## NCERT Solutions Class 10 Maths Chapter 2 Polynomials

### NCERT Solutions

#### Exercise 2.1

**Question.1. The graphs of y = p(x) are given in fig. below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.****(i) ****Solution:** None, because graph is not intersecting the at X-axis at any point.

**(ii) ****Solution:** 1, because graph intersects X-axis at only one point.

**(iii) ****Solution:** 3, because graph intersects X-axis at three points.

**(iv) ****Solution:** 2, because graph intersects X-axis at two points.

**(v) ****Solution:** 4, because graph intersects X-axis at four points.

**(vi) ****Solution:** 3 zeroes, because graph intersects X-axis at three points.

##### Exercise 2.2

**Question.1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients :****(i) x^{2}-2x-8 ****Solution:** Let p(x) = x^{2}-2x-8

= x^{2}-4x+2x-8

= x(x-4)+2(x-4)

= (x – 4)(x + 2)

For finding the zeroes

Put, p(x) = 0

= (x – 4)(x + 2) = 0

x = 4, – 2

**Verification:** Now, sum of zeroes = 4 + (– 2)

= {\frac{2}{1}}

= {\frac{-(Coefficient of x)}{(Coefficient of x^{2})}}

Product of zeroes = 4 \times (– 2)

= {\frac{-8}{1}}

= {\frac{Constant term}{Coefficient of x^{2}}}

Hence, relationship between the zeroes and the coefficients is true.

**(ii) 4s^{2}-4s+1 ****Solution:** Let p(s) = 4s^{2}-4s+1

= 4s^{2}-2s-2s+1

= 2s(2s-1)-1(2s-1)

= (2s – 1) (2s – 1)

= (2s-1)^{2}

For finding the zeroes, put p(s) = 0.

2s – 1 = 0

s = \frac{1}{2}

Hence, zeroes of p(s) are \frac{1}{2} and 12 \frac{1}{2} .

**Verification:** Now, sum of zeroes = \frac{1}{2} + \frac{1}{2}

= 1 = \frac{4}{4} = {\frac{-(Coefficient of s)}{(Coefficient of s^{2})}}

Product of zeroes = \frac{1}{2} \times \frac{1}{2}

= \frac{1}{4}

= {\frac{Constant term}{Coefficient of s^{2}}}

Hence, relationship between the zeroes and the coefficients is true.

**(iii) 6x^{2}-3-7x ****Solution:** Let p(x) = 6x^{2}-3-7x = 6x^{2}-7x-3

= 6x^{2}-9x+2x-3

= 3x(2x-3)+1(2x-3)

= (2x – 3) (3x + 1)

For finding the zeroes put p(x) = 0

(2x – 3) (3x + 1) = 0

x = {\frac{3}{2}}, {\frac{-1}{3}}

Hence, zeroes of the polynomial are {\frac{3}{2}} and {\frac{-1}{3}}.

**Verification:** Now sum of zeroes = {\frac{3}{2}} + {\frac{-1}{3}}

= {\frac{9-2}{6}}

= {\frac{7}{6}}

= {\frac{-(Coefficient of x)}{(Coefficient of x^{2})}}

Product of zeroes = {\frac{3}{2}} \times {\frac{(-1)}{3}}

= {\frac{-3}{6}}

= {\frac{Constant term}{Coefficient of x^{2}}}

**(iv) 4u^{2}+8u ****Solution:** Let p(u) = 4u^{2}+8u

= 4u (u + 2)

For finding the zeroes put p(u) = 0,

4u (u + 2) = 0

u = 0, – 2

Hence, zeroes are 0 and – 2.

**Verification:** Now sum of zeroes = 0 + (– 2)

= {\frac{-2}{1}} \times {\frac{4}{4}}

= {\frac{-8}{4}}

= {\frac{-(Coefficient of u)}{(Coefficient of u^{2})}}

Product of zeroes = 0 \times (– 2)

= {\frac{0}{1}} \times {\frac{4}{4}}

= {\frac{0}{4}}

= {\frac{Constant term}{Coefficient of u^{2}}}

Hence, relationship between the zeroes and the coefficients is true.

**(v) t^{2}-15 ****Solution:** Let p(t) = t^{2}-15 .

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

= (t-\sqrt{15})(t+\sqrt{15}) [∵ a^{2}-b^{2} = (a + b) (a – b)]

For finding the zeroes put p(t) = 0,

(t-\sqrt{15})(t+\sqrt{15}) = 0

t = -\sqrt{15}) , \sqrt{15}

**Verification:** Sum of the zeroes = -\sqrt{15}) + \sqrt{15}

= {\frac{0}{1}}

= = {\frac{-(Coefficient of t)}{(Coefficient of t^{2})}}

Product of zeroes = -\sqrt{15}) \times \sqrt{15}

= {-\frac{15}{1}}

= {\frac{Constant term}{Coefficient of t^{2}}}

Hence, relationship between the zeroes and the coefficients is true.

**(vi) 3x^{2}-x-4 ****Solution:** Let p(x) = 3x^{2}-x-4

= 3x^{2}-4x+3x-4

= x(3x – 4) + 1 (3x – 4)

= (x + 1) (3x – 4)

For finding the zeroes put p(x) = 0,

(x + 1) (3x – 4) = 0

x = -1, {\frac{4}{3}}

**Verification:** Sum of zeroes = -1 + {\frac{4}{3}}

= {\frac{1}{3}}

= = {\frac{-(Coefficient of x)}{(Coefficient of x^{2})}}

Product of zeroes = -1 \times {\frac{4}{3}}

= {\frac{-4}{3}}

= {\frac{Constant term}{Coefficient of x^{2}}}

Hence, relationship between the zeroes and the coefficients is true.

**Question.2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.****(i) {\frac{1}{4}}, – 1** **Solution:** Let quadratic polynomial is p(x) whose zeroes are α and β.

Given that : α + β = 14 and αβ = – 1

So, required polynomial = k[x^{2} – (α + β)x + αβ]

= k\left[x^{2}-\frac{1}{4}x-1\right]

If k = 4, then 4x^{2}-x-4 .

**(ii) \sqrt{2}, {\frac{1}{3}}****Solution:** Let quadratic polynomial is ax^{2}+bx+c whose zeroes are α and β.

Given that: α + β = \sqrt{2} and αβ = 13

So, required polynomial = k[x^{2} – (α + β)x + αβ]

= k\left[x^{2}-\sqrt{2}x+{\frac{1}{3}}\right]

If k = 3, then 3x^{2}-3 \sqrt{2}x+1 .

**(iii) 0, \sqrt{5} ****Solution:** Let quadratic polynomial is ax^{2}+bx+c whose zeroes are α and β.

Given that: α + β = 0 and αβ = \sqrt{5}

Hence, required polynomial = k[x^{2} – (α + β)x + αβ]

= k\left[x^{2}-0.x+ \sqrt{5} \right]

= k\left[x^{2}+ \sqrt{5} \right]

If k = 1, then x^{2}+ \sqrt{5} .

**(iv) 1, 1****Solution:** Let quadratic polynomial is ax^{2}+bx+c whose zeroes are α and β.

Given that : α + β = 1 and αβ = 1

Hence, required polynomial = k[x^{2} – (α + β)x + αβ]

= k\left[x^{2}-x+1 \right]

If k = 1, then x^{2}-x+1 .

**(v) {\frac{-1}{4}}, {\frac{1}{4}} ****Solution:** Let quadratic polynomial is ax^{2}+bx+c whose zeroes are α and β.

Given that: α + β = -{\frac{1}{3}} and αβ = {\frac{1}{3}}

So, required polynomial = k[x^{2} – (α + β)x + αβ]

= k\left[x^{2}+\frac{1}{4}x+\frac{1}{4} \right]

If k = 4, then 4x^{2}+x+1 .

**(vi) 4, 1****Solution:** Let quadratic polynomial is ax^{2}+bx+c whose zeroes are α and β.

Given that: α + β = 4 and αβ = 1

Hence, required polynomial = k[x^{2} – (α + β)x + αβ]

= k\left[x^{2}-4x+1 \right].