# AP 9th Class Maths Important Questions Chapter 2 Polynomials and Factorisation

These AP 9th Class Maths Important Questions 2nd Lesson Polynomials and Factorisation will help students prepare well for the exams.

## AP State Syllabus 9th Class Maths 2nd Lesson Important Questions and Answers Polynomials and Factorisation

Question 1.
Find the value of p ($$\frac{2}{3}$$) for p(y) = 2y3 – y2 – 13y – 6.
Solution:
Given polynomial
p(x) = 2y3 – y2 – 13y – 6

Question 2.
If ‘2’ is a zero of the polynomial P(x) = 4x2 – 3x + 5a then find the value of a.
Solution:
Given polynomial P(x) = 4x2 – 3x + 5a
the zero of the polynomial is ‘2’, we know that p(2) = 0
P(x) = 0.
4(2)2 – 3(2) + 5a = 0
16 – 6 + 5a = 0
5a = -10
a = $$\frac{-10}{5}$$ = -2
a = -2.

Question 3.
Applying a suitable identity find the product of (x – y), (x + y) and (x2 + y2).
Solution:
(x + y) (x – y) (x2 + y2) = (x2 – y2)
(x2 + y2) = (x2)2 – (y2)2 = x4 – y4.

Question 4.
Fill In the blanks given in the table with suitable answers.

Solution:

Question 5.
Find the remainder when x3 + 1 is divided by (x + 1) by using division method.
Solution:

Remainder = 0

Question 6.
What are the possible polynomial expressions for dimensions of the cuboid whose volume is x3 – x ?
Solution:
Volume of cuboid = x3 – x
By splitting x3 – x as factors
x3 – x = x(x2 – 1) = x(x + 1) (x – 1)
Possible dimensions of cuboid are x, (x + 1) and (x – 1)

Question 7.
Write the polynomial in ‘x’ whose zeroes are 1, 2 and – 1.
Solution:
The given zeroes of polynomials in ‘x’ are : 1, 2 and – 1.
The factors of the polynomial are : x – 1, x – 2 and x + 1
The required polynomial is : (x – 1) (x – 2) (x + 1)
= (x2 – 1) (x – 2) = x3 – 2x2 – x + 2.

Question 8.
The polynomial p(x) = x4 – 2x3 + 3x2 – ax + 3a – 7 when divided by x + 1 leave the remainder 19. Find the value of “a”. Also find the remainder when p(x) is divided by x + 2.
Solution:
Given polynomial = p(x) = x4 – 2x3 + 3x2 – ax + 3a – 7
When p(x) is divided by (x + 1) leaves the remainder is 19.
∴ p(-1) = 19
p(-1) = (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + 3a – 7 = 19
⇒ 1 + 2 + 3 + a + 3a – 7 = 19
⇒ 4a – 1 = 19
⇒ 4a = 20
⇒ a = 20 ÷ 4 = 5
p(x) = x4 – 2x3 + 3x2 – 5x + 8
p(x) is divided by (x + 2), then p(-2). P(-2) = (-2)4 – 2(-2)3 + 3(-2)2 – 5(-2) + 8
= 16 + 16 + 12 + 10 + 8 = 62
Required remainder = 62.

Question 9.
In ABC, E and F are mid points of sides AB and AC respectively then prove that i) EF // BC and ii) EF = $$\frac{1}{2}$$ BC
Solution:
Given : B and F are mid points of AB and AC.

R.T.P. : i) EF // BC, ii) EF = $$\frac{1}{2}$$ BC
Construction: GC // AB, extend EF upto G.
Proof:
ΔAEF ΔCGF
∠AFE = ∠CFG (Vertically opposite angles)
AF = FC
∠EAF = ∠GCF (Alternate angles)
∴ ΔAEF ≅ ΔCGF
∴ CG = BE and CG // BF (Constrution)
∴ EBCG is a parallelogram.

Question 10.

i) The point belongs to Q3
if) The abscissa of the point C
iii) The point lie on X – axis
iv) The coordinates of origin
v) The point satisfy x > 0, y < 0
vi) The point satisfy x – y = 1
vii) The position of point B
viii) The Quadrant contain (3, – 2)
Solution:
i) D
ii) 3
iii) F, H
iv) 0,0
v) C (3, – 2)
vi) A (2, 1)
vii) Positive Y- axis
viii) Q4

Question 11.
i) Verify that
p3 + q3 + r3 – 3pqr = (p + q + r)
(p2 + q2 + r2 – pq – qr – rp)

ii) If a + b + c = 0, then prove that a3 + b3 + c3 = 3abc.
Solution:
i) RHS = (p + q + r)
(p2 + q2 + r2 – pq – qr – rp)
= p(p2 + q2 + r2 – pq – qr – rp) + q (p2 + q2 + r2 – pq – qr – rp) + r (p2 + q2 + r2 – pq – qr – rp)
= p3 + pq2 + pr2 – p2q – pqr – p2r + p2q + q3 + qr2 – pq2 – q2r – pqr + p2r + q2r + r3 – pqr – qr2 – r2p
= p3 + q3</sup + r3 – pqr – pqr – pqr
= p3 + q3 + r3 – 3pqr

ii) Given a + b + c = 0 ⇒ a + b = -c ……………..(1)

Cubing on both sides
(a + b)3 = (-c)3
a3 + b3 + 3ab (a + b) = -c3
a3 + b3 + 3ab (-c) =-c3 {From(l)}
a3 + b3 – 3abc = -c3
a3 + b3 + c3 – 3abc

Question 12.
If both (x – 2) and (x – $$\frac{1}{2}$$) are factors of px2 + 5x + q, show that p = q.
Solution:
Let f(x) = px2 + 5x + q, if (x – 2) is the factor of f(x)
⇒ f(2) = p(2)2 + 5 × 2 + q = 0
⇒ 4p + 10 + q = 0 ……………. (1)
If (x – $$\frac{1}{2}$$ ) is the factor of f(x)
f($$\frac{1}{2}$$) = P($$\frac{1}{2}$$)2 + 5 × $$\frac{1}{2}$$ + q = 0
⇒ $$\frac{\mathrm{p}}{4}+\frac{5}{2}$$ + q = 0
⇒ p + 10 + 4q = 0 ……………. (2)
from (1) and (2)
⇒ 4p + 10 + q = p + 10 + 4q
⇒ 4p + q = p + 4q
⇒ 4p – p = 4q – q
⇒ 3p = 3q
⇒ p = q

Question 13.
If ax2 + bx + c and bx2 + ax +-c have a common factor (x + 1) then show that a = b and c =’0.
Solution:
Given that (x + 1) is the factor of f(x) = ax2 + bx + c
⇒ f(- 1) = a(- 1)2 + b(- 1) + c = 0
⇒ a – b + c = 0 …………….. (1)
(x + 1) is the factor of f(x) = bx2 +, ax + c
f(- 1) = b(- 1)2 + a(- 1) + c = 0
⇒ b – a + c = 0 …………… (2)
from (1) and (2)
a – b + c = b – a + c
⇒ a – b = b – a
⇒ a + a = b + b
⇒ 2a = 2b
∴ a = b
if a = b, from equation (1)
c = 0