Telangana SCERT Solution Class IX (9) Math Chapter 2 Polynomials and Factorisation Exercise 2.2
(1) Given, p(x) = 4x2 – 5x + 3
(i) Now, x = 0
∴ p(o) = 4×02 – 5 x 0 + 3
= 3
(ii) Now, x = -1
∴p(1) = 4x(-1)2 – 5(-1) + 3
= 4×1 + 5 + 3
= 12
(iii) No, x = 2
∴ p(2) = 4×22 – 5×2 + 3
= 16 – 10 + 3
= 9
(2) (i) p(x) = x2 – x + 1
∴ p(0) = 02 – 0 + 1
= 1
P(1) = 12 -1 + 1
= 1
P(2) = 22 – 2 + 1
= 3
(ii) p(y) = 2 + y + 2y2 – y3
∴ p(0) = 2 + 0 + 2 x 02 – 03
= 2
P(1) = 2 + 1 + 2 x 12 – 13
= 3 + 2 – 1
= 4
P(2) = 2 + 2 + 2 x 22 – 23
= 2 + 2 + 8 – 8
= 4
(iii) p(z) = z3
∴ p(0) = 03
= 0
P(1) = 13
= 1
P(2) = 23
= 8.
(IV) P(t) = (t – 1) (t + 1)
∴ p(0) = (0 -1) (0+1)
= -1 x +1
= -1
P(1) = (1 – 1) (1 + 1)
= 0 x 2
= 0
P(2) = (2-1) (2+1)
= 1 x 3
= 3
(v) p(x) = x2 – 3x + 2
∴ p(0) = 02 – 3×0 + 2
= 2
P(1) = 12 – 3×1 + 2
= 0
P(2) = 22 – 3×2 + 2
= 4 – 6 + 2
= 0
(iii) p(x) = x2 – 1; x = ±1
P(±1) = (±1)2 – 1
= 1 – 1
= 0
∴ x = ±1 is the zero of the above given polynomial.
(iv) p(x) = (x – 1) (x + 2) ; x = -1, -2
P(-1) = (-1 -1) (-1 + 2)
= -2 x 1
= -2 ≠0
P(-2) = (-2 – 1) (-2 + 2)
= -3 x 0
= 0
∴ x = -1 is not the zero of the given polynomial but x = -2 is.
(iv) p(y) = y2 ; y = 0
P(0) = 02
= 0
∴ y = 0 is the zero of the above given polynomial.
(4) (i) f(x) = x+2
∴ f(x) = 0
Or, x + 2 = 0
Or, x = -2
∴ x = -2 is the zero of the above given polynomial.
(ii) to given,
f(x) = x – 2
∴ f(x) = 0
Or, x – 2 = 0
Or, x = 2
∴ x = 2 is the zero of the given polynomial.
(iii) given f(x) = 2x + 3
∴ f(x) = 0
Or, 2x + 3 = 0
Or, 2x = -3
Or, x = -3/2
∴ x = -3/2 is the zero of the given polynomial.
(iv) f(x) = 2x – 3
∴ f(x) = 0
Or, 2x – 3 = 0
Or, x = 3/2
∴ x = 3/2 is the zero of the above given polynomial.
(v) given, f(x) = x2
Or, ∴ f(x) = 0
Or, x2 = 0
Or, x = 0
∴ x = 0 is the zero of the above given polynomial.
(vi) f(x) = px, p ≠ 0
f(x) = 0
Or, px = 0
∴ x = 0 [∵ p ≠ 0]
∴ x = 0 is the zero of the given polynomial.
(vii) f(x) = px + q, p ≠ 0, p and q are the real number.
f(x) = 0
or, px + q = 0
Or, px = -q
Or, x = -q/p
∴ x = -q/p is the zero of the above given polynomial.
(5) Given, p(x) = 2x2 – 3x + 7a
Or, p(2) = 2x2 -3x + 7a = 0
∴2 x 22 – 3 x 2 + 7a = 0
Or, 2 x 4 – 6 + 7a = 0
Or, 2 + 7a = 0
Or, 7a = -2
Or, a = -2/7
(6) Given, f(x) = 2x3 – 3x2 + ax + b
Or, f(0) = 2 x 03 – 3 x 02 + 0 x 0 + b
= b
Given, f(0) = 0
∴ b = 0 ….(i)
Now,
f(1) = 2 x 13 – 3 x12 + a x 1 + b
= 2 – 3 + a + b
= a + b -3
Given, f(1) = 0
Or, a + b – 3 = 0
Or, a + b – 3 = 0 [putting the value of b from eqn(i)]
Or, a = 3
∴ a = 3 and b = 0.
The last answer is a=1, and b=0. A is not equal to 3. Please correct it..
Number Please
everything is fine and awsome….but its better to make the written answers is the pics clear before posting it as ur posting on the internet where many students use this info its important to make it clear and neat.Its confusing alot.
Hi Mam Thank You for giving us time. Here only little of topics are in handwritten format. We have place it as because, Students will understand where the crossing have took place.