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## Chapter 2: Polynomials NCERT Solutions

#### TEXTBOOK’S EXERCISE – 2.1

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

Sol. (i) From the graph, it is clear that it does not intersect the x-axis. Hence, it has no zero.

(ii) From the graph, it is clear that it intersects the x-axis at one point. Hence, it has one zero.

(iii) From the graph, it is clear that it intersects the x-axis at three points. Hence, it has three zeroes.

(iv) From the graph, it is clear that it intersects the x-axis at two points. Hence, it has two zeroes.

(v) From the graph, it is clear that it intersects the x-axis at four points. Hence, it has four zeroes.

(vi) From the graph, it is clear that it intersects the x-axis at three points. Hence, it has three zeroes.

TEXTBOOK’S EXERCISE – 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x2 – 2x – 8

(ii) 4s2 – 4s + 1

(iii) 6x2 – 3 – 7x

(iv) 4u2 + 8u

(v) t2 – 15

(vi) 3x2 – x – 4

X2 – 2x – 8 = x2–4x + 2x – 8

=X( x–4) + 2(x–4)

=(x–4) (x–2)

The value of x2–2x – 8 is zero, if

(x – 4) = 0 or (x + 2) = 0

Then, x = 4 and x = – 2.

So, zeroes of x2 – 2x – 8 are –2 and 4

Thus, sum of zeroes = (–2) + (4) = 2

Product of zeroes = (–2) (4) = – 8

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

4s2 – 4s+ 1= 4s2– 2s-2s+1

=2s(2s-1) – (2s-1)

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

This value of 4s2 – 4s+ 1 is zero, if (2s-1) = 0 or (2s-1)=0

Product of zeroes

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

6x2 – 3- 7x = 6x2 – 7x -3 = 6x2 – 9x +2x – 3

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

The value of 6x2-3 -7x is zero, if

(2x-3) =0 or (3x+1)=0

Then x = 3/2 or x = −1/3

So, zeroes of 6x2 – 3- 7x are 3/2 and 1/3 . Thus, sum of zeroes

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

Quadratic polynomial is 4u2 + 8u = 4u (u+2)

The value of 4u2 +8u is zero, if

if 4u = 0 or u + 2 = 0.

Then u = 0 or u = – 2.

So, zeroes of 4u2 + 8u are 0 and –2.

Thus, sum of zeroes = 0 + (–2) = – 2

Product of zeroes = (0) (–2) = 0

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

The value of t2 – 15 is zero if

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

3x2 – x – 4 = 3x2 + 3x – 4x – 4

= 3x (x + 1) 4(x + 1)

= (x + 1) (3x – 4)

The value of 3x2 – x – 4 is zero, if

(x + 1) = 0 or 3x – 4 = 0

Then x = – 1 or x = 43

So, zeroes of 3x2 – x – 4 are –1 and 43.

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