NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

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) \(3x^2 - x - 4\)

Answer :

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) \(4,1\)

Answer :

(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)
\(\Rightarrow \) \(x^2 - 4x + 1\)

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) \(p(x) = x^4 - 5x + 6 , g(x)= 2 - x^2\)

Answer :

(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) \(x^2 - 3x + 1 , x^5 - 4x^3 + x^2 + 3x + 1\)

Answer :

(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 \(-\sqrt{{5}\over{3}}\).

Answer :

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).

Answer :

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

Answer :

(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) = \(5\)

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) \(x^3 - 4x^2 + 5x - 2 ; 2,1,1\)

Answer :

(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.

Answer :

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 \(b\).

Answer :

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.

Answer :

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 \(a\).

Answer :

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):

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