# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials  Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 10 Maths Chapter 2 Polynomials Exercise 2.1

Q1 ) 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) (ii) (iii) (iv) (v) (vi) NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

(i) Graph does not meets x-axis at all, so it does not have any zero.
(ii) Graph meets x-axis 1 time, so it has one zero.
(iii) Graph meets x-axis 3 times, so it has three zero.
(iv) Graph meets x-axis 2 times, so it has two zero.
(v) Graph meets x-axis 4 times, so it has four zero.
(vi) Graph meets x-axis 3 times, so it has three zero.

## NCERT solutions for class 10 Maths Chapter 2 Polynomials Exercise 2.2

Q1 ) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.
(i) $$x^2 - 2x - 8$$
(ii) $$4s^2 - 4s + 1$$
(iii) $$6x^2 - 3 - 7x$$
(iv) $$4u^2 + 8u$$
(v) $$t^2 - 15$$
(vi) $$3x^2 - x - 4$$

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

General form of equation is $$ax^2 + bx + c$$
where, a = Coefficient of $$x^2$$
b = Coefficient of x
c = Constant term

Once we'll find the zeroes of the polynomial we can verify it using the following rules

Sum of zeroes = $${{-b}\over {a}}$$
Product of zeroes = $${{c}\over {a}}$$

(i) $$x^2 - 2x - 8$$
=$$x^2 - 4x + 2x - 8$$
= $$x(x - 4) + 2(x - 4)$$
= $$(x + 2)(x - 4)$$

So, zeroes are -2 and 4
Now, Verification:

Sum of the zeroes = 2
= $${{-b}\over {a}}$$ = $${{-(-2)}\over {1}}$$ = 2
Product of the zeroes = -8
= $${{c}\over {a}}$$ = $${{-8}\over {1}}$$ = -8

(ii) $$4s^2 - 4s + 1$$
=$$4s^2 - 2s - 2s + 18$$
= $$2s(2s - 1) - 1(2s - 1)$$
= $$(2s - 1)(2s - 1)$$

So, zeroes are $${{1}\over {2}}$$ and $${{1}\over {2}}$$
Now, Verification:

Sum of the zeroes = 1
= $${{-b}\over {a}}$$ = $${{-(-4)}\over {4}}$$ = 1
Product of the zeroes = $${{1}\over {4}}$$
= $${{c}\over {a}}$$ = $${{1}\over {4}}$$

(iii) $$6x^2 - 7x - 3$$
=$$6x^2 - 9x + 2x -3$$
= $$3x(2x - 3) + 1(2x - 3)$$
= $$(3x + 1)(2x - 3)$$

So, zeroes are $${{-1}\over {3}}$$ and $${{3}\over {2}}$$
Now, Verification:
Sum of the zeroes
= $${{-1}\over {3}} + {{3}\over {2}}$$ = $${{7}\over {6}}$$
= $${{-b}\over {a}}$$ = $${{-(-7)}\over {6}}$$
Product of the zeroes
= $${{-1}\over {3}} × {{3}\over {2}}$$ = $${{-1}\over {2}}$$
= $${{c}\over {a}}$$ = $${{-3}\over {6}}$$ = $${{-1}\over {2}}$$

(iv) $$4u^2 + 8u$$
=$$4u(u + 2)$$
=$$4u = 0$$
=$$u = 0$$
=$$u = -2$$

So, zeroes are 0 and -2
Now, Verification:

Sum of the zeroes = -2
= $${{-b}\over {a}}$$ = $${{-(8)}\over {4}}$$ = -2
Product of the zeroes = 0
= $${{c}\over {a}}$$ = $${{0}\over {4}}$$ = 0

(v) $$t^2 - 15$$
= $$t^2 - (\sqrt{15})^2$$
= $$(t + \sqrt{15})(t -\sqrt{15})$$

So, zeroes are $$-\sqrt{15}$$ and $$\sqrt{15}$$
Now, Verification:
Sum of the zeroes = 0
= $${{-b}\over {a}}$$ = $${{-(0)}\over {1}}$$ = 0
Product of the zeroes = -15
= $${{c}\over {a}}$$ = $${{-15}\over {1}}$$ = -15

(vi) $$3x^2 - x - 4$$
=$$3x^2 - 4x + 3x -4$$
= $$x(3x - 4) + 1(3x - 4)$$
= $$(3x - 4)(x + 1)$$

So, zeroes are $${{4}\over {3}}$$ and -1
Now, Verification:
Sum of the zeroes
= $${{4}\over {3}} + (-1)$$ = $${{1}\over {3}}$$
= $${{-b}\over {a}}$$ = $${{-(-1)}\over {3}}$$ = $${{1}\over {3}}$$
Product of the zeroes
= $${{4}\over {3}} × (-1)$$ = $${{-4}\over {3}}$$
= $${{c}\over {a}}$$ = $${{-4}\over {3}}$$

Q2 ) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $${{1} \over {4}} , -1$$
(ii) $$\sqrt{2}, {{1}\over{3}}$$
(iii) $$0, \sqrt{5}$$
(iv) $$1,1$$
(v) $${{-1} \over {4}},{{1} \over {4}}$$
(vi) $$4,1$$

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

(i) $${{1} \over {4}} , -1$$
$$\Rightarrow$$Sum of the zeroes = $${{1} \over {4}}$$ = $${{-b}\over {a}}$$
$$\Rightarrow$$ Product of the zeroes = -1 = $${{c}\over {a}}$$
Polynomial can be formed by : $$x^2$$ - (sum of zeroes)x + (product of zeroes)
$$\Rightarrow$$$$x^2 - {{1} \over {4}}x + (-1)$$ => $$4x^2 - x - 4$$

(ii) $$\sqrt{2} , {{1}\over{3}}$$
$$\Rightarrow$$Sum of the zeroes = $$\sqrt{2}$$ = $${{-b}\over {a}}$$
$$\Rightarrow$$ Product of the zeroes = $${{1} \over {3}}$$ = $${{c}\over {a}}$$
Polynomial can be formed by : $$x^2$$ - (sum of zeroes)x + (product of zeroes)
$$\Rightarrow$$$$x^2 - \sqrt{2} x + {{1} \over {3}}$$ $$\Rightarrow$$ $$3x^2 - 3\sqrt{2}x + 1$$

(iii) $$0 , \sqrt{5}$$
$$\Rightarrow$$Sum of the zeroes = 0 = $${{-b}\over {a}}$$
$$\Rightarrow$$ Product of the zeroes = $$\sqrt{5}$$ = $${{c}\over {a}}$$
Polynomial can be formed by : $$x^2$$ - (sum of zeroes)x + (product of zeroes)
$$\Rightarrow$$$$x^2 - 0x + ( \sqrt{5})$$ => $$x^2 + \sqrt{5}$$

(iv) $$1 , 1$$
$$\Rightarrow$$Sum of the zeroes = 1 = $${{-b}\over {a}}$$
$$\Rightarrow$$ Product of the zeroes = 1 = $${{c}\over {a}}$$
Polynomial can be formed by : $$x^2$$ - (sum of zeroes)x + (product of zeroes)
$$\Rightarrow$$ $$x^2 - 1x + 1$$

(v) $${{-1} \over {4}} , {{1} \over {4}}$$
$$\Rightarrow$$ Sum of the zeroes = $${{-1} \over {4}}$$ = $${{-b}\over {a}}$$
$$\Rightarrow$$ Product of the zeroes = $${{1} \over {4}}$$ = $${{c}\over {a}}$$
Polynomial can be formed by : $$x^2$$ - (sum of zeroes)x + (product of zeroes)
$$\Rightarrow$$ $$x^2 - {{-1} \over {4}}x + {{1} \over {4}}$$ $$\Rightarrow$$ $$4x^2 + x + 1$$

(vi) $$4 , 1$$
$$\Rightarrow$$ Sum of the zeroes = 4 = $${{-b}\over {a}}$$
$$\Rightarrow$$ Product of the zeroes = 1 = $${{c}\over {a}}$$
Polynomial can be formed by : $$x^2$$ - (sum of zeroes)x + (product of zeroes)
$$\Rightarrow$$ $$x^2 - 4x + 1$$

## NCERT solutions for class 10 Maths Chapter 2 Polynomials Exercise 2.3

Q1 ) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following.
(i) $$p(x) = x^3 - 3x^2 + 5x - 3 , g(x)= x^2 - 2$$
(ii) $$p(x) = x^4 - 3x^2 + 4x + 5 , g(x)= x^2 - x + 1$$
(iii) $$p(x) = x^4 - 5x + 6 , g(x)= 2 - x^2$$

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

(i) Solving it using long division:
$$\begin{array}{rrrr|ll} x^3 & - 3x^2 & + 5x & - 3 & x^2 - 2\\ -x^3 & & + 2x & & x - 3 \\ \hline & -3x^2 & +7x & -3\\ & \phantom{-}3x^2 & & -6 & & & & \\ \hline & & +7x & -9 \\ \hline \end{array}$$
Here, Quotient = $$x - 3$$ and Remainder = $$7x - 9$$.

(ii) Solving it using long division:
$$\begin{array}{rrrr|ll} x^4 & - 3x^2 & + 4x & + 5 & x^2 - x + 1\\ x^3 -x^4 & -x^2 & & & x^2 + x - 3 \\ \hline x^3 & - 4x^2 & + 4x & + 5\\ \phantom{-}x^3 & + x^2 & -x & & & \\ \hline & -3x^2 & + 3x & +5 \\ & +3x^2 & -3x & +3 \\ \hline & & & 8 \\ \hline \end{array}$$
Here, Quotient = $$x^2 + x - 3$$ and Remainder = $$8$$.

(iii) Solving it using long division:
$$\begin{array}{rrrr|ll} &x^4 & - 5x &+ 6 &-x^2 + 2\\ &2x^2 - x^4 & & & -x^2 - 2 \\ \hline & 2x^2 & - 5x & + 6\\ & \phantom{-}-2x^2 & & + 4 & & & & \\ \hline & & -5x &+10 \\ \hline \end{array}$$
Here, Quotient = $$-x^2 - 2$$ and Remainder = $$-5x + 10$$.

Q2 ) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.
(i) $$t^2 - 3 , 2t^4 + 3t^3 -2t^2 - 9t -12$$
(ii) $$x^2 + 3x + 1 , 3x^4 + 5x^3 - 7x^2 + 2x + 2$$
(iii) $$x^2 - 3x + 1 , x^5 - 4x^3 + x^2 + 3x + 1$$

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

(i) Solving it using long division:
$$\begin{array}{rrrr|ll} 3t^3 + 2t^4 & - 2t^2 & - 9t & -12 & t^2 - 3 \\-2t^4 & +6t^2 & & & 2t^2 + 3t + 4 \\ \hline 3t^3 & + 4t^2 & -9t & -12\\ \phantom{-}-3t^3 & & + 9t & & & \\ \hline & 4t^2 & & -12 \\ & -4t^2 & & +12 \\ \hline & & & 0\\ \hline \end{array}$$
Here, Remainder = 0 so the first polynomial $$t^2 - 3$$ is a factor of second polynomial $$2t^4 + 3t^3 -2t^2 - 9t -12$$.

(ii) Solving it using long division:
$$\begin{array}{rrrr|ll} 3x^4 + 5x^3 & - 7x^2 & + 2x & +2 & x^2 + 3x + 1 \\-3x^4 - 9x^3 & -3x^2 & & & 3x^2 - 4x + 2 \\ \hline -4x^3 & - 10x^2 & +2x & +2\\ \phantom{-}+4x^3 & +12x^2 & + 4x & & & \\ \hline & 2x^2 & +6x & +2 \\ & -2x^2 & -6x & -2 \\ \hline & & & 0\\ \hline \end{array}$$
Here, Remainder = 0 so the first polynomial $$x^2 + 3x + 1$$ is a factor of second polynomial $$3x^4 + 5x^3 - 7x^2 + 2x + 2$$.

(iii) Solving it using long division:
$$\begin{array}{rrrr|ll} x^5 - 4x^3 & + x^2 & + 3x & +1 & x^3 - 3x + 1\\ -x^5 + 3x^3 & -x^2 & & & x^2 - 1 \\ \hline & -x^3 & +3x & + 1\\ & \phantom{-} +x^3 & -3x & + 1 & & & & \\ \hline & & & 2 \\ \hline \end{array}$$
Here, Remainder = 2 so the first polynomial $$x^2 - 3x + 1$$ is not a factor of second polynomial $$x^5 - 4x^3 + x^2 + 3x + 1$$.

Q3 ) Obtain all other zeroes of $$3x^4 + 6x^3 - 2x^2 -10x - 5$$ if two of its zeroes are $$\sqrt{{5}\over{3}}$$ and $$-\sqrt{{5}\over{3}}$$.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Two zeroes are $$\sqrt{{5}\over{3}}$$ and $$-\sqrt{{5}\over{3}}$$.

$$=> (x -\sqrt{{5}\over{3}}) ( x + \sqrt{{5}\over{3}}) = x ^2 - {{5}\over {3}}$$
= $${{1}\over {3}} × (3x^2 - 5)$$

Both $${{1}\over {3}}$$ and $$(3x^2 - 5)$$ are the factors of $$3x^4 + 6x^3 - 2x^2 -10x - 5$$.

So, we'll divide the given polynomial by $$3x^2 - 5$$

Solving it using long division:

$$\begin{array}{rrrr|ll} 6x^3 + 3x^4 & -2x^2 & -10x & -5 & 3x^2 - 5 \\ -3x^4 & +5x^2 & & & x^2 + 2x + 1 \\ \hline 6x^3 & +3x^2 & -10x & -5\\ \phantom{-}-6x^3 & & + 10x & & & \\ \hline & 3x^2 & & -5 \\ & -3x^2 & & +5 \\ \hline & & & 0\\ \hline \end{array}$$
$$3x^4 + 6x^3 - 2x^2 -10x - 5$$
=> $$(3x^2 - 5) (x^2 + 2x + 1)$$
=> $$(3x^2 - 5) (x(x + 1) + 1(x + 1))$$
=> $$(3x^2 - 5) (x + 1)(x + 1)$$
So the other two zeroes are x = -1 and -1.

Q4 ) On dividing $$x^3 - 3x^2 + x + 2$$ by a polynomial g(x), the quotient and remainder were $$x-2$$ and $$-2x+4$$ respectively. Find g(x).

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Here, p(x) = $$x^3 - 3x^2 + x + 2$$,
q(x) = $$x-2$$,
r(x) = $$-2x+4$$

=>According to polynomial division algorithm,

=>p(x) = g(x).q(x) + r(x)
=>$$x^3 - 3x^2 + x + 2$$ = g(x).$$(x-2)$$ + $$(-2x+4)$$
=> $$x^3 - 3x^2 + x + 2 + 2x - 4$$ = g(x).$$(x-2)$$
=> $$x^3 - 3x^2 + 3x - 2$$ = g(x).$$(x-2)$$
=> $$g(x) = {{x^3 - 3x^2 + 3x - 2}\over{(x-2)}}$$

Solving it using long division:

$$\begin{array}{rrrr|ll} x^3 & -3x^2 & +3x & -2 & x - 2 \\ -x^3 & +2x^2 & & & x^2 - x + 1 \\ \hline & -x^2 & +3x & -2\\ \phantom{-} & +x^2 & -2x & & & \\ \hline & & x & -2 \\ & & -x & + 2 \\ \hline & & & 0\\ \hline \end{array}$$

=>g(x) = $$x^2 - x + 1$$

Q5 ) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

(i) p(x) = $$2x^2 + 2x + 8$$
=>q(x) = $$x^2 + x + 4$$
=> g(x) = $$2$$ and r(x) = $$0$$

(ii) p(x) = $$x^3 + x^2 + x + 1$$
=>q(x) = $$x + 1$$
=> g(x) = $$x^2 - 1$$ and r(x) = $$2x + 2$$

(iii) p(x) = $$x^3 - x^2 + 2x + 3$$
=>q(x) = $$x^2 + 2$$
=> g(x) = $$x - 1$$ and r(x) = $$5$$

## NCERT solutions for class 10 Maths Chapter 2 Polynomials Exercise 2.4

Q1 ) Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) $$2x^3 + x^2 - 5x + 2 ; {{1}\over{2}},1,-2$$
(ii) $$x^3 - 4x^2 + 5x - 2 ; 2,1,1$$

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

(i)The general form of cubic equation is $$ax^3 + bx^2 + cx + d$$.

Here, $$a = 2,b = 1,c = -5,d = 2$$

On substituting the value of zeroes in the given equation, p(x)= $$2x^3 + x^2 - 5x + 2$$

p$$({{1}\over{2}})$$= $$2({{1}\over{2}})^3 + ({{1}\over{2}})^2 + -5({{1}\over{2}}) + 2$$
=>$${{1}\over{4}} + {{1}\over{4}} - {{5}\over{2}} + 2 = 0$$
p$$(1)$$= $$2(1)^3 + (1)^2 + -5(1) + 2$$
=>$$2 + 1 - 5 + 2 = 0$$
p$$(-2)$$= $$2(-2)^3 + (-2)^2 + -5(-2) + 2$$
=>$$-16 + 4 + 10 + 2 = 0$$

Hence, these zeroes satisfies the the given equation.

Now, we'll check whether the zeroes satisfies the following equations:

Let the three zeroes be p,q and r
$$p + q + r$$ = $${{-b}\over{a}}$$............(i)
$$pq + qr + rp$$ = $${{c}\over{a}}$$..........(ii)
$$pqr$$ = $${{-d}\over{a}}$$.................(iii)
On checking equation (i):=> $${{1}\over{2}} + 1 + (-2)$$
=>$${{3}\over{2}} - 2$$
=> $${{-1}\over{2}} = {{-b}\over{a}}$$

On checking equation (ii):=> $${{1}\over{2}} × 1 + 1 × (-2) + (-2) × {{1}\over{2}}$$
=>$${{1}\over{2}} - 2 - 1$$
=> $${{-5}\over{2}} = {{c}\over{a}}$$
On checking equation (iii):=> $${{1}\over{2}} × 1 × (-2)$$
=>$${{-2}\over{2}}$$
=> $$-1 = {{-d}\over{a}}$$

Hence it is verified the numbers given alongside of the cubic polynomials are their zeroes.

(ii) The general form of cubic equation is $$ax^3 + bx^2 + cx + d$$.

Here, $$a = 1,b = -4,c = 5,d = -2$$

On substituting the value of zeroes in the given equation, p(x)= $$x^3 - 4x^2 + 5x - 2$$

p$$(2)$$= $$2^3 - 4(2)^2 + 5(2) - 2$$
=>$$8 - 16 + 10 - 2 = 0$$
p$$(1)$$= $$1^3 - 4(1)^2 + 5(1) - 2$$
=>$$1 - 4 + 5 - 2 = 0$$

Hence, these zeroes satisfies the the given equation.

Now, we'll check whether the zeroes satisfies the following equations:

Let the three zeroes be p,q and r
$$p + q + r$$ = $${{-b}\over{a}}$$............(i)
$$pq + qr + rp$$ = $${{c}\over{a}}$$..........(ii)
$$pqr$$ = $${{-d}\over{a}}$$.................(iii)

On checking equation (i):=> $$2 + 1 + 1$$
=>$$4$$
=> $${{-(-4)}\over{1}} = {{-b}\over{a}}$$

On checking equation (ii):=> $$2 × 1 + 1 × 1 + 1 × 2$$

=>$$2 + 1 + 2 = 5$$
=> $${{5}\over{1}} = {{c}\over{a}}$$
On checking equation (iii):=> $$2 × 1 × 1$$
=>$$2$$
=> $${{-(-2)}\over{1}}= {{-d}\over{a}}$$

Hence it is verified the numbers given alongside of the cubic polynomials are their zeroes.

Q2 ) Find a cubic polynomial with the sum, the sum of the product of its zeroes taken two at a time and the product of its zeroes are $$2, -7,-14$$ respectively.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

The general form of cubic equation is $$ax^3 + bx^2 + cx + d$$

Let the three zeroes be p,q and r
Here:
$$p + q + r$$ = $${{-b}\over{a}} = 2$$ ............(i)
$$pq + qr + rp$$ = $${{c}\over{a}} = -7$$..........(ii)
$$pqr$$ = $${{-d}\over{a}} = -14$$.................(iii)

On equating (i), (ii) and (iii) => $$a = 1,b = -2,c = -7,d = 14$$

Putting the values of a,b,c and d in general form

=> $$x^3 - 2x^2 - 7x + 14$$

Q3 ) If the zeroes of the polynomial $$x^3 - 3x^2 + x + 1$$ are $$a-b , a , a+b$$ find $$a$$ and $$b$$.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Let the three zeroes be p,q and r

We know:

$$p + q + r$$ = $${{-b}\over{a}} =$$ ............(i)
$$pq + qr + rp$$ = $${{c}\over{a}}$$..........(ii)
$$pqr$$ = $${{-d}\over{a}}$$.................(iii)

On checking equation (i):=> $$a - b + a + a + b = 3a$$

=>$$3a = {{-b}\over{a}} = {{-(-3)}\over{1}}$$
=> $$3a = 3$$
=> $$a = 1$$

On checking equation (ii):=> $$(a - b) × a + a × (a + b) + (a + b) × (a - b)$$

=>$$a^2 - ab + a^2 + ab + a^2 - b^2 = {{c}\over{a}} = {{1}\over{1}} = 1$$
=> $$3a^2 - b^2 = 1$$
=> $$3(1)^2 - b^2 = 1 (a = 1)$$
=>$$3 - b^2 = 1$$
=> $$b^2 = 4$$
=> $$b = +2, -2$$

Therefore, $$a = 1, b = +2, -2$$

Q4 ) If the two zeroes of the polynomial $$x^4- 6x^3 - 26x^2 -138x - 35$$ are $$2 + \sqrt{3}$$ and $$2 - \sqrt{3}$$ find other zeroes.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Two zeroes are$$2 + \sqrt{3}$$ and $$2 - \sqrt{3}$$.

$$=> (x -(2 + \sqrt{3})) (x - (2 - \sqrt{3}))$$
$$= (x - 2 - \sqrt{3}) (x - 2 + \sqrt{3})$$
$$=> (x - 2)^2 - ( \sqrt{3})^2$$
$$= x^2 - 4x + 1$$

$$x^2 - 4x + 1$$ the factors of $$x^4- 6x^3 - 26x^2 -138x - 35$$.

So, we'll divide the given polynomial by $$x^2 - 4x + 1$$ Solving it using long division:

$$\begin{array}{rrrr|ll} x^4 -6x^3 & -26x^2 & +138x & -35 & x^2 -4x +1 \\ -x^4 + 4x^3 & -x^2 & & & x^2 -2x - 35 \\ \hline -2x^3 & -27x^2 & +138x & -35\\ \phantom{-}+2x^3 & -8x^2 & + 2x & & & \\ \hline & -35x^2 & +140x & -35 \\ & +35x^2 & -140x & +35 \\ \hline & & & 0\\ \hline \end{array}$$

$$x^4- 6x^3 - 26x^2 -138x - 35$$
=> $$( x^2 -4x +1) (x^2 -2x - 35)$$
=> $$( x^2 -4x +1) (x^2 -7x + 5x - 35)$$
=> $$(x^2 -4x +1) (x(x - 7) + 5(x - 7))$$
=> $$(x^2 -4x +1) (x + 5)(x - 7)$$

Thus, $$x + 5 = 0$$ and $$x - 7 = 0$$ are the solutions.

So, the other two zeroes are $$x = -5 , 7$$.

Q5 ) If the polynomial $$x^4- 6x^3 + 16x^2 -25x + 10$$ is divided by another polynomial $$x^2 -2x + k$$ the remainder comes out to be $$x +a$$ find $$k$$ and $$a$$.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Solving it using long division:

$$\begin{array}{rrrr|ll} x^4 -6x^3 & +16x^2 & -25x & +10 & x^2 -2x + k\\ -x^4 + 2x^3 & -kx^2 & & & x^2 -4x + (8-k) \\ \hline -4x^3 & +(16-k)x^2 & -25x & +10 \\ \phantom{-}+4x^3 & -8x^2 & + 4kx & & & \\ \hline & (8-k)x^2 & +(4k-25)x & +10 \\ & -(8-k)x^2 & +2(8-k)x & -(8-k)k \\ \hline & (2k-9)x & -(8-k)k & +10\\ \hline \end{array}$$

Here remainder = $$(2k-9)x -(8-k)k +10$$...........(i)

Given remainder = $$x + a$$..........................(ii)

On comparing the coefficients x in equation (i) and (ii):

=>$$2k - 9 = 1$$
=>$$2k = 10$$
=>$$k = 5$$

On comparing the a in equation (i) and (ii):

=>$$-(8-k)k + 10 = a$$
=>$$-(8-5)5 + 10 = a$$
=>$$a = -15 + 10 = -5$$

##### FAQs Related to NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
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