ML Aggarwal Solution Class 10 Chapter 6 Factorization test

 Test

Question 1

Find the remainder when $2 x^{3}-3 x^{2}+4 x+7$ is divided by

(i) x – 2

(ii) x + 3

(iii) 2x + 1

Sol :

$2 x^{3}-3 x^{2}+4 x+7$

(i) Let x – 2 = 0, then x = 2

Substituting value of x in f(x)

$f(2)=2(2)^{3}-3(2)^{2}+4(2)+7$

= 2 × 8 – 3 × 4 + 4 × 2 + 7

= 16 – 12 + 8 + 7 = 19

Remainder = 19

(ii) Let x + 3 = 0, then x = – 3

Substituting the value of x in f(x)

$f(-3)=2(-3)^{3}-3(-3)^{2}+4(-3)+7$

$=2 \times(-27)-3(9)+4(-3)+7$

=-54-27-12+7=-93+7=-86

∴Remainder=-86

(iii) Let 2x+1=0, then 2x=-1

$\Rightarrow x=-\frac{1}{2}$

Now substituting the value of x in f(x)

$f\left(-\frac{1}{2}\right)=2\left(-\frac{1}{2}\right)^{3}-3\left(-\frac{1}{2}\right)^{2}+4\left(-\frac{1}{2}\right)+7$

$=2\left(-\frac{1}{8}\right)-3\left(\frac{1}{4}\right)+4\left(-\frac{1}{2}\right)+7$

$=-\frac{1}{4}-\frac{3}{4}-2+7$=-1-2+7=4

∴Remainder=4

Question 2

When $2 x^{3}-9 x^{2}+10 x-p$ is divided by (x + 1), the remainder is – 24.Find the value of p.

Sol :

Let x + 1 = 0 then x = -1

Substituting the value of x in f(x)

$f(x)=2 x^{3}-9 x^{2}+10 x-p$

$f(-1)=2(-1)^{3}-9(-1)^{2}+10(-1)-p$

=-2-9-10-p=-21-p

∵Remainder=-24

∴-21-p=-24

-p=-24+21=-3

∴p=3

Question 3

If (2x – 3) is a factor of $6 x^{2}+x+a$, find the value of a. With this value of a, factorise the given expression.

Sol :

Let 2x – 3 = 0 then 2x = 3

⇒ $x=\frac{3}{2}$

Substituting the value of x in f(x)

$f\left(\frac{3}{2}\right)=6\left(\frac{3}{2}\right)^{2}+\frac{3}{2}+a=6 \times \frac{9}{4}+\frac{3}{2}+a$

$=\frac{27}{2}+\frac{3}{2}+a=\frac{30}{2}+a=15+a$

∴Remainder=0

∴15+a=0

a=-15

Now , f(x) will be $6 x^{2}+x-15$

Dividing $6 x^{2}+x-15$ by 2x-3, we get

Figure to be added

$\therefore 6 x^{2}+x-15=(2 x-3)(3 x+5)$

Question 4

When $3 x^{2}-5 x+p$ is divided by (x – 2), the remainder is 3. Find the value of p. Also factorise the polynomial $3 x^{2}-5 x+p-3$

Sol :

$f(x)=3 x^{2}-5 x+p$

Let (x – 2) = 0, then x = 2

$f(2)=3(2)^{2}-5(2)+p$

= 3×4–10+p

=12–10+p

=2+p

∵Remainder=3

∴2+p=3

p=3-2=1

Hence p=1

Now, f(x)$=3 x^{2}-5 x+p-3$

$=3 x^{2}-5 x+1-3=3 x^{2}-5 x-2$

Dividing by (x-2), we get

Figure to be added

$3 x^{2}-5 x-2=(x-2)(3 x+1)$

Question 5

Prove that (5x + 4) is a factor of $5 x^{3}+4 x^{2}-5 x-4$. Hence factorize the given polynomial completely.

Sol :

$f(x)=5 x^{3}+4 x^{2}-5 x-4$

Let 5x + 4 = 0, then 5x = -4

$\Rightarrow x=\frac{-4}{2}$

$\therefore f\left(-\frac{4}{5}\right)=5\left(-\frac{4}{5}\right)^{3}+4\left(-\frac{4}{5}\right)^{2}-5\left(-\frac{4}{5}\right)-4$

$=\cdot 5 \times\left(-\frac{64}{125}\right)+4 \times \frac{16}{25}+4-4$

$=-\frac{64}{25}+\frac{64}{25}+4-4=0$

$\because f\left(-\frac{4}{5}\right)=0$

∴(5x+4) is a factor f(x)

Now dividing f(x) by 5 x+4, we get

$5 x^{3}+4 x^{2}-5 x-4$

$=(5 x+4)\left(x^{2}-1\right)=(5 x+4)\left\{(x)^{2}-(1)^{2}\right\}$

=(5 x+4)(x+1)(x-1)

Figure to be added

Question 6

Use factor theorem to factorise the following polynomials completely:

(i) $4 x^{3}+4 x^{2}-9 x-9$

(ii) $x^{3}-19 x-30$

Sol :

(i) $f(x)=4 x^{3}+4 x^{2}-9 x-9$

Let x = -1, then

$f(-1)=4(-1)^{3}+4(-1)^{2}-9(-1)-9$

=4(-1)+4(1)+9-9

=-4+4+9-9

=13-13=0

∴(x+1) is a factor of f(x)

Now dividing f(x) by x+1, we get 

$f(x)=4 x^{3}+4 x^{2}-9 x-9$

$=(x+1)\left(4 x^{2}-9\right)$

$=(x+1)\left\{(2 x)^{2}-(3)^{2}\right\}$

=(x+1)(2 x+3)(2 x-3)

Figure to be added

(ii) $f(x)=x^{3}-19 x-30$

Let x=-2, then

$f(-2)=(-2)^{3}-19(-2)-30$

=-8+38-30=38-38=0

∴(x+2) is a factor of f(x)

Now dividing f(x) by (x+2), we get

$f(x)=x^{3}-19 x-30$

$=(x+2)\left(x^{2}-2 x-15\right)=(x+2)\left\{\left(x^{2}-5 x+3 x-15\right\}\right.$

$=(x+2)\{x(x-5)+3(x-5)\}$

$=(x+2)(x-5)(x+3)$

Figure to be added

Question 7

If $x^{3}-2 x^{2}+p x+q$ has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, factorize the given polynomial completely.

Sol :

$f(x)=x^{3}-2 x^{2}+p x+q$

(x + 2) is a factor

$f(-2)=(-2)^{3}-2(-2)^{2}+p(-2)+q$

$=-8-2 \times 4-2 p+q$

=-8-8-2p+q

=-16-2 p+q

∵(x+2) is a factor of f(x)

∴f(-2)=0

$\Rightarrow-16-2 p+q=0$

$\Rightarrow \quad 2 p-q=-16$..(i)

Again, let x+1=0, then x=-1

$\therefore f(-1)=(-1)^{3}-2(-1)^{2}+p(-1)+q$

$=-1-2 \times 1-p+q=-1-2-p+q=-3-p+q$

∵Remainder=9, then

-3-p+q=9 

$\Rightarrow -p+q=9+3=12$

-p+q=12...(ii)

Adding (i) and (ii)

p=-4

Substituting the value of p in (ii)

-(-4)+q=12

4+q=12 

⇒ q=12-4=8

∴p=p=-4, q=8

∴$f(x)=x^{3}-2 x^{2}-4 x+8$

Dividing f(x) by (x+2), we get

$f(x)=(x+2)\left(x^{2}-4 x+4\right)$

$=(x+2)\left\{(x)^{2}-2 \times x(-2)+(2)^{2}\right\}$

$=(x+2)(x-2)^{2}$

Figure to be added

Question 8

If (x + 3) and (x – 4) are factors of $x^{3}+a x^{2}-b x+24$, find the values of a and b: With these values of a and b, factorise the given expression.

Sol :

$f(x)=x^{3}+a x^{2}-b x+24$

Let x + 3 = 0, then x = -3

Substituting the value of x in f(x)

$f(-3)=(-3)^{3}+a(-3)^{2}-b(-3)+24$

=-27+9 a+3 b+24

=9 a+3 b-3

∵x+3 is a factor

∴Remainder=0

∴9a+3b-3=0

⇒3a+b-1=0 (dividing by 3)

⇒3a+b=1..(i)

Again, let x-4=0, then x=4

Substituting the value of x in f(x)

$f(x)=(4)^{3}+a(4)^{2}-b(4)+24$

=64+16a-4b+24

=16a-4b+88

∵x-4 is a factor

∴Remainder=0

16a-4b+88=0

⇒16a-4b=-88 (dividing by 4)

⇒4a-b=-2..(ii)

Adding (i) and (ii)

7a=-21

⇒a=-3

Substituting the value of a in (i)

3(-3)+b=1

⇒-9+b=1

⇒b=1+9=10

∴a=-3, b=10

Now , f(x) will be 

$f(x)=x^{3}-3 x^{2}-10 x+24$

∵x+3 and x-4 are factors of f(x)

∴Dividing f(x) by (x+3)(x-4)

or $x^{2}-x-12$

Figure to be added

$x^{3}-3 x^{2}-10 x+24$

$=\left(x^{2}-x-12\right)(x-2)$

=(x+3)(x-4)(x-2)

Question 9

If $2 x^{3}+a x^{2}-11 x+b$ leaves remainder 0 and 42 when divided by (x – 2) and (x – 3) respectively, find the values of a and b. With these values of a and b, factorize the given expression.

Sol :

$f(x)=2 x^{3}+a x^{2}-11 x+b$

Let x – 2 = 0, then x = 2,

Substituting the value of x in f(x)

$f(2)=2(2)^{3}+a(2)^{2}-11(2)+b$

=2×8+4a-22+b

=16+4a-22+b

=4a+b-6

∵Remainder=0

∴4a+b-6=0

⇒4a+b=6..(i)

Again, let x-3=0, then x=3

Substituting the value of x in f(x)

$f(3)=2(3)^{3}+a(3)^{2}-11 \times 3+b$

=2×27+9a-33+b

=54+9a-33+b

⇒9a+b+21

∵Remainder=42

∴9a+b+21=42

⇒9a+b=42-21

⇒9a+b=21..(ii)

Subtracting (i) from (ii)

5a=15 

$\Rightarrow a=\frac{15}{5}=3$

Substituting the value of a in (i)

4(3)+b=6

12+b=6

b=6-12

b=-6

∴f(x) will be $2 x^{3}+3 x^{2}-11 x-6$

∵x-2 is a factor (as remainder=0)

∴Dividing f(x) by x-2, we get

Figure to be added

$\therefore 2 x^{3}+3 x^{2}-11 x-6$

$=(x-2)\left(2 x^{2}+7 x+3\right)$

$=(x-2)\left[2 x^{2}+6 x+x+3\right]$

=(x-2)[2x(x+3)+1(x+3)]

=(x-2)(x+3)(2x+1)

Question 10

If (2x + 1) is a factor of both the expressions $2 x^{2}-5 x+p$ and $2 x^{2}+5 x+ q$, find the value of p and q. Hence find the other factors of both the polynomials.

Sol :

Let 2x + 1 = 0, then 2x = -1

$x=-\frac{1}{2}$

Substituting the value of x in

$f(x)=2 x^{2}-5 x+p$

$f\left(-\frac{1}{2}\right)=2\left(\frac{-1}{2}\right)^{2}-5\left(\frac{-1}{2}\right)+p$

$=2 \times \frac{1}{4}+\frac{5}{2}+p$

$=\frac{1}{2}+\frac{5}{2}+p$

=3+p

∵2x+1 is the factor of p(x)

∴Remainder=0

⇒3+p=0

⇒p=-3

Again, substituting the value of x in q(x)

$q(x)=2 x^{2}+5 x+q$

$q\left(-\frac{1}{2}\right)=2\left(-\frac{1}{2}\right)^{2}+5\left(-\frac{1}{2}\right)+q=2 \times \frac{1}{4}-\frac{5}{2}+q$

$=\frac{1}{2}-\frac{5}{2}+q=-\frac{4}{2}+q=q-2$

∵2x+1 is the factor of q(x)

∴Remainder=0

⇒q-2=0

⇒q=2

Hence, p=-3, q=2

Now (i) ∵2x+1 is the factor of p(x)

$=2 x^{2}-5 x-3$

∴Dividing p(x) by 2x+1,

Figure to be added

$\therefore 2 x^{2}-5 x-3=(2 x+1)(x-3)$

(ii) ∵ 2x+1 is the factor of $q(x)=2 x^{2}+5 x+2$

∴Dividing q(x) by 2x+1,

Figure to be added

$\therefore 2 x^{2}+5 x+2$

=(2 x+1)(x+2)

Question 11

When a polynomial f(x) is divided by (x – 1), the remainder is 5 and when it is,, divided by (x – 2), the remainder is 7. Find – the remainder when it is divided by (x – 1) (x – 2).

Sol :

When f(x) is divided by (x – 1),

Remainder = 5

Let x – 1 = 0 ⇒ x = 1

∴f(1)=5

When divided by (x-2), 

Remainder=7

Let x-2=0

x=2

∴f(x)=(x-1)(x-2)q(x)+ax+b

Where q(x) is the quotient and ax+b is remainder

Putting x=1, we get,

f(1)=(1-1)(1-2)q(1)+a×1+b

=0+a+b

=a+b

and x=2, then

f(1)=(1-1)(1-2) q(1)+a×1+b

=0+a+b=a+b

∴a+b=5..(i)

2a+b=7..(ii)

Subtracting , we get

-a=-2

a=2

Subtracting the value of a in (5)

2+b=5 

⇒b=5-2=3

∴a=2, b=3

∴Remainder=ax+b

=2x+3