# SSC > Polynomials

Explore popular questions from Polynomials for SSC. This collection covers Polynomials previous year SSC questions hand picked by experienced teachers.

General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension
Q 1.

Correct2

Incorrect-0.5

Which of the following gives the correct factors of {tex} 4x^{2}- \left(x^{2}-3\right)^{2} {/tex}?
I. {tex} \left(x-2\right) \left(x-1\right) {/tex}
II. {tex} \left(x-1\right) \left(x+3\right) {/tex}
III. {tex} \left(3-x\right) \left(x+1\right) {/tex}
IV. {tex} \left(2x-1\right) \left(2x+3\right) {/tex}
Select the correct answers from the codes given below :

A

I and III

B

II and IV

II and III

D

I, II and III

##### Explanation

{tex} 4x^{2}- \left(x^{2}-3\right)^{2} = \left(2x\right)^{2}- \left(x^{2}-3\right)^{2} {/tex}

= {tex} \left(2x+x^{2}-3\right) \left(2x-x^{2}+3\right) {/tex}

= {tex} \left(x^{2}+2x-3\right) \left(2x-x^{2}+3\right) {/tex}

= {tex} \left(x+3\right) \left(x-1\right) \left(3-x\right) \left(x+1\right) {/tex}

Q 2.

Correct2

Incorrect-0.5

The remainder when {tex} 2x^{3}-3x^{2}+4x-1 {/tex} is divided by x - 1 is

A

-10

2

C

1

D

10

##### Explanation

When {tex} f \left(x\right)=2x^{3}-3x^{2}+4x-1 {/tex} is divided by {tex} x-1 {/tex}

Remainder = {tex} f \left(1\right) {/tex}

= {tex} 2.1^{3}-3.1^{2}+4.1-1 {/tex}

= 2

Q 3.

Correct2

Incorrect-0.5

If {tex} x+\frac{1}{x}=\sqrt{3} {/tex}, then {tex} x^{3}+\frac{1}{x^{3}} {/tex} is equal to

0

B

1

C

3

D

{tex} \left(x+1\right) {/tex}

##### Explanation

Given : {tex} x+\frac{1}{3}=\sqrt{3} {/tex}

Taking cube on both sides, we have

{tex} \left(x+\frac{1}{x}\right)^{3}= \left(\sqrt{3}\right)^{3} {/tex}

=> {tex} x^{3} + \frac{1}{x^{3}} + 3x^{2}\frac{1}{x} + 3x\frac{1}{x^{2}} = 3\sqrt{3} {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}}+3 \left(x+\frac{1}{x}\right)=3\sqrt{3} {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}}+3 \times \sqrt{3} = 3\sqrt{3} {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}} = 0 {/tex}

Q 4.

Correct2

Incorrect-0.5

One of the factors of {tex} x^{3}+6x^{2}+11x+6 {/tex} is

A

{tex} \left(x-\frac{1}{2}\right) {/tex}

B

{tex} \left(x+\frac{1}{2}\right) {/tex}

C

{tex} \left(x-1\right) {/tex}

{tex} \left(x+1\right) {/tex}

##### Explanation

{tex} f \left(x\right)=x^{3}+6x^{2}+11x+6 {/tex}

But x = -1 makes {tex} f \left(x\right)=0 {/tex}, therefore {tex} x+1 {/tex} must be factor of {tex} f \left(x\right) {/tex}.

Q 5.

Correct2

Incorrect-0.5

If polynomial {tex} x^{3}-3x^{2}+kx+42 {/tex} is divisible by {tex} x+3 {/tex}, then value of k will be

A

4

B

14

-4

D

-14

##### Explanation

Since{tex} f(x) = x^{3} + 3x^{2} + Kx + 42 {/tex} is divisible by

{tex} x+3=x- \left(-3\right) {/tex}.

{tex} \therefore f\left(-3\right)=0 {/tex}

=> {tex} \left(-3\right)^{3}-3 \left(-3\right)^{2}+k \left(-3\right)+42 = 0 {/tex}

=> {tex} -27-27-3k+42 = 0 {/tex}

=> 3k + 12 = 0

=> k = -4

Q 6.

Correct2

Incorrect-0.5

H.C.F. and L.C.M. of polynomials {tex} \left(x-2\right) {/tex}, {tex} x^{2}-4 {/tex} are respectively

A

1, {tex} x - 2 {/tex}

B

{tex} x - 2 {/tex}, 8

C

{tex} x - 2 {/tex}, {tex} x + 2 {/tex}

{tex} x - 2 {/tex}, {tex} x^{2}-4 {/tex}

##### Explanation

{tex} x^{2}-4x = \left(x+2\right) \left(x-2\right) {/tex}

{tex} \therefore L.C.M.\ of\ \left(x-2\right)\ and\ \left(x^{2}-4\right) {/tex}

= {tex} 1 \times \left(x-2\right) \left(x+2\right) = x^{2}-4 {/tex}

and H.F.C. of {tex} \left(x-2\right)\ and\ \left(x^{2}-4\right) = x-2 {/tex}

Q 7.

Correct2

Incorrect-0.5

If {tex} a+b+c=0 {/tex}, thena {tex} a^{3}+b^{3}+c^{3} {/tex} is equal to

A

{tex} a^{2} \left(b+c\right)+b^{2} \left(c+a\right)+c^{2} \left(a+b\right) {/tex}

B

{tex} 3 \left(b+c\right) \left(c+a\right) \left(a+b\right) {/tex}

{tex} 3 abc {/tex}

D

{tex} 6 a^{2}b^{2}c^{2} {/tex}

##### Explanation

Given: {tex} a+b+c=0 {/tex}

=> {tex} a+b = -c {/tex}

Taking cube on both sides, we have

{tex} \left(a+b\right)^{3} = \left(-c\right)^{3} {/tex}

=> {tex} a^{3}+b^{3}+3ab \left(a+b\right)=-c^{3} {/tex}

=> {tex} a^{3}+b^{3}+3ab \left(-c\right)=-c^{3} {/tex}

=> {tex} a^{3}+b^{3}+c^{3}=3abc {/tex}

Q 8.

Correct2

Incorrect-0.5

If {tex} a^{m^{n}}= \left(a^{m}\right)^{n} {/tex}, then the value of m is

A

n

B

{tex} \frac{n}{n^{n-1}} {/tex}

{tex} \frac{1}{n^{n-1}} {/tex}

D

{tex} n^{\frac{n-1}{n}} {/tex}

##### Explanation

Given: {tex} a^{m^{n}} = \left(a^{m}\right)^{n} = a^{m \times n} {/tex}

{tex} \therefore m^{n} = mn {/tex}

{tex} m^{n-1} = n {/tex}

{tex} \therefore m = n^{\frac{1}{n-1}} {/tex}

Q 9.

Correct2

Incorrect-0.5

Value of {tex} \left(X^{b-c}\right)^{b+c},\ \left(X^{c-a}\right)^{c+a},\ \left(X^{a-b}\right)^{a+b} {/tex} is

A

0

B

X

1

D

X+1

##### Explanation

{tex} \left(X^{b-c}\right)^{b+c}. \left(X^{c-a}\right)^{c+a}. \left(X^{a-b}\right)^{a+b} {/tex}

= {tex} \left(X\right)^{ \left(b-c\right) \left(b+c\right)}. \left(X\right)^{ \left(c-a\right) \left(c+a\right) }. \left(X\right)^{ \left(a-b\right) \left(a+b\right) } {/tex}

{tex} X^{b^{2}-c^{2}}.X^{c^{2}-a^{2}}.X^{a^{2}-b^{2}} {/tex}

= {tex} X^{b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2}} {/tex}

={tex} X^{0} = 1 {/tex}

Q 10.

Correct2

Incorrect-0.5

If {tex} 2^{x}=4^{y}=8^{z} and \frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}=\frac{24}{7} {/tex}, then value of z is

A

{tex} \frac{7}{16} {/tex}

B

{tex} \frac{7}{32} {/tex}

{tex} \frac{7}{48} {/tex}

D

{tex} \frac{7}{46} {/tex}

##### Explanation

Given : {tex} 2^{x} = 4^{y} = 8^{z} {/tex}

=> {tex} 2^{x} = 2^{2y} = 2^{3z} {/tex}

=> {tex} x = 2y = 3z = k\ (say) {/tex}

From {tex} \frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}=\frac{24}{7} {/tex}

{tex} \frac{1}{2k}+\frac{1}{2k}+\frac{1}{2k}=\frac{24}{7} {/tex}

=> {tex} \frac{3}{2k}=\frac{24}{7} {/tex}

=> {tex} k = \frac{7}{16} {/tex}

{tex} \therefore z = \frac{k}{3} = \frac{7}{16 \times 3} = \frac{7}{48} {/tex}

Q 11.

Correct2

Incorrect-0.5

Which one of the following is a polynomial?

A

{tex} x^{3}+\sqrt{3x^{2}}+4\sqrt{x} {/tex}

B

{tex} x^{6}+2\sqrt{x^{2}}+x^{-1} {/tex}

C

{tex} x^{5}+x^{4}+x^{-1}+x^{-5} {/tex}

{tex} x^{3}+\sqrt{5x^{2}}+x+\sqrt{3} {/tex}

Q 12.

Correct2

Incorrect-0.5

If {tex} X^{y}=Y^{z} {/tex}, then {tex} \left(\frac{Y}{X}\right)^{\frac{x}{y}} {/tex} equals

A

{tex} X^{\frac{x}{y}} {/tex}

B

{tex} X^{ \left(\frac{x}{y}\right)-1 } {/tex}

C

{tex} X^{\frac{y}{x}} {/tex}

{tex} X^{1-\frac{x}{y}} {/tex}

##### Explanation

{tex} X^{y} = Y^{x} {/tex}

{tex} \left(X\right)^{\frac{y}{x}} = Y {/tex}

{tex} \left(\frac{Y}{X}\right)^{\frac{x}{y}} = \frac{ \left(X^{\frac{y}{x}}\right)^{\frac{x}{y}} }{ \left(X\right)^{\frac{x}{y}} } {/tex}

={tex} \frac{X}{X^{\frac{x}{y}}} {/tex}

= {tex} \left(X\right)^{1-\frac{x}{y}} {/tex}

Q 13.

Correct2

Incorrect-0.5

If {tex} x+\frac{1}{x}=3 {/tex}, then value of {tex} x^{3}+\frac{1}{x^{3}} {/tex} is

18

B

24

C

27

D

36

##### Explanation

Given: {tex} x+\frac{1}{x}=3 {/tex}

Taking cube on both sides, we have

{tex} \left(x+\frac{1}{x}\right)^{3}= \left(3\right)^{3} {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}}+3. \left(x+\frac{1}{x}\right)=27 {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}}+3 \times 3 = 27 {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}} = 18 {/tex}

Q 14.

Correct2

Incorrect-0.5

If {tex} x^{5}-9x^{2}+12x-14 {/tex} is divided by {tex} x-3 {/tex}, then the remainder is

A

1

B

2

C

56

184

##### Explanation

If {tex} f \left(x\right) {/tex} is divided by {tex} x-3 {/tex}, then remainder will be {tex} f \left(3\right) {/tex}

Therefore, Remainder = {tex} f \left(3\right) {/tex}

={tex} 3^{5}-9 \times 3^{2}+12 \times 3-14 {/tex}

= {tex} 243 - 81 + 36 - 14 = 184 {/tex}

Q 15.

Correct2

Incorrect-0.5

If polynomial {tex} f \left(x\right) {/tex} is such that {tex} f \left(-2\right)=0 {/tex}, then which of the following is always a factor of {tex} f \left(x\right) {/tex}?

A

2x

B

2-x

x+2

D

x-2

Q 16.

Correct2

Incorrect-0.5

If {tex} x^{2}+ax+b {/tex} leaves the same remainder 5 when divided by x-1 or x+1, then values of 'a' and 'b' are respectively.

0 and 4

B

3 and 0

C

0 and 3

D

4 and 0

##### Explanation

When {tex} f \left(x\right) = x^{2}+ax+b {/tex} is divided by {tex} x-1 {/tex}, then remainder will be 5.

or {tex} f \left(1\right)=5 {/tex}

Therefore, 1 + a + b = 5

=> a + b = 4 ...(i)

When {tex} f \left(x\right) = x^{2}+ax+b {/tex} is divided by {tex} x+1 {/tex}, then remainder will be 5.

or {tex} f \left(-1\right)=5 {/tex}

Therefore, 1 - a + b = 5

=> -a + b = 4 ...(ii)

Solving equations (i) and (ii), we get

a = 0, b = 4

Q 17.

Correct2

Incorrect-0.5

Product of zeroes of the polynomial {tex} x^{3}-6x^{2}+11x-6 {/tex} is

A

11

B

-6

C

1

6

##### Explanation

Given polynomial is

{tex} x^{3}-6x^{2}+11x-6 {/tex}

{tex} Product\ of\ the\ zeros\ =

- \frac{constant\ term}{coefficient\ of\ x^{3}} {/tex}

= {tex} - \frac{ \left(-6\right) }{1} = 6 {/tex}

Q 18.

Correct2

Incorrect-0.5

H.C.F. of {tex} x^{5}+2x^{4}+x^{3}\ and\ x^{7}-x^{5} {/tex} is

A

x

B

{tex} x \left(x+1\right) {/tex}

C

{tex} x^{3} {/tex}

{tex} x^{3} \left(x+1\right) {/tex}

##### Explanation

{tex} x^{5}+2x^{4}+x^{3} = x^{3} \left(x^{2}+2x+1\right) = x^{3} \left(x+1\right)^{2} {/tex}

and {tex} x^{7}-x^{5} = x^{5} \left(x^{2}-1\right) {/tex}

= {tex} x^{5} \left(x+1\right) \left(x-1\right) {/tex}

{tex} \therefore H.C.F.\ =\ x^{3} \left(x+1\right) {/tex}

Q 19.

Correct2

Incorrect-0.5

H.C.F. of two polynomials is a - b, for the same polynomials, L.C.M. is {tex} \left(a^{2}-b^{2}\right) \left(a^{2}+ab+b^{2}\right) {/tex}. If one of the polynomials is {tex} a^{3}-b^{3} {/tex}, then other will be

{tex} a^{2}-b^{2} {/tex}

B

{tex} a^{2}+ab+b^{2} {/tex}

C

{tex} a^{2}+b^{2} {/tex}

D

{tex} \left(a+b\right) {/tex}

##### Explanation

Product of polynomials = H.C.F. X L.C.M.

Therefore, Other polynomials

= {tex} \frac{ \left(a-b\right) \times \left(a^{2}-b^{2}\right) \left(a^{2}+ab+b^{2}\right) }{a^{3}-b^{3}} {/tex}

= {tex} \frac{ \left(a-b\right) \times \left(a^{2}-b^{2}\right) \left(a^{2}+ab+b^{2}\right) }{ \left(a-b\right) \left(a^{2}+ab+b^{2}\right)} {/tex}

= {tex} a^{2} - b^{2} {/tex}

Q 20.

Correct2

Incorrect-0.5

If {tex} f \left(x\right) = 2x^{2}+\sqrt{2}x+8 {/tex}, then {tex} f \left(x\right) {/tex} is a polynomial over

real numbers

B

irrational numbers

C

rational number

D

positive rationals

Q 21.

Correct2

Incorrect-0.5

Factors of {tex} 12x^{2}+11x+2 {/tex} are

{tex} \left(3x+2\right) \left(4x+1\right) {/tex}

B

{tex} \left(3x-2\right) \left(4x-1\right) {/tex}

C

{tex} \left(4x+2\right) \left(3x+1\right) {/tex}

D

{tex} \left(x+3\right) \left(x-4\right) \left(x-1\right) {/tex}

##### Explanation

Given polynomial

= {tex} 12x^{2}+11x+2 {/tex}

= {tex} 12x^{2}+5x+3x+2 {/tex}

= {tex} 4x \left(3x+2\right)+1 \left(3x+2\right) {/tex}

= {tex} \left(4x+1\right) \left(3x+2\right) {/tex}

Q 22.

Correct2

Incorrect-0.5

H.C.F. of the polynomials {tex} 2x^{3}-3x^{2}-11x+6\ and\ 2x^{2}+x-1 {/tex} is

A

{tex} 3x-1 {/tex}

{tex} 2x-1 {/tex}

C

{tex} x - \frac{1}{2} {/tex}

D

{tex} 2x-5 {/tex}

##### Explanation

{tex} 2x^{2}+x-1 = 2x^{2}+2x-x-1 {/tex}

= {tex} 2x \left(x+1\right)-1 \left(x+1\right) {/tex}

= {tex} \left(2x-1\right) \left(x+1\right) {/tex} ...(i)

and {tex} 2x^{3}-3x^{2}-11x+6 {/tex}

= {tex} \left(2x-1\right) \left(x^{2}-x-6\right) {/tex}

= {tex} \left(2x-1\right) \left(x^{2}-3x+2x-6\right) {/tex}

= {tex} \left(2x-1\right)\left[ x \left(x-3\right)+2 \left(x-3\right) \right] {/tex}

= {tex} \left(2x-1\right) \left(x+2\right) \left(x-3\right) {/tex} ...(ii)

From equations (i) and (ii), we get

H.C.F. = {tex} \left(2x - 1\right) {/tex}

Q 23.

Correct2

Incorrect-0.5

L.C.M. of {tex} x^{4}+x^{2}+1,\ x^{4}-x^{2}-2x-1,\ x^{6}-1 {/tex} is

A

{tex} \left(x^{6}+1\right) \left(x^{4}+x^{2}+1\right) {/tex}

B

{tex} \left(x^{6}-1\right) \left(x^{2}+x+1\right) {/tex}

C

{tex} \left(x^{6}+1\right) \left(x^{2}-x-1\right) {/tex}

{tex} \left(x^{6}-1\right) \left(x^{2}-x-1\right) {/tex}

##### Explanation

{tex} x^{4}+x^{2}+1 = x^{4}+2x^{2}+1-x^{2} {/tex}
= {tex} \left(x^{2}+1\right)^{2}-x^{2} {/tex}
= {tex} \left(x^{2}+1+x\right) \left(x^{2}+1-x\right) {/tex} ...(i)
{tex} x^{4}-x^{2}-2x-1 = x^{4}- \left(x^{2}+2x+1\right) {/tex}
= {tex} \left(x^{2}\right)^{2}- \left(x+1\right)^{2} {/tex}
= {tex} \left(x^{2}+x+1\right) \left(x^{2}-x-1\right) {/tex} ...(ii)
{tex} x^{6}-1 = \left(x^{3}-1\right) \left(x^{3}+1\right) {/tex}
= {tex} \left(x-1\right) \left(x^{2}+x+1\right) \left(x+1\right) \left(x^{2}-x+1\right) {/tex} ...(iii)
L.C.M. = {tex} \left(x^{3}-1\right) \left(x^{3}+1\right) \left(x^{2}-x-1\right) {/tex}
= {tex} \left(x^{6}-1\right) \left(x^{2}-x-1\right) {/tex}

Q 24.

Correct2

Incorrect-0.5

If a four-digited perfect square number is such that the number formed by the first two digits and the number formed by the last two digits are also perfect squares; then four-digited number is

A

6416

B

3616

1681

D

1664

##### Explanation

{tex} 1681 = 41^{2} {/tex}

Q 25.

Correct2

Incorrect-0.5

If {tex} x^{9}-3 {/tex} is divided by {tex} x^{3}-1 {/tex}, then the remainder will be

A

2

-2

C

1

D

-1