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) = 2y^{3} – y^{2} – 13y – 6.

Solution:

Given polynomial

p(x) = 2y^{3} – y^{2} – 13y – 6

Question 2.

If ‘2’ is a zero of the polynomial P(x) = 4x^{2} – 3x + 5a then find the value of a.

Solution:

Given polynomial P(x) = 4x^{2} – 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 (x^{2} + y^{2}).

Solution:

(x + y) (x – y) (x^{2} + y^{2}) = (x^{2} – y^{2})

(x^{2} + y^{2}) = (x^{2})^{2} – (y^{2})^{2} = x^{4} – y^{4}.

Question 4.

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

Solution:

Question 5.

Find the remainder when x^{3} + 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 x^{3} – x ?

Solution:

Volume of cuboid = x^{3} – x

By splitting x^{3} – x as factors

x^{3} – x = x(x^{2} – 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)

= (x^{2} – 1) (x – 2) = x^{3} – 2x^{2} – x + 2.

Question 8.

The polynomial p(x) = x^{4} – 2x^{3} + 3x^{2} – 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) = x^{4} – 2x^{3} + 3x^{2} – 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) = x^{4} – 2x^{3} + 3x^{2} – 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.

Read the following table and answer the following questions given below.

i) The point belongs to Q_{3}

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) Q_{4}

Question 11.

i) Verify that

p^{3} + q^{3} + r^{3} – 3pqr = (p + q + r)

(p^{2} + q^{2} + r^{2} – pq – qr – rp)

ii) If a + b + c = 0, then prove that a^{3} + b^{3} + c^{3} = 3abc.

Solution:

i) RHS = (p + q + r)

(p^{2} + q^{2} + r^{2} – pq – qr – rp)

= p(p^{2} + q^{2} + r^{2} – pq – qr – rp) + q (p^{2} + q^{2} + r^{2} – pq – qr – rp) + r (p^{2} + q^{2} + r^{2} – pq – qr – rp)

= p^{3} + pq^{2} + pr^{2} – p^{2}q – pqr – p^{2}r + p^{2}q + q^{3} + qr^{2} – pq^{2} – q^{2}r – pqr + p^{2}r + q^{2}r + r^{3} – pqr – qr^{2} – r^{2}p

= p^{3} + q^{3</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}

a^{3} + b^{3} + 3ab (a + b) = -c^{3}

a3 + b3 + 3ab (-c) =-c3 {From(l)}

a^{3} + b^{3} – 3abc = -c^{3}

a^{3} + b^{3} + c^{3} – 3abc

Question 12.

If both (x – 2) and (x – \(\frac{1}{2}\)) are factors of px^{2} + 5x + q, show that p = q.

Solution:

Let f(x) = px^{2} + 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 ax^{2} + bx + c and bx^{2} + 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) = ax^{2} + bx + c

⇒ f(- 1) = a(- 1)^{2} + b(- 1) + c = 0

⇒ a – b + c = 0 …………….. (1)

(x + 1) is the factor of f(x) = bx^{2} +, 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