ML Aggarwal Solution Class 10 Chapter 6 Factorization MCQs

MCQs

Question 1

When $x^{3}-3 x^{2}+5 x-7$ is divided by x – 2,then the remainder is

(a) 0

(b) 1

(c) 2

(d) – 1

Sol :

$f(x)=x^{3}-3 x^{2}+5 x-7$

g(x) = x – 2, if x – 2 = 0, then x = 2

Remainder will be

$\therefore f(2)=(2)^{3}-3(2)^{2}+5 \times 2-7$

=8-12+10-7=18-19=-1

∴Remainder=-1

Ans (d)

Question 2

When $2 x^{3}-x^{2}-3 x+5$ is divided by 2x + 1, then the remainder is

(a) 6

(b) – 6

(c) – 3

(d) 0

Sol :

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

g(x) = 2x + 1

Let 2x+1=0, then $x=\frac{-1}{2}$

Then remainder will be 

$f\left(\frac{-1}{2}\right)=2\left(\frac{-1}{2}\right)^{3}-\left(\frac{-1}{2}\right)^{2}-3\left(\frac{-1}{2}\right)+5$

$=2 \times \frac{-1}{8}-\frac{1}{4}+\frac{3}{2}+5$

$=\frac{-1}{4}-\frac{1}{4}+\frac{3}{2}+5$

$=\frac{-1-1+6+20}{4}=\frac{24}{4}=6$

∴Remainder=6

Ans (a)

Question 3

If on dividing $4 x^{2}-3 k x+5$ by x+2, the remainder is – 3 then the value of k is

(a) 4

(b) – 4

(c) 3

(d) – 3

Sol :

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

g(x)=x+2

Remainder = –3

Let x+2=0, then x=– 2

Now remainder will be

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

=16+6k+5

=21+6k

∴21+6k=-3

6k=-3-21=-24

$\Rightarrow k=\frac{-24}{6}=-4$

∴k=-4

Ans (b)

Question 4

If on dividing $2 x^{3}+6 x^{2}-(2 k-7) x+5$ by x+3, the remainder is k – 1 then the value of k is

(a) 2

(b) – 2

(c) – 3

(d) 3

Sol :

$f(x)=2 x^{3}+6 x^{2}-(2 k-7) x+5$

g(x) = x + 3

Remainder = k – 1

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

∴Remainder will be 

$f(-3)=2(-3)^{3}+6(-3)^{2}-(2 k-7)(-3)+5$

=-54+54+3(2 k-7)+5

=-54+54+6 k-21+5

=6 k-16

∴6k-16=k-1

6k-k=-1+16

⇒5k=15

⇒$k=\frac{15}{5}=3$

∴k=3

Ans (d)

Question 5

If x + 1 is a factor of 3x3 + kx2 + 7x + 4, then the value of k is

(a) – 1

(b) 0

(c) 6

(d) 10

Solution:

f(x) = 3x3 + kx2 + 7x + 4

g(x) = x + 1

Remainder = 0

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

$f(-1)=3(-1)^{3}+k(-1)^{2}+7(-1)+4$

=-3+k-7+4=k-6

∵Remainder=0

∴k-6=0

⇒k=6

Ans (c)