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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.2

Question 1.
Find the value of the polynomial f(y) = 6y – 3y2 + 3 at
(i) y = 1
(ii) y = -1
(iii) y = 0
Solution:
(i) When y = 1
f(y) = 6y – 3y2 + 3
f(1) = 6(1) – 3(1)2 + 3
= 6 – 3 + 3 = 6

(ii) When y = – 1
f(y) = 6y – 3y2 + 3
f(-1) = 6(-1) – 3(-1)2 + 3
= – 6 – 3 + 3
= – 6

(iii) When y = 0
f(y) = 6y – 3y2 + 3
f(0) = 6(0) – 3(0)2 + 3
= 0 – 0 + 3
= 3 Question 2.
If p(x) = x2 – 2√2x + 1, find p(2√2).
Solution:
p(x) = x2 – 2√2x + 1
p(2√2) = (2√2)2 – 2√2 (2√2) + 1
= 8 – 8 + 1
= 0 + 1
= 1

Question 3.
Find the zeros of the polynomial in each of the following.
(i) P(x) = x – 3
Solution:
p( 3) = 3 – 3
= 0
p(3) is the zero of p(x)

(ii) p(x) = 2x + 5
Solution: = -5 + 5
= 2(0)
= 0
Hence –$$\frac{5}{2}$$ is the zero of p(x). (iii) q(y) = 2y – 3
Solution: = 2 × 0
= 0
Hence $$\frac{3}{2}$$ is the zero of q(y).

(iv) f(z) = 8z
Solution:
f(0) = 8 × 0
= 0
Hence 0 is the zero of f(z)

(v) p(x) = ax when a ≠ 0
Solution:
p(0) = a(0)
= 0
Hence, 0 is the zero of p(x)

(vi) h(x) = ax + b, a ≠ 0, a, b∈R
Solution: Hence –$$\frac{b}{a}$$ is the zero of h(x). Question 4.
Find the roots of the polynomial equations.
(i) 5x – 6 = 0
Solution:
5x = 6
x = $$\frac{6}{5}$$
$$\frac{6}{5}$$ is the root of the polynomial.

(ii) x + 3 = 0
Solution:
x = -3
-3 is the root of the polynomial. (iii) 10x + 9 = 0
Solution:
10x = -9
x = –$$\frac{9}{10}$$
–$$\frac{9}{10}$$ is the root of the polynomial.

(iv) 9x – 4 = 0
Solution:
9x = 4
x = $$\frac{4}{9}$$
$$\frac{4}{9}$$ is the root of the polynomial. Question 5.
Verify whether the following are zeros of the polynomial, indicated against them,or not.
(i) p(x) = 2x – 1, x = $$\frac{1}{2}$$
Solution:
p ($$\frac{1}{2}$$) = 2($$\frac{1}{2}$$) – 1
= 1 – 1
= 0
∴ $$\frac{1}{2}$$ is the zero of the polynomial.

(ii) p(x) = x3 – 1, x = 1
Solution:
p(1) = 13 – 1
= 1 – 1
= 0
∴ 1 is the zero of the polynomial (iii) p(x) = ax + b, x = $$\frac{-b}{a}$$
Solution:
p($$\frac{-b}{a}$$) = a($$\frac{-b}{a}$$) + b
= -b + b
= 0
∴ $$\frac{-b}{a}$$ is the zero of the polynomial. a

(iv) p(x) = (x + 3) (x – 4); x = -3, x = 4
Solution:
P(-3) = (-3 + 3) (-3 – 4)
= (0) (-7)
= 0
P( 4) = (4 + 3) (4 – 4)
= (7) (0)
= 0
∴ -3 and 4 are the zeros of the polynomial. Question 6.
Find the number of zeros of the following polynomials represented by their graphs. Solution:
(i) Number of zeros = 2 (The curve is intersecting the x-axis at 2 points)
(ii) Number of zeros = 3 (The curve is intersecting the x-axis at 3 points)
(iii) Number of zeros = 0 (The curve is not intersecting the x-axis)
(iv) Number of zeros = 1 (The curve is intersecting at the origin)
(v) Number of zeros = 1 (The curve is intersecting the x-axis at one point)