polynomials Exercise 6.4 below and it will be beneficial for them.

## RD Sharma Class 9 Factorisation of Polynomials Exercise 6.4 Solutions

In each of the following, use factor theorem to find whether polynomial $\mathrm{g}(\mathrm{x})$ is a factor of polynomial $\mathrm{f}(\mathrm{x})$, or not : $(1-7)$

Q1. $f(x)=x^{3}-6 x^{2}+11 x-6, g(x)=x-3$
Sol:
Here, $f(x)=x^{3}-6 x^{2}+11 x-6$
$g(x)=x-3$
To prove that $g(x)$ is the factor of $f(x)$,
we should show $=>f(3)=0$
here,$x-3=0$
$\Rightarrow x=3$
Substitute the value of $x$ in $f(x)$
$f(3)=3^{3}-6 *(3)^{2}+11(3)-6$
$=27-(6 * 9)+33-6$
$=27-54+33-6$
$=60-60$
$=0$
Since, the result is $0 \mathrm{~g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$

Q2. $\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{4}+17 \mathrm{x}^{3}+9 \mathrm{x}^{2}-7 \mathrm{x}-10, \mathrm{~g}(\mathrm{x})=\mathrm{x}+5$
Sol:
Here,$f(x)=3 x^{4}+17 x^{3}+9 x^{2}-7 x-10$
$g(x)=x+5$
To prove that $g(x)$ is the factor of $f(x)$,
we should show $\Rightarrow f(-5)=0$
here,$x+5=0$
$\Rightarrow x=-5$
Substitute the value of $x$ in $f(x)$
$f(-5)=3(-5)^{4}+17(-5)^{3}+9(-5)^{2}-7(-5)-10$
$=(3 * 625)+(12 *(-125))+(9 * 25)+35-10$
$=1875-2125+225+35-10$
$=2135-2135$
$=0$
Since, the result is $0 \mathrm{~g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$

Q3. $f(x)=x^{5}+3 x^{4}-x^{3}-3 x^{2}+5 x+15, g(x)=x+3$
Sol:
Here, $f(x)=x^{5}+3 x^{4}-x^{3}-3 x^{2}+5 x+15$
$g(x)=x+3$
To prove that $g(x)$ is the factor of $f(x)$,
we should show $\Rightarrow f(-3)=0$
here, $x+3=0$
$\Rightarrow x=-3$
Substitute the value of $x$ in $f(x)$
$f(-3)=(-3)^{5}+3(-3)^{4}-(-3)^{3}-3(-3)^{2}+5(-3)+15$
$=-243+243+27-27-15+15$
$=0$
Since, the result is $0 \mathrm{~g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$

Q4. $f(x)=x^{3}-6 x^{2}-19 x+84, g(x)=x-7$
Sol:
Here, $f(x)=x^{3}-6 x^{2}-19 x+84$
$g(x)=x-7$
To prove that $g(x)$ is the factor of $f(x)$,
we should show $=>f(7)=0$
here, $x-7=0$
$\Rightarrow x=7$
Substitute the value of $x$ in $f(x)$
$\mathrm{f}(7)=7^{3}-6(7)^{2}-19(7)+84$
$=343-(6 * 49)-(19 * 7)+84$
$=342-294-133+84$
$=427-427$
$=0$
Since, the result is $0 \mathrm{~g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$

Q5. $f(x)=3 x^{3}+x^{2}-20 x+12, g(x)=3 x-2$
Sol:
Here,$f(x)=3 x^{3}+x^{2}-20 x+12$
$g(x)=3 x-2$
To prove that $g(x)$ is the factor of $f(x)$,
we should show $\Rightarrow f\left(\frac{2}{3}\right)=0$
here, $3 x-2=0$
$\Rightarrow 3 x=2$
$\Rightarrow x=\frac{2}{3}$
Substitute the value of $x$ in $f(x)$
$f\left(\frac{2}{3}\right)=3\left(\frac{2}{3}\right)^{\wedge}\{3\}+\left(\frac{2}{3}\right)^{2}-20\left(\frac{2}{3}\right)+12$
$=3\left(\frac{8}{27}\right)+\frac{4}{9}-\frac{40}{3}+12$
$=\frac{8}{9}+\frac{4}{9}-\frac{40}{3}+12$
$=\frac{12}{9}-\frac{40}{3}+12$
Taking L.C.M
$=\frac{12-120+108}{9}$
$=\frac{120-120}{9}$
$=0$
Since, the result is $0 \mathrm{~g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$

Q6. $f(x)=2 x^{3}-9 x^{2}+x+13, g(x)=3-2 x$
Sol:
Here, $f(x)=2 x^{3}-9 x^{2}+x+13$
$g(x)=3-2 x$
To prove that $g(x)$ is the factor of $f(x)$,
To prove that $\mathrm{g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$,
we should show $\Rightarrow f\left(\frac{3}{2}\right)=0$
here, $3-2 x=0$
$\Rightarrow-2 x=-3$
$\Rightarrow 2 x=3$
$\Rightarrow x=\frac{3}{2}$
Substitute the value of $x$ in $f(x)$
$f\left(\frac{3}{2}\right)=2\left(\frac{3}{2}\right)^{3}-9\left(\frac{3}{2}\right)^{2}+\left(\frac{3}{2}\right)+13$
$=2\left(\frac{27}{8}\right)-9\left(\frac{9}{4}\right)+\frac{3}{2}+12$
$=\left(\frac{27}{4}\right)-\left(\frac{81}{4}\right)+\frac{3}{2}+12$
Taking L.C.M
$=\frac{21-81+6+48}{4}$
$=\frac{81-81}{4}$
$=0$
Since, the result is $0 \mathrm{~g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$

Q7. $f(x)=x^{3}-6 x^{2}+11 x-6, g(x)=x^{2}-3 x+2$
Sol :
Here, $f(x)=x^{3}-6 x^{2}+11 x-6$
$g(x)=x^{2}-3 x+2$
First we need to find the factors of $x^{2}-3 x+2$
$\Rightarrow x^{2}-2 x-x+2$
$\Rightarrow x(x-2)-1(x-2)$
$\Rightarrow(x-1)$ and $(x-2)$ are the factors
To prove that $g(x)$ is the factor of $f(x)$,
The results of $f(1)$ and $f(2)$ should be zero
Let,$x-1=0$
$x=1$
substitute the value of $x$ in $f(x)$
$f(1)=1^{3}-6(1)^{2}+11(1)-6$
$=1-6+11-6$
$=12-12$
$=0$
Let,$x-2=0$
$x=2$
substitute the value of $x$ in $f(x)$
$f(2)=2^{3}-6(2)^{2}+11(2)-6$
$=8-(6 * 4)+22-6$
$=8-24+22-6$
$=30-30$
$=0$
Since, the results are $0 \mathrm{~g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$

Q8. Show that $(x-2),(x+3)$ and $(x-4)$ are the factors of $x^{3}-3 x^{2}-10 x+24$
Sol :
Here, $f(x)=x^{3}-3 x^{2}-10 x+24$
The factors given are $(x-2),(x+3)$ and $(x-4)$
To prove that $\mathrm{g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$,
The results of $f(2), f(-3)$ and $f(4)$ should be zero
Let, $x-2=0$
$\Rightarrow x=2$
Substitute the value of $x$ in $f(x)$
$f(2)=2^{3}-3(2)^{2}-10(2)+24$
$=8-(3 * 4)-20+24$
$=8-12-20+24$
$=32-32$
$=0$
Let,$x+3=0$
$\Rightarrow x=-3$
Substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(-3)=(-3)^{3}-3(-3)^{2}-10(-3)+24$
$=-27-3(9)+30+24$
$=-27-27+30+24$
$=54-54$
$=0$
Let,$x-4=0$
$\Rightarrow x=4$
Substitute the value of $x$ in $f(x)$
$f(4)=(4)^{3}-3(4)^{2}-10(4)+24$
$=64-\left(3^{*} 16\right)-40+24$
$=64-48-40+24$
$=84-84$
$=0$
Since, the results are $0 \mathrm{~g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$

Q9. Show that $(x+4),(x-3)$ and $(x-7)$ are the factors of $x^{3}-6 x^{2}-19 x+84$
Sol:
Here, $f(x)=x^{3}-6 x^{2}-19 x+84$
The factors given are $(x+4),(x-3)$ and $(x-7)$
To prove that $g(x)$ is the factor of $f(x)$,
The results of $f(-4), f(3)$ and $f(7)$ should be zero
Let, $x+4=0$
$\Rightarrow x=-4$
Substitute the value of $x$ in $f(x)$
$f(-4)=(-4)^{3}-6(-4)^{2}-19(-4)+84$
$=-64-(6 * 16)-(19 *(-4))+84$
$=-64-96+76+84$
$=160-160$
$=0$
Let, $x-3=0$
$\Rightarrow x=3$
Substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(3)=(3)^{3}-6(3)^{2}-19(3)+84$
$=27-(6 * 9)-(19 * 3)+84$
$=27-54-57+84$
$=111-111$
$=0$
Let, $x-7=0$
$\Rightarrow x=7$
Substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(7)=(7)^{3}-6(7)^{2}-19(7)+84$
$=343-(6 * 49)-(19 * 7)+84$
$=343-294-133+84$
$=427-427$
$=0$
Since, the results are $0 \mathrm{~g}(\mathrm{x})$ is the factor of $\mathrm{f}(\mathrm{x})$

Q10. For what value of a is $(x-5)$ a factor of $x^{3}-3 x^{2}+a x-10$
Sol:
Here, $f(x)=x^{3}-3 x^{2}+a x-10$
By factor theorem
If $(x-5)$ is the factor of $f(x)$ then,$f(5)=0$
$\Rightarrow x-5=0$
$\Rightarrow x=5$
Substitute the value of $x$ in $f(x)$
$f(5)=5^{3}-3(5)^{2}+a(5)-10$
$=125-(3 * 25)+5 a-10$
$=125-75+5 a-10$
$=5 a+40$
Equate $\mathrm{f}(5)$ to zero
$f(5)=0$
$\Rightarrow 5 a+40=0$
$\Rightarrow 5 a=-40$
$\Rightarrow a=\frac{-40}{5}$
$=-8$
When $a=-8,(x-5)$ will be factor of $f(x)$

Q11. Find the value of a such that $(x-4)$ is a factor of $5 x^{3}-7 x^{2}-a x-28$
Sol:
Here, $f(x)=5 x^{3}-7 x^{2}-a x-28$
By factor theorem
If $(x-4)$ is the factor of $f(x)$ then,$f(4)=0$
$\Rightarrow x-4=0$
$\Rightarrow x=4$
Substitute the value of $x$ in $f(x)$
$f(4)=5(4)^{3}-7(4)^{2}-a(4)-28$
$=5(64)-7(16)-4 a-28$
$=320-112-4 a-28$
$=180-4$
Equate $f(4)$ to zero, to find a
$f(4)=0$
$\Rightarrow 180-4 a=0$
$\Rightarrow-4 a=-180$
$\Rightarrow 4 a=180$
$\Rightarrow a=\frac{180}{4}$
$\Rightarrow a=45$
When $a=45,(x-4)$ will be factor of $f(x)$

Q12. Find the value of $a$, if $(x+2)$ is a factor of $4 x^{4}+2 x^{3}-3 x^{2}+8 x+5 a$
Sol:
Here, $f(x)=4 x^{4}+2 x^{3}-3 x^{2}+8 x+5 a$
By factor theorem
If $(x+2)$ is the factor of $f(x)$ then,$f(-2)=0$
$\Rightarrow x+2=0$
$\Rightarrow x=-2$
Substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(-2)=4(-2)^{4}+2(-2)^{3}-3(-2)^{2}+8(-2)+5 a$
$=4(16)+2(-8)-3(4)-16+5 a$
$=64-16-12-16+5 a$
$=5 a+20$
equate $\mathrm{f}(-2)$ to zero
$f(-2)=0$
$\Rightarrow 5 a+20=0$
$\Rightarrow 5 a=-20$
$\Rightarrow a=\frac{-20}{5}$
$\Rightarrow a=-4$
When $a=-4,(x+2)$ will be factor of $f(x)$

Q13. Find the value of $k$ if $x-3$ is a factor of $k^{2} x^{3}-k x^{2}+3 k x-k$
Sol:
Let $f(x)=k^{2} x^{3}-k x^{2}+3 k x-k$
From factor theorem if $x-3$ is the factor of $f(x)$ then $f(3)=0$
$\Rightarrow x-3=0$
$\Rightarrow x=3$
Substitute the value of $x$ in $f(x)$
$f(3)=k^{2}(3)^{3}-k(3)^{2}+3 k(3)-k$
$=27 \mathrm{k}^{2}-9 \mathrm{k}+9 \mathrm{k}-\mathrm{k}$
$=27 \mathrm{k}^{2}-\mathrm{k}$
$=\mathrm{k}(27 \mathrm{k}-1)$
Equate $f(3)$ to zero, to find $k$
$\Rightarrow f(3)=0$
$\Rightarrow \mathrm{k}(27 \mathrm{k}-1)=0$
$\Rightarrow \mathrm{k}=0$ and $27 \mathrm{k}-1=0$
$\Rightarrow k=0$ and $27 k=1$
$\Rightarrow k=0$ and $k=\frac{1}{27}$
When $k=0$ and $\frac{1}{27},(x-3)$ will be the factor of $f(x)$

Q14. Find the values of a and $b$, if $x^{2}-4$ is a factor of $a x^{4}+2 x^{3}-3 x^{2}+b x-4$
Sol:
Given,$f(x)=a x^{4}+2 x^{3}-3 x^{2}+b x-4$
$g(x)=x^{2}-4$
first we need to find the factors of $g(x)$
$\Rightarrow \mathrm{X}^{2}-4$
$\Rightarrow \mathrm{X}^{2}=4$
$\Rightarrow x=\sqrt{4}$
$\Rightarrow x=\pm 2$
$(x-2)$ and $(x+2)$ are the factors
By factor therorem if $(x-2)$ and $(x+2)$ are the factors of $f(x)$ the result of $f(2)$ and $f(-2)$ should be zero
Let, $x-2=0$
$\Rightarrow x=2$
Substitute the value of $x$ in $f(x)$
$f(2)=a(2)^{4}+2(2)^{3}-3(2)^{2}+b(2)-4$
$=16 a+2(8)-3(4)+2 b-4$|
$=16 a+2 b+16-12-4$
$=16 a+2 b$
Equate $\mathrm{f}(2)$ to zero
$\Rightarrow 16 a+2 b=0$
$\Rightarrow 2(8 a+b)=0$
$\Rightarrow 8 a+b=0—-1$
Let,$x+2=0$
$\Rightarrow x=-2$
Substitute the value of $x$ in $f(x)$
$f(-2)=a(-2)^{4}+2(-2)^{3}-3(-2)^{2}+b(-2)-4$
$=16 a+2(-8)-3(4)-2 b-4$
$=16 a-2 b-16-12-4$
$=16 a-2 b-32$
$=16 a-2 b-32$
Equate $f(2)$ to zero
$\Rightarrow 16 a-2 b-32=0$
$\Rightarrow 2(8 a-b)=32$
$\Rightarrow 8 a-b=16—-2$
Solve equation 1 and 2
$8 a+b=0$
$8 a-b=16$
$16 a=16$
$a=1$
substitute a value in eq 1
$8(1)+b=0$
$\Rightarrow b=-8$
The values are $\mathrm{a}=1$ and $\mathrm{b}=-8$

Q15. Find $\alpha, \beta$ if $(x+1)$ and $(x+2)$ are the factors of $x^{3}+3 x^{2}-2 \alpha x+\beta$
Sol:
Given, $f(x)=x^{3}+3 x^{2}-2 \alpha x+\beta$ and the factors are $(x+1)$ and $(x+2)$
From factor theorem, if they are tha factors of $\mathrm{f}(\mathrm{x})$ then results of $\mathrm{f}(-2)$ and $\mathrm{f}(-1)$ should be zero
Let,$x+1=0$
$\Rightarrow x=-1$
Substitute value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(-1)=(-1)^{3}+3(-1)^{2}-2 \alpha(-1)+\beta$
$=-1+3+2 \alpha+\beta$
$=2 \alpha+\beta+2—-1$
Let,$x+2=0$
$\Rightarrow x=-2$
Substitute value of $x$ in $f(x)$
$f(-2)=(-2)^{3}+3(-2)^{2}-2 \alpha(-2)+\beta$
$=-8+12+4 \alpha+\beta$
$=4 \alpha+\beta+4—–2$
Solving 1 and 2 i.e $(1-2)$
$\Rightarrow 2 \alpha+\beta+2-(4 \alpha+\beta+4)=0$
$\Rightarrow-2 \alpha-2=0$
$\Rightarrow 2 \alpha=-2$
$\Rightarrow \alpha=-1$
Substitute $\alpha=-1$ in equation
$\Rightarrow 2(-1)+\beta=-2$
$\Rightarrow \beta=-2+2$
$\Rightarrow \beta=0$
The values are $\alpha=-1$ and $\beta=0$

Q16. Find the values of $p$ and $q$ so that $x^{4}+p x^{3}+2 x^{2}-3 x+q$ is divisible by $\left(x^{2}-1\right)$
Sol:
Here, $f(x)=x^{4}+p x^{3}+2 x^{2}-3 x+q$
$g(x)=x^{2}-1$
first, we need to find the factors of $\mathrm{x}^{2}-1$
$\Rightarrow x^{2}-1=0$
$\Rightarrow x^{2}=1$
$\Rightarrow x=\pm 1$
$\Rightarrow(x+1)$ and $(x-1)$
From factor theorem, if $x=1,-1$ are the factors of $f(x)$ then $f(1)=0$ and $f(-1)=0$
Let us take, $x+1$
$\Rightarrow x+1=0$
$\Rightarrow x=-1$
Substitute the value of $x$ in $f(x)$
$f(-1)=(-1)^{4}+p(-1)^{3}+2(-1)^{2}-3(-1)+q$
$=1-p+2+3+q$
$=-p+q+6—-1$
Let us take, $x-1$
$\Rightarrow x-1=0$
$\Rightarrow x=1$
Substitute the value of $x$ in $f(x)$
$f(1)=(1)^{4}+p(1)^{3}+2(1)^{2}-3(1)+q$
$=1+p+2-3+q$
$=p+q—-2$
Solve equations 1 and 2
$-p+q=-6$
$p+q=0$
$2 q=-6$
$q=-3$
substitute q value in equation 2
$p+q=0$
$p-3=0$
$p=3$
the values of are $p=3$ and $q=-3$

Q17. Find the values of a and $b$ so that $(x+1)$ and $(x-1)$ are the factors of $x^{4}+a x^{3}-3 x^{2}+2 x+b$
Sol:
Here, $f(x)=x^{4}+a x^{3}-3 x^{2}+2 x+b$
The factors are $(x+1)$ and $(x-1)$
From factor theorem, if $x=1,-1$ are the factors of $f(x)$ then $f(1)=0$ and $f(-1)=0$
Let, us take $x+1$
$\Rightarrow x+1=0$
$\Rightarrow x=-1$
Substitute value of $x$ in $f(x)$
$f(-1)=(-1)^{4}+a(-1)^{3}-3(-1)^{2}+2(-1)+b$
$=1-a-3-2+b$
$=-a+b-4—1$
Let, us take $x-1$
$\Rightarrow x-1=0$
$\Rightarrow x=1$
Substitute value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(1)=(1)^{4}+a(1)^{3}-3(1)^{2}+2(1)+b$
$=1+a-3+2+b$
$=a+b—2$
Solve equations 1 and 2
$-a+b=4$
$a+b=0$
$2 b=4$
$b=2$
substitute value of $\mathrm{b}$ in eq 2
$a+2=0$
$a=-2$
the values are $a=-2$ and $b=2$

Q18. If $x^{3}+a x^{2}-b x+10$ is divisible by $x^{3}-3 x+2$, find the values of a and $b$
Sol:
Here, $f(x)=x^{3}+a x^{2}-b x+10$
$g(x)=x^{3}-3 x+2$
first, we need to find the factors of $\mathrm{g}(\mathrm{x})$
$g(x)=x^{3}-3 x+2$
$=x^{3}-2 x-x+2$
$=x(x-2)-1(x-2)$
$=(x-1)$ and $(x-2)$ are the factors
From factor theorem, if $x=1,2$ are the factors of $f(x)$ then $f(1)=0$ and $f(2)=0$
Let, us take $x-1$
$\Rightarrow x-1=0$
$\Rightarrow x=1$
Substitute the value of $x$ in $f(x)$
$f(1)=1^{3}+a(1)^{2}-b(1)+10$
$=1+a-b+10$
$=a-b+11—1$
Let, us take $x-2$
$\Rightarrow x-2=0$
$\Rightarrow x=2$
Substitute the value of $x$ in $f(x)$
$f(2)=2^{3}+a(2)^{2}-b(2)+10$
$=8+4 a-2 b+10$
$=4 a-2 b+18$
Equate $\mathrm{f}(2)$ to zero
$\Rightarrow 4 a-2 b+18=0$
$\Rightarrow 2(2 a-b+9)=0$
$\Rightarrow 2 a-b+9—-2$
Solve 1 and 2
$a-b=-11$
$2 a-b=-9$
$(-)(+)(+)$
$-a=-2$
$a=2$
substitute a value in eq 1
$\Rightarrow 2-b=-11$
$\Rightarrow-b=-11-2$
$\Rightarrow-b=-13$
$\Rightarrow b=13$
The values are $\mathrm{a}=2$ and $\mathrm{b}=13$

Q19. If both $(x+1)$ and $(x-1)$ are the factors of $a x^{3}+x^{2}-2 x+b$, Find the values of $a$ and $b$
Sol:
Here, $f(x)=a x^{3}+x^{2}-2 x+b$
$(x+1)$ and $(x-1)$ are the factors
From factor theorem , if $x=1,-1$ are the factors of $f(x)$ then $f(1)=0$ and $f(-1)=0$
Let, $x-1=0$
$\Rightarrow x=-1$
Substitute $x$ value in $f(x)$
$f(1)=a(1)^{3}+(1)^{2}-2(1)+b$
$=a+1-2+b$
$=a+b-1—-1$
Let, $x+1=0$
$\Rightarrow x=-1$
Substitute $x$ value in $f(x)$
$f(-1)=a(-1)^{3}+(-1)^{2}-2(-1)+b$
$=-a+1+2+b$
$=-a+b+3—-2$
Solve equations 1 and 2
$a+b=1$
$-a+b=-3$
$2 b=-2$
$\Rightarrow b=-1$
substitute $b$ value in eq 1
$\Rightarrow a-1=1$
$\Rightarrow a=1+1$
$\Rightarrow a=2$
The values are $\mathrm{a}=2$ and $\mathrm{b}=-1$

Q20. What must be added to $\mathrm{x}^{3}-3 \mathrm{x}^{2}-12 \mathrm{x}+19$ so that the result is exactly divisible by $\mathrm{x}^{2}+\mathrm{x}-6$
Sol:
Here,$p(x)=x^{3}-3 x^{2}-12 x+19$
$g(x)=x^{2}+x-6$
by division algorithm, when $p(x)$ is divided by $g(x)$, the remainder wil be a linear expression in $x$
let, $r(x)=a x+b$ is added to $p(x)$
$\Rightarrow f(x)=p(x)+r(x)$
$=x^{3}-3 x^{2}-12 x+19+a x+b$
$f(x)=x^{3}-3 x^{2}+x(a-12)+19+b$
We know that,$g(x)=x^{2}+x-6$
First, find the factors for $\mathrm{g}(\mathrm{x})$
$g(x)=x^{2}+3 x-2 x-6$
$=x(x+3)-2(x+3)$
$=(x+3)(x-2)$ are the factors
From, factor theorem when $(x+3)$ and $(x-2)$ are the factors of $f(x)$ the $f(-3)=0$ and $f(2)=0$
Let, $x+3=0$
$\Rightarrow x=-3$
Substitute the value of $x$ in $f(x)$
$f(-3)=(-3)^{3}-3(-3)^{2}+(-3)(a-12)+19+b$
$=-27-27-3 a+24+19+b$
$=-3 a+b+1—1$
Let, $x-2=0$
$\Rightarrow x=2$
Substitute the value of $x$ in $f(x)$
$f(2)=(2)^{3}-3(2)^{2}+(2)(a-12)+19+b$
$=8-12+2 a-24+b$
$=2 a+b-9—2$
Solve equations 1 and 2
$-3 a+b=-1$
$2 a+b=9$
$(-)(-)(-)$
$-5 a=-10$
$a=2$
substitute the value of a in eq 1
$\Rightarrow-3(2)+b=-1$
$\Rightarrow-6+b=-1$
$\Rightarrow b=-1+6$
$\Rightarrow b=5$
$\therefore r(x)=a x+b$
$=2 x+5$
$\therefore \mathrm{x}^{3}-3 \mathrm{x}^{2}-12 \mathrm{x}+19$ is divided by $\mathrm{x}^{2}+\mathrm{x}-6$ when it is added by $2 \mathrm{x}+5$

Q21. What must be added to $x^{3}-6 x^{2}-15 x+80$ so that the result is exactly divisible by $x^{2}+x-12$
Sol:
Let, $p(x)=x^{3}-6 x^{2}-15 x+80$
$q(x)=x^{2}+x-12$
by division algorithm, when $p(x)$ is divided by $q(x)$ the remainder is a linear expression in $x$.
so, let $\mathrm{r}(\mathrm{x})=\mathrm{ax}+\mathrm{b}$ is subtracted from $\mathrm{p}(\mathrm{x})$, so that $\mathrm{p}(\mathrm{x})-\mathrm{q}(\mathrm{x})$ is divisible by $\mathrm{q}(\mathrm{x})$
let $f(x)=p(x)-q(x)$
$q(x)=x^{2}+x-12$
$=x^{2}+4 x-3 x-12$
$=x(x+4)(-3)(x+4)$
$=(x+4),(x-3)$
clearly, $(x-3)$ and $(x+4)$ are factors of $q(x)$
so, $f(x)$ wiil be divisible by $q(x)$ if $(x-3)$ and $(x+4)$ are factors of $q(x)$
from, factor theorem
$f(-4)=0$ and $f(3)=0$
$\Rightarrow f(3)=3^{3}-6(3)^{2}-3(a+15)+80-b=0$
$=27-54-3 a-45+80-b$
$=-3 a-b+8—1$
Similarly,
$f(-4)=0$
$\Rightarrow f(-4) \Rightarrow(-4)^{3}-6(-4)^{2}-(-4)(a+15)+80-b=0$
$\Rightarrow-64-96-4 a+60+80-b=0$
$\Rightarrow 4 a-b-20=0—-2$
Substract eq 1 and 2
$\Rightarrow 4 a-b-20-8+3 a+b=0$
$\Rightarrow 7 a-28=0$
$\Rightarrow a=\frac{28}{7}$
$\Rightarrow a=4$
Put $a=4$ in eq 1
$\Rightarrow-3(4)-b=-8$
$\Rightarrow-b-12=-8$
$\Rightarrow-b=-8+12$
$\Rightarrow b=-4$
Substitute a and $\mathrm{b}$ values in $\mathrm{r}(\mathrm{x})$
$\Rightarrow r(x)=a x+b$
$=4 x-4$
Hence, $p(x)$ is divisible by $q(x)$, if $r(x)=4 x-4$ is subtracted from it

Q22. What must be added to $3 \mathrm{x}^{3}+\mathrm{x}^{2}-22 \mathrm{x}+9$ so that the result is exactly divisible by $3 \mathrm{x}^{2}+7 \mathrm{x}-6$
Sol:
Let, $p(x)=3 x^{3}+x^{2}-22 x+9$ and $q(x)=3 x^{2}+7 x-6$
By division theorem, when $p(x)$ is divided by $q(x)$, the remainder is a linear equation in $x$.
Let, $r(x)=a x+b$ is added to $p(x)$, so that $p(x)+r(x)$ is divisible by $q(x)$
$f(x)=p(x)+r(x)$
$\Rightarrow f(x)=3 x^{3}+x^{2}-22 x+9(a x+b)$
$\Rightarrow=3 x^{3}+x^{2}+x(a-22)+b+9$
We know that,
$q(x)=3 x^{2}+7 x-6$
$=3 x^{2}+9 x-2 x-6$
$=3 x(x+3)-2(x+3)$
$=(3 x-2)(x+3)$9
So, $f(x)$ is divided by $q(x)$ if $(3 x-2)$ and $(x+3)$ are the factors of $f(x)$
From, factor theorem
$f\left(\frac{2}{3}\right)=0$ and $f(-3)=0$
let, $3 x-2=0$
$3 x=2$ruI
$x=\frac{2}{3}$
$\Rightarrow f\left(\frac{2}{3}\right)=3\left(\frac{2}{3}\right)^{3}+\left(\frac{2}{3}\right)^{\wedge}\{2\}+\left(\frac{2}{3}\right)(a-22)+b+9$
$=3\left(\frac{8}{27}\right)+\frac{4}{9}+\frac{2}{3} a-\frac{44}{3}+b+9$
$=\frac{12}{9}+\frac{2}{3} \mathrm{a}-\frac{44}{3}+\mathrm{b}+9$
$=\frac{12+6 \mathrm{a}-132+9 \mathrm{~b}+81}{9}$
Equate to zero
$\Rightarrow \frac{12+6 a-132+9 b+81}{9}=0$
$\Rightarrow 6 a+9 b-39=0$
$\Rightarrow 3(2 a+3 b-13)=0$
$\Rightarrow 2 a+3 b-13=0—-1$
Similarly,
Let, $x+3=0$
$\Rightarrow x=-3$
$\Rightarrow \mathrm{f}(-3)=3(-3)^{3}+(-3)^{2}+(-3)(\mathrm{a}-22)+\mathrm{b}+9$
$=-81+9-3 a+66+b+9$
$=-3 a+b+3$
Equate to zero
$-3 a+b+3=0$Multiply by 3
$-9 a+3 b+9=0—2$
Substact eq 1 from 2
$\Rightarrow-9 a+3 b+9-2 a-3 b+13=0$
$\Rightarrow-11 a+22=0$
$\Rightarrow-11 a=-22$
$\Rightarrow a=\frac{22}{11}$
$\Rightarrow a=2$
Substitute a value in eq 1
$\Rightarrow-3(2)+b=-3$
$\Rightarrow-6+b=-3$
$\Rightarrow b=-3+6$
$\Rightarrow b=3$
Put the values in $\mathrm{r}(\mathrm{x})$
$r(x)=a x+b$
$=2 x+3$
Hence, $p(x)$ is divisible by $q(x)$, if $r(x)=2 x+3$ is added to it

Q23. If $x-2$ is a factor of each of the following two polynomials, find the value of a in each case :
1. $x^{3}-2 a x^{2}+a x-1$
2. $x^{5}-3 x^{4}-a x^{3}+3 a x^{2}+2 a x+4$
Sol:
(1) let $f(x)=x^{3}-2 a x^{2}+a x-1$
from factor theorem
if $(x-2)$ is the factor of $f(x)$ the $f(2)=0$let, $x-2=0$
$\Rightarrow x=2$
Substitute $x$ value in $f(x)$9$f(2)=2^{3}-2 a(2)^{2}+a(2)-1$
$=8-8 a+2 a-1$
$=-6 a+7$
Equate $\mathrm{f}(2)$ to zero
$\Rightarrow-6 a+7=0$
$\Rightarrow-6 a=-7$
$\Rightarrow a=\frac{7}{6}$
When,$(x-2)$ is the factor of $f(x)$ then $a=\frac{7}{6}$
(2) Let, $f(x)=x^{5}-3 x^{4}-a x^{3}+3 a x^{2}+2 a x+4$
from factor theorem
if $(x-2)$ is the factor of $f(x)$ the $f(2)=0$
let, $x-2=0$
$\Rightarrow x=2$
Substitute $x$ value in $f(x)$
$f(2)=2^{5}-3(2)^{4}-a(2)^{3}+3 a(2)^{2}+2 a(2)+4$
$=32-48-8 a+12+4 a+4$
$=8 a-12$
Equate $f(2)$ to zero
$\Rightarrow 8 a-12=0$
$\Rightarrow 8 a=12$
$\Rightarrow a=\frac{12}{8}$
=$\frac{3}{2}$
So, when $(x-2)$ is a factor of $f(x)$ then $a=\frac{3}{2}$

Q24. In each of the following two polynomials, find the value of $a$, if $(x-a)$ is a factor :
1. $x^{6}-a x^{5}+x^{4}-a x^{3}+3 x-a+2$
2. $x^{5}-a^{2} x^{3}+2 x+a+1$
Sol:
(1) $x^{6}-a x^{5}+x^{4}-a x^{3}+3 x-a+2$
let, $f(x)=x^{6}-a x^{5}+x^{4}-a x^{3}+3 x-a+2$
here, $x-a=0$
$\Rightarrow x=a$
Substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(a)=a^{6}-a(a)^{5}+(a)^{4}-a(a)^{3}+3 (a)-a+2$
$=a^{6}-a^{6}+(a)^{4}-a^{4}+3(a)-a+2$$=2 a+2$
Equate to zero
$\Rightarrow 2 a+2=0$
$\Rightarrow 2(a+1)=0$
$\Rightarrow a=-1$
So, when $(x-a)$ is a factor of $f(x)$ then $a=-1$
(2) $x^{5}-a^{2} x^{3}+2 x+a+1$
let, $f(x)=x^{5}-a^{2} x^{3}+2 x+a+1$
here,$x-a=0$
$\Rightarrow x=a$
Substitute the value of $x$ in $f(x)$
$f(a)=a^{5}-a^{2} a^{3}+2(a)+a+1$
$=a^{5}-a^{5}+2 a+a+1$
$=3 a+1$
Equate to zero+
$\Rightarrow 3 a+1=0$
$\Rightarrow 3 a=-1$
$\Rightarrow a=\frac{-1}{3}$
So, when $(x-a)$ is a factor of $f(x)$ then $a=\frac{-1}{3}$

Q25. In each of the following two polynomials, find the value of a, if $(x+a)$ is a factor :
1. $x^{3}+a x^{2}-2 x+a+4$
2. $x^{4}-a^{2} x^{2}+3 x-a$
Sol:
(1) $x^{3}+a x^{2}-2 x+a+4$
let, $f(x)=x^{3}+a x^{2}-2 x+a+4$
here, $x+a=0$
$\Rightarrow x=-a$
Substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(-a)=(-a)^{3}+a(-a)^{2}-2(-a)+a+4$
$=(-a)^{3}+a^{3}-2(-a)+a+4$
$=3 a+4$
Equate to zero
$\Rightarrow 3 a+4=0$
$\Rightarrow 3 a=-4$
$\Rightarrow a=\frac{-4}{3}$
So, when $(x+a)$ is a factor of $f(x)$ then $a=\frac{-4}{3}$
(2) $x^{4}-a^{2} x^{2}+3 x-a$
let, $f(x)=x^{4}-a^{2} x^{2}+3 x-a$
here, $x+a=0$
$\Rightarrow x=-a$
Substitute the value of $x$ in $f(x)$
$f(-a)=(-a)^{4}-a^{2}(-a)^{2}+3(-a)-a$
$=a^{4}-a^{4}-3(a)-a$
$=-4 a$
Equate to zero
$\Rightarrow-4 a=0$w
$\Rightarrow a=0$
So, when $(x+a)$ is a factor of $f(x)$ then $a=0$

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