Important Questions for CBSE Class 10 Maths Chapter 2 - Polynomials 2024-25

Very Short Answer Questions (1 Mark)

1. The graphs of $y=p(x)$ are given to us, for some polynomials $p(x)$. Find the number of zeroes of $p(x)$, in each case.

i.

Ans: The graph does not meet the $x-$ axis at any point. Hence, it does not have any zero.

ii.

Ans: The graph meets the $x-$ axis at only $1$ point. Thus, the polynomial $p(x)$ corresponding to the given graph has only $1$ zero.

iii.

Ans: The graph meets the $x-$ axis at $3$ points. Thus, the polynomial $p(x)$ corresponding to the given graph has $3$ zeroes.

iv.

Ans: The graph meets the $x-$ axis at $2$ points. Thus, the polynomial $p(x)$ corresponding to the given graph has $2$ zeroes.

v.

Ans: The graph meets the $x-$ axis at $3$ points. Thus, the polynomial $p(x)$ corresponding to the given graph has $3$ zeroes.

vi.

Ans: The graph meets the $x-$ axis at $4$ points. Thus, the polynomial $p(x)$ corresponding to the given graph has $4$ zeroes.

2. Which of the following is polynomial?

1. $x^2-6\sqrt{x}+2$  

2.  $\sqrt{x}+{\displaystyle \frac{1}{\sqrt{x}}}$ 

3. $${\displaystyle \frac{5}{x^2-3x+1}}$$ 

4. none of these 

Ans: d. none of these

3. Polynomial $2x^4+3x^3-5x^2-5x^2+9x+1$  is a

1. linear polynomial

2. quadratic polynomial

3. cubic polynomial

4. bi-quadratic polynomial

Ans: d. bi-quadratic polynomial

4. If $\alpha$ and $\beta$  are zeros of $x^2+5x+8$  , then the value of $(\alpha +\beta )$  is           

1. $5$

2. $-5$

3.  $8$

4. $-8$

Ans:  b. $-5$

5. The sum and product of the zeros of a quadratic polynomial are $2$  and $-15$ respectively. The quadratic polynomial is    

1. $x^2-2x+15$ 

2. $x^2-2x-15$ 

3. $x^2+2x+15$ 

4. $x^2+2x+15$ 

Ans: b. $$x^2-2x-15$$

6. If $p(x)=2x^2-3x+5$,$3x+5$ , then the value of $p(-1)$ is equal to                 

1. $7$ 

2. $8$ 

3. $9$ 

4. $10$ 

Ans: d. $$10$$

7. Zeros of $p(x)=x^2-2x-3$     

1. $3$ and $1$ 

2. $3$ and $-1$ 

3. $-3$ and $-1$ 

4. $1$ and $-3$ 

Ans: b. $$3$$ and $$-1$$

8. If $\alpha$ and $\beta$ are the zeros of $2x^2+5x-10$ , then the value of $\alpha \beta$ is     

1. $-{\displaystyle \frac{5}{2}}$ 

2.  $5$ 

3. $-5$ 

4.  ${\displaystyle \frac{2}{5}}$ 

Ans: c. $$-5$$

9. A quadratic polynomial, the sum and product of whose zeros are $0$ and $\sqrt{5}$                                     

1. $x^2+\sqrt{5}$ 

2. $x^2-\sqrt{5}$  

3. $x^2-5$ 

4. none of these

Ans: a. $$x^2+\sqrt{5}$$

10. Which of the following is polynomial?    

1. $x^2-6\sqrt{x}+2$  

2. $\sqrt{x}+{\displaystyle \frac{1}{\sqrt{x}}}$ 

3. $${\displaystyle \frac{5}{x^2-3x+1}}$$ 

4. none of these 

Ans: d. none of these

11. Polynomial $2x^4+3x^3-5x^2-5x^2+9x+1$ is a  

1. linear polynomial

2. quadratic polynomial

3. cubic polynomial

4. bi-quadratic polynomial

Ans: d. bi-quadratic polynomial

12. If $\alpha$ and $\beta$ are zeros of $x^2+5x+8$ , then the value of $(\alpha +\beta )$ is   

1. $5$

2. $-5$

3. $8$

4. $-8$

Ans:  b. $$-5$$

13. The sum and product of the zeros of a quadratic polynomial are $2$  and $-15$ respectively. The quadratic polynomial is    

1. $x^2-2x+15$ 

2. $x^2-2x-15$ 

3. $x^2+2x+15$ 

4. $x^2+2x+15$ 

Ans: b. $$x^2-2x-15$$

Short Answer Questions                                                    (2 Marks)

1. Find the quadratic polynomial where sum and product of the zeros are $a$ and${\displaystyle \frac{1}{a}}$.                           

Ans: A quadratic polynomial is given by: $x^2-$(sum of zeros)$x+$(product of zeros)

Given, the sum of zeros as $a$ and the product of zeros as ${\displaystyle \frac{1}{a}}$.

Thus, the quadratic polynomial is $x^2-ax+{\displaystyle \frac{1}{a}}$, i.e., ${\displaystyle \frac{1}{a}}[a{x}^2-a^{2}x+1]$.

2. If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^2-x-4$ , find the value of ${\displaystyle \frac{\text{1}}{\alpha }}\text{+}{\displaystyle \frac{\text{1}}{\beta }}\text{- }alpha\beta$.

Ans: Given, $f(x)=x^2-x-4$ and the zeros of the polynomial are $\alpha$ and $\beta$.

Therefore, the sum of the zeros,  $\alpha \text{ + }\beta =-\left({\displaystyle \frac{-1}{1}}\right)=1$ 

The product of the zeros, $\alpha \beta ={\displaystyle \frac{-4}{1}}=-4$ 

The value of ${\displaystyle \frac{\text{1}}{\alpha }}\text{+}{\displaystyle \frac{\text{1}}{\beta }}\text{- }\alpha \beta \text{ =}{\displaystyle \frac{\alpha \text{ + }\beta }{\alpha \beta }}\text{- }\alpha \beta$ is

$\Rightarrow {\displaystyle \frac{\text{1}}{\alpha }}\text{+}{\displaystyle \frac{\text{1}}{\beta }}\text{- }\alpha \beta ={\displaystyle \frac{1}{-4}}-(-4)$

$\Rightarrow -{\displaystyle \frac{1}{4}}+4={\displaystyle \frac{15}{4}}$

Thus, ${\displaystyle \frac{\text{1}}{\alpha }}\text{+}{\displaystyle \frac{\text{1}}{\beta }}\text{- }\alpha \beta ={\displaystyle \frac{15}{4}}$.

3. If the square of the difference of the zeros of the quadratic polynomial $f\left(x\right)={\text{x}}^2+px+45$ is equal to $144$ , find the value of $p$.               

Ans: It is given that ${\text{( }\alpha \text{ - }\beta \text{ )}}^{\text{2}}=144$

From the given quadratic polynomial, $f(x)=x^2+px+45$,

The sum of the roots, $\alpha \text{ + }\beta =-p$ and the product of the roots, $\alpha \beta =45$. 

Since, ${\text{( }\alpha \text{ - }\beta \text{ )}}^{\text{2}}=144$

$$\Rightarrow {\alpha }^{\text{2}}\text{+}{\beta }^{\text{2}}\text{-2 }\alpha \beta =144$$ 

$$\Rightarrow {\text{( }\alpha \text{ + }\beta \text{ )}}^{\text{2}}\text{- 4 }\alpha \beta =144$$

Substituting, $\alpha \text{ + }\beta =-p$ and $\alpha \beta =45$, we get:

$$\Rightarrow {(-p)}^2-4\times 45=144$$

$$\Rightarrow p^2=144+180$$

$$\Rightarrow p^2=324$$

$$\therefore p=\pm 18$$.

4. Divide $(6x^3-26x-21+x^2)$ by $(-7+3x)$.  

Ans: Dividing $(6x^3-26x-21+x^2)$ by $(-7+3x)$, we obtain:

$$\begin{array}{cc}& 3x-7\stackrel{2x^2+5x+3}{\overline{)6x^3+x^2-26x-21}}\\ & \underset{―}{6x^3-14x^2}\\ & 15x^2-26x-21\\ & \underset{―}{15x^2-35x}\\ & 9x-21\\ & \underset{―}{9x-21}\\ & \underset{―}{0}\end{array}$$

Thus, the quotient is:

Quotient $=2x^2+5x+3$. 

5. Find the value of ‘$k$ ’ such that the quadratic polynomial $x^2-(k+6)x+2(2k+1)$ has sum of the zeros is half of their product.

Ans: It is given that, Sum of the zeros $={\displaystyle \frac{1}{2}}\times$ Product of the zeros.

From the given quadratic polynomial, $x^2-(k+6)x+2(2k+1)$, the sum of the zeros is $(k+6)$ and product of the zeros is $2(2k+1)$.

Hence,

$\Rightarrow (k+6)={\displaystyle \frac{1}{2}}[2(2k+1)]$

$\Rightarrow k+6=2k+1$

$\Rightarrow k=5$.

6. If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left(x\right)=x^2-p(x+1)-c$ , show that $\left(\alpha \text{ +1}\right)\left(\beta \text{ +1}\right)=1-c$.             

Ans: It is given that $\alpha$ and $\beta$ are the roots of the quadratic polynomial $f(x)=x^2-p(x+1)-c$.

$f(x)=x^2-px-(p+c)$

$\therefore \alpha \text{ + }\beta \text{ = p }$ 

And $\alpha \beta =-(p+c)$

Thus,

$\Rightarrow \text{( }\alpha \text{ +1)( }\beta \text{ +1)= }\alpha \beta \text{ +( }\alpha \text{ + }\beta \text{ )}+1$

$\Rightarrow \text{( }\alpha \text{ +1)( }\beta \text{ +1)}=-p-c+p+1$

$\Rightarrow \text{( }\alpha \text{ +1)( }\beta \text{ +1)}=1-c$.

7. If the sum of the zeros of the quadratic polynomial $f(t)=k{t}^2+2t+3k$ is equal to their product, find the value of ‘$k$

Ans: The given quadratic polynomial is, $f(t)=k{t}^2+2t+3k$.

It is given that, sum of the zeros = product of the zeros

Hence,

$\Rightarrow {\displaystyle \frac{-2}{k}}={\displaystyle \frac{3k}{k}}$

$\Rightarrow k=-{\displaystyle \frac{2}{3}}$ .

8. Divide $(x^4-5x+6)$ by $(2-x^2)$. 

Ans: Dividing $(x^4-5x+6)$ by $(2-x^2)$, we obtain:

$$\begin{array}{cc}& 2-x^2\stackrel{-x^2-2}{\overline{)x^4-5x+6}}\\ & \underset{―}{x^4-2x^2}\\ & 2x^2-5x+6\\ & \underset{―}{2x^2-4}\\ & \underset{―}{-5x+10}\end{array}$$

$\therefore$ Quotient $=-x^2-2$

Remainder $=-5x+10$

9. Find the zeros of the polynomial $p(x)=4\sqrt{3}{x}^2+5x-2\sqrt{3}$ and verify the relationship between the zeros and its coefficients.

Ans: The given polynomial is, $p(x)=4\sqrt{3}{x}^2+5x-2\sqrt{3}$.

It can be written as: $p(x)=4\sqrt{3}{x}^2+8x-3x-2\sqrt{3}$.

Hence,

$\Rightarrow p(x)=4x(\sqrt{3}x+2)-\sqrt{3}(\sqrt{3}x+2)$

$\Rightarrow p(x)=(4x-\sqrt{3})(\sqrt{3}x+2)$

Thus, the zeros are $4x-\sqrt{3}=0$ and $\sqrt{3}x+2=0$ 

$\Rightarrow x={\displaystyle \frac{\sqrt{3}}{4}}$ and $x=-{\displaystyle \frac{2}{\sqrt{3}}}$

We know, sum of the zeros $={\displaystyle \frac{-\text{ Coefficient of }x}{\text{ Coefficient of }{x}^2}}$

$=[{\displaystyle \frac{\sqrt{3}}{4}}+{\displaystyle \frac{(-2)}{\sqrt{3}}}]={\displaystyle \frac{-5}{4\sqrt{3}}}$ 

Similarly, the product of the zeros $={\displaystyle \frac{\text{Constant term}}{\text{ Coefficient of }{x}^2}}$

$\Rightarrow {\displaystyle \frac{-2\sqrt{3}}{4\sqrt{3}}}={\displaystyle \frac{-1}{2}}$ .

10. Find the value of ‘$k$’ so that the zeros of the quadratic polynomial $3x^2-kx+14$ are in the ratio $7:6$.

Ans: Let the zeros of the quadratic polynomial, $3x^2-kx+14$ be $$7p$$ and $$6p$$.

$\therefore 7p+6p={\displaystyle \frac{-(-k)}{3}}={\displaystyle \frac{k}{3}}$ 

$\Rightarrow 39p=k$

Also, $7p\times 6p={\displaystyle \frac{14}{3}}$ 

$\Rightarrow 42p^2={\displaystyle \frac{14}{3}}$

$\Rightarrow p=3$ 

Thus,

$\Rightarrow k=39\times 3$

$\Rightarrow k=117$.

11. If one zero of the quadratic polynomial $f(x)=4x^2-8kx-9$ is negative of the other, find the value of ‘$k$’.

Ans: It is given that one zero of the quadratic polynomial $f(x)=4x^2-8kx-9$ is the negative of the other,

Let us take one zero to be $\alpha$, then the other is $-\alpha$. 

Hence, the sum of the zeros $=0$

$\Rightarrow {\displaystyle \frac{8k}{4}}=0$

$\Rightarrow k=0$

12. Check whether the polynomial $(t^2-3)$ is a factor of the polynomial $2t^4+3t^3-2t^2-9t-12$ by Division method.                                                

Ans: To determine whether the polynomial, $(t^2-3)$ is a factor of the polynomial $2t^4+3t^3-2t^2-9t-12$ by division method, divide $2t^4+3t^3-2t^2-9t-12$ from $(t^2-3)$. Thus, we have

$$\begin{array}{cc}& t^3-3\stackrel{2t^2+3t+4}{\overline{)2t^4+3t^3-2t^2-9t-12}}\\ & \underset{―}{2t^4-6t^2}\\ & 3t^3+4t^2-9t-12\\ & \underset{―}{3t^3-9t}\\ & 4t^2-12\\ & \underset{―}{4t^2-12}\\ & \underset{―}{0}\end{array}$$  

$\therefore (t^2-3)$ is the factor of given polynomial $2t^4+3t^3-2t^2-9t-12$.

Short Answer Questions (3 Marks)

1. Apply division algorithm to find the quotient $q\left(x\right)$ and remainder $r\left(x\right)$ an dividing $f\left(x\right)$ by $g\left(x\right)$ where $f(x)=x^3-6x^2+11x-6$ , $g(x)=x^2+x+1$. 

Ans: According to the division algorithm, $f(x)=g(x)\times q(x)+r(x)$.

Thus, dividing $x^3-6x^2+11x-6$ from $x^2+x+1$, we obtain:

$$\begin{array}{cc}& x^2+x+1\stackrel{x-7}{\overline{)x^3-6x^2+11x-6}}\\ & \underset{―}{x^3+x^2+x}\\ & -7x^2+10x-6\\ & \underset{―}{-7x^2-7x-7}\\ & \underset{―}{17x+1}\end{array}$$

Hence,

$(x^3-6x^2+11x-6)=x^2+2x+1(x-7)+(17x+1)$ .

2. If two zeros of the polynomial $x^4-6x^3-26x^2+138x-35$ are $2\pm \sqrt{3}$ , find the other zeros.

Ans: It is given that the two zeros of the polynomial $x^4-6x^3-26x^2+138x-35$ are $2+\sqrt{3}$ and $2-\sqrt{3}$.

Therefore, the sum of zeros is $2+\sqrt{3}+2-\sqrt{3}=4$

and product of the zeros is $1$. 

Hence,$(x^2-4x+1)$ is the factor of $x^4-6x^3-26x^2+138x-35$.

So, the other factors can be determined by:

$\begin{array}{cc}& x^2-4x+1\stackrel{x^2-2x-35}{\overline{)x^4-6x^3-26x^2+138x-35}}\\ & \underset{―}{x^4-4x^3+x^2}\\ & -2x^3-27x^2+138x-35\\ & \underset{―}{-2x^3+8x^2-2x}\\ & -35x^2+140x-35\\ & \underset{―}{-35x^2+140x-35}\\ & \underset{―}{0}\end{array}$

Now,

$\Rightarrow x^2-2x-35=x^2-7x+5x-35$ 

$\Rightarrow x(x-7)+5(x-7)$

$\Rightarrow x^2-2x-35=(x+5)(x-7)$

Thus, the zeros are

$x=7$ and $x=-5$

So, the other two zeros are $7$ and $-5$.

3. What must be subtracted from the polynomial $f(x)=x^4+2x^3-13x^2-12x+21$ so that the resulting polynomial is exactly divisible by $g(x)=x^2-2x+3$. 

Ans: According to the division algorithm, $f(x)=g(x)\times q(x)+r(x)$ .

Thus, $f(x)-r(x)=g(x)\times q(x)$

Dividing $f(x)=x^4+2x^3-13x^2-12x+21$ from $g(x)=x^2-2x+3$, we obtain:

$$\begin{array}{cc}& x^2-4x+3\stackrel{x^2+6x+8}{\overline{)x^4+2x^3-13x^2-12x+21}}\\ & underline{x}^4-4x^3+3x^2\\ & 6x^3-16x^2-12x+21\\ & hspace2.5cm\underset{―}{6x^3-24x^2+18x}\\ & 8x^2-30x+21\\ & hspace5cm\underset{―}{8x^2-32x+24}\\ & \underset{―}{2x-3}\end{array}$$

Thus, we must be subtract $(2x-3)$ from $f(x)=x^4+2x^3-13x^2-12x+21$ so that the resulting polynomial is exactly divisible by $g(x)=x^2-2x+3$.

4. What must be added to $6x^5+5x^4+11x^3-3x^2+x+5$ so that it may be exactly divisible by $3x^2-2x+4$?

Ans: Dividing $6x^5+5x^4+11x^3-3x^2+x+5$ from $$3x^2-2x+4$$, we obtain:

$$\begin{array}{cc}& quad3x^2-2x+4\stackrel{2x^3+3x^2+3x+3}{\overline{)6x^5+5x^4+11x^3-3x^{+}x+5}}\\ & \underset{―}{6x^5-4x^4+8x^3}\\ & 9x^4+3x^3-3x^2+x+5\\ & \underset{―}{9x^4-6x^3+12x^2}\\ & 9x^3-15x^2+x+5\\ & \underset{―}{9x^3-6x^2+12x}\\ & -9x^2-11x+5\\ & \underset{―}{-9x^2-6x+12}\\ & \underset{―}{-17x-7}\end{array}$$

So, we must add $$(3x^2-2x+4)-(-17x-7)$$ , i.e.,

$$\Rightarrow 3x^2-2x+4+17x+7$$

$$\Rightarrow 3x^2+15x+13$$.

Hence, we would add $$3x^2+15x+13$$ to $6x^5+5x^4+11x^3-3x^2+x+5$ so that it is exactly divisible by $$3x^2-2x+4$$.

5. Find all the zeros of the polynomial $f(x)=2x^4-3x^3-3x^2+6x-2$ , if being given that two of its zeros are $\sqrt{2}$ and $\text{-}\sqrt{2}$.

Ans: The two zeros of the polynomial $$f(x)=2x^4-3x^3-3x^2+6x-2$$ are $\sqrt{2}$ and $-\sqrt{2}$.

$$\therefore (x-\sqrt{2})(x+\sqrt{2})$$ or $${(x-2)}^2$$ is the factor of the given polynomial.

$$\begin{array}{cc}& x^2-2\stackrel{2x-3x+1}{\overline{)2x^4-3x^3-3x^2+6x-2}}\\ & \underset{―}{2x^4-4x^2}\\ & -3x^3+x^2+6x-2\\ & \underset{―}{-3x^3+6x}\\ & x^2-2\\ & \underset{―}{x^2-2}\\ & \underset{―}{0}\end{array}$$ 

The quotient is $$q(x)=2x^2-3x+1$$. Thus,

$$q(x)=2x^2-3x+1$$ 

$$\Rightarrow 2x^2-2x-x+1$$

$$\Rightarrow 2x(x-1)-1(x-1)$$

$$q(x)=(2x-1)(x-1)$$

Hence, the other two zeroes are

$$x=1$$ and $$x={\displaystyle \frac{1}{2}}$$.

6. On dividing $x^3-3x^2+x+2$ by a polynomial $g(x)$ the quotient and the remainder were $(x-2)$ and $-2x+4$ respectively, find $g(x)$

Ans: According to the division algorithm, $f(x)=g(x)\times q(x)+r(x)$.

$\therefore g(x)={\displaystyle \frac{p(x)-r(x)}{q(x)}}$.

Substituting for $p\left(x\right)=x^3-3x^2+x+2$, $q(x)=(x-2)$ and $r(x)=-2x+4$, we get:

$g\left(x\right)={\displaystyle \frac{x^3-3x^2+x+2+2x-4}{x-2}}$ 

Thus,

$$\begin{array}{cc}& x-2\stackrel{x^2-x+1}{\overline{)x^3-3x^2+3x-2}}\\ & \underset{―}{x^3-2x^2}\\ & -x^2+3x-2\\ & \underset{―}{-x^2+2x}\\ & x-2\\ & \underset{―}{x-2}\\ & \underset{―}{0}\end{array}$$

Hence, $g(x)=x^2-x+1$.

7. Find all zeros of $f(x)=2x^3-7x^2+3x+6$ if its two zeros are $-\sqrt{{\displaystyle \frac{3}{2}}}$ and $\sqrt{{\displaystyle \frac{3}{2}}}$ . 

Ans: The two zeros of the given polynomial, $f(x)=2x^4-2x^3-7x^2+3x+6$ are  $-\sqrt{{\displaystyle \frac{3}{2}}}$ and $\sqrt{{\displaystyle \frac{3}{2}}}$ 

$$\therefore (x+\sqrt{{\displaystyle \frac{3}{2}}})(x-\sqrt{{\displaystyle \frac{3}{2}}})={\displaystyle \frac{1}{2}}(2x^2-3)$$ 

So, $(2x^2-3)$ is the factor of $f(x)$.

Dividing  $f(x)=2x^4-2x^3-7x^2+3x+6$ from $(2x^2-3)$, we get:

$$\begin{array}{cc}& 2x^2-3\stackrel{x^2-x-2}{\overline{)2x^4-2x^3-7x^2+3x+6}}\\ & \underset{―}{2x^4-3x^2}\\ & -2x^3-4x^2+3x+6\\ & \underset{―}{-2x^3+3x}\\ & -4x^2+6\\ & \underset{―}{-4x^2+6}\\ & \underset{―}{0}\end{array}$$

Thus, we have

$g(x)=x^2-x-2$ 

$\Rightarrow x^2-2x+x-2$

$\Rightarrow x(x-2)+1(x-2)$

$g\left(x\right)=(x+1)(x-2)$

Hence, the other two zeros are

$x=-1$ and 

$x=2$ 

8. Obtain all zeros of the polynomial $f(x)=2x^4+x^3-14x^2-19x-6$ , if two of its zeros are $-2$ and $-1$.  

Ans: The two zeros of the given polynomial, $f(x)=2x^4+x^3-14x^2-19x-6$ are  $-2$ and $-1$. 

$\therefore (x+2)(x+1)=x^2+3x+2$ 

So, $x^2+3x+2$ is the factor of $f\left(x\right)$.

Dividing $f(x)=2x^4+x^3-14x^2-19x-6$ from $x^2+3x+2$, we get:

$$\begin{array}{cc}& x^2-3x+2\stackrel{2x^2-5x-3}{\overline{)2x^4+x^3-14x^2-19x-6}}\\ & \underset{―}{2x^4+6x^3+4x^2}\\ & -5x^3-18x^2-19x-6\\ & \underset{―}{-5x^3-15x^2-10x}\\ & -3x^2-9x-6\\ & \underset{―}{-3x^2-9x-6}\\ & \underset{―}{0}\end{array}$$ 

Thus, we have

$g(x)=2x^2-5x-3$ 

$\Rightarrow \text{ 2}{x}^2-6x+x-3$

$\Rightarrow 2x(x-3)+1(x-3)$

$g\left(x\right)=(x-3)(2x+1)$

Hence, the other two zeros are

$x=3$ and 

$x=-{\displaystyle \frac{1}{2}}$ 

Other two zeros are $3$ and $-{\displaystyle \frac{1}{2}}$ .

9. Obtain all other zeros of $3x^4+6x^3-2x^2-10x-5$ if its two zeros are $-\sqrt{{\displaystyle \frac{5}{3}}}$ and $\sqrt{{\displaystyle \frac{5}{3}}}$.

Ans: The two zeros of the given polynomial, $3x^4+6x^3-2x^2-10x-5$ are  $-\sqrt{{\displaystyle \frac{5}{3}}}$ and $\sqrt{{\displaystyle \frac{5}{3}}}$. 

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

So, $(3x^2-5)$ is the factor of $f\left(x\right)$.

Dividing $3x^4+6x^3-2x^2-10x-5$ from $(3x^2-5)$, we get: