## 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].