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

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

## AP Board Class 9 Maths 2nd Lesson Polynomials Important Questions

### 9th Class Maths Polynomials 2 Marks Important Questions

Question 1.

If – 2 is a zero of the polynomial

p(x) = 2x^{2} – x + a, find the value of ‘a’.

Solution:

p(x) = 2x^{2} – x + a. As the zero the polynomial is – 2, we know that p(-2) = 0.

2x^{2} – x + a = 0

2(-2)^{2} -(-2) + a = 0

8 + 2 + a = 0

10 + a = 0

⇒ a = – 10

Question 2.

State the Remainder Theorem.

Solution:

Let p(x) be any polynomial of degree greater than or equal to one and let ’a’ be any real number.

If p(x) is divided by the linear polynomial (x – a), then the remainder is P(a).

Question 3.

If 2(a^{2} + b^{2}) = (a + b)^{2}, then show that a = b.

Solution:

2(a^{2} + b^{2}) = (a + b)^{2}, (given)

⇒ 2a^{2} + 2b^{2} = a^{2} + b^{2} + 2ab

⇒ a^{2} + b^{2} – 2ab = 0

⇒ (a – b)^{2} = 0

⇒ a – b = 0

⇒ a = b

Question 4.

If a + b + c = 12 and ab + be + ca = 47, find a^{2} + b^{2} + c^{2}.

Solution:

We know that

(a + b+ c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

⇒ a^{2} + b^{2} + c^{2} = (a + b + c)^{2} – 2(ab + bc + ca)

= (12)^{2} – 2(47)

= 144 – 94 = 50

Question 5.

Find the remainder when x^{3} – ax^{2} + 6x – a is divided by x – a.

Solution:

Let p(x) = x^{3} – ax^{2} + 6x – a

⇒ x – a = 0

⇒ x = a

∴ Remainder = p(a)

= (a)^{3} – a(a)^{2} + 6(a) – a

= a^{3} – a^{3} + 6a – a

= 5a

Question 6.

Check whether 7 + 3x is a factor of 3x^{3} + 7x.

Solution:

7 + 3x will be a factor of 3x^{3} + 7x only if

7 + 3x divides 3x^{3} + 7x leaving no remainder.

Let p(x) = 3x^{3} + 7x

7 + 3x = 0

⇒ 3x = – 7

⇒ x = \(\frac{-7}{3}\)

∴ Remainder = P(\(\frac{-7}{3}\))

= 3(\(\frac{-7}{3}\))^{3} + 7(\(\frac{-7}{3}\))

= \(\frac{-343}{3}\) – \(\frac{49}{3}\) = \(\frac{-490}{3}\) ≠ 0

∴ 7 + 3x is not a factor of 3x^{3} + 7x.

Question 7.

Using suitable identity, find the value of \frac{87^3+13^3}{87^2-87 \times 13+13^2}\(\)

Solution:

\(\frac{87^3+13^3}{87^2-87 \times 13+13^2}\) = \(\) (∵ a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

= 87 + 13

= 100

Question 8.

Using suitable identity find (x + y + z)^{2} – (x – y – z)^{2}.

Solution:

(x + y + z)^{2} – (x – y – z)^{2}

= (x + y + z)^{2} – [x + (- y) + (- z)]^{2}

= (x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2xz) – [x^{2} + (-y)^{2} + (-z)^{2} + 2x(-y) + 2(-y) (- z) + 2(- z)x]

= (x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx) – (x^{2} + y^{2} + z^{2} – 2xy + 2yz – 2zx)

= 4xy + 4zx

= 4x(y + z)

Question 9.

Simplify: (2x + y)^{3} + (2x – y)^{3}

Solution:

(2x + y)^{3} + (2x – y)^{3}

= [(2x)^{3} + (y)^{3} + 3(2x)(y)(2x + y)] + [(2x)^{3} – (y)^{3} – 3(2x)(y) (2x – y)]

= (8x^{3} + y^{3} + 12x^{2}y + 6xy^{2}) + (8x^{3} – y^{3} – 12x^{2}y + 6xy^{2})

= 16x^{3} + 12xy^{2}

= 4x(4x^{2} + 3y^{2})

Question 10.

If x + a is a factor of ma – nx – 3x^{2}, then prove that a = \(\frac{m+n}{3}\).

Solution:

Let p(x) = ma – nx – 3x^{2}

If x + a is a factor of p(x), then by factor theorem,

p(-a) = 0 [∵ x + a = 0 ⇒ x = -a]

⇒ ma – n(- a) – 3(- a)^{2} = 0

⇒ ma + na – 3a^{2} = 0

⇒ m + n – 3a = 0

⇒ m + n = 3a

∴ m = \(\frac{m+n}{3}\)

Hence proved.

Question 11.

Find the sum of die remainders when the polynomial p(x) = x^{3} – 3x^{2} + 4x – 9 is divided by (x – 1) and (x + 2).

Solution:

p(x) = x^{3} – 3x^{2} + 4x – 9

By remainder theorem, remainder when p(x) is divided by (x – 1).

= p(1) [∵ x – 1 = 0 ⇒ x = 1]

= (1)^{3} – 3(1)^{2} + 4(1) – 9

= 1 – 3 + 4 – 9 = – 7

By remainder theorem, remaindgr when p(x) is divided by (x + 2).

= p(-2) [∵ x + 2 = 0 ⇒ x = – 2]

= (- 2)^{3} – 3(- 2)^{2} + 4(- 2) – 9

= – 8 – 12 – 8 – 9

= -37

∴ Sum of the two remainders

= (-7) + (-37)

= -44

Question 12.

Factorise :

i) 343a^{3} – 729b^{3}

Solution:

343a^{3} – 729b^{3}

= (7a)^{3} – (9b)^{3}

= (7a – 9b) [(7a)^{2} + (9b)^{2} + (7a)(9b)]

[∵ x^{3} – y^{3} = (x – y) (x^{2} + y^{2} + xy)]

= (7a – 9b) (49a^{2} + 81b^{2} + 63ab)

ii) 25x^{3}y – 121xy^{3}

Solution:

25x^{3}y – 121xy^{3}

= xy (25x^{2} – 121y^{2})

= xy[(5x)^{2} – (11y)^{2}]

= xy(5x – 11y) (5x + 11y)

iii) 8(x + y)^{3} + 27(x – y)^{3}

Solution:

8(x + y)^{3} + 27(x – y)^{3}

= [2(x + y)]^{3} + [3(x – y)]^{3}

= [2(x + y) + 3(x – y)] × [(2(x + y))^{2} + (3(x – y))^{2}] – [2(x + y)][3(x – y)]]

[∵ x^{3} + y^{3} = (x + y)(x^{2} + y^{2} – xy)]

= (5x – y) x (4x^{2} + 4y^{2} + 8xy + 9x^{2} + 9y^{2} – 18xy – 6x^{2} + 6y^{2}]

= (5x – y)(7x^{2} + 19y^{2} – 10xy)

Question 13.

If x + \(\frac{1}{x}\) = 4, find the value of x^{3} + \(\frac{1}{x^3}\).

Solution:

We know that,

(x + \(\frac{1}{x}\))^{3} = x^{3} + \(\frac{1}{x^3}\) + 3(x)(\(\frac{1}{x}\))(x + \(\frac{1}{x}\))

(x + \(\frac{1}{x}\))^{3} = x^{3} + \(\frac{1}{x^3}\) + 3(x + \(\frac{1}{x}\))

(4)^{3} = x^{3} + \(\frac{1}{x^3}\) + 3(4)

64 = x^{3} + \(\frac{1}{x^3}\) + 12

x^{3} + \(\frac{1}{x^3}\) = 64 – 12

⇒ x + \(\frac{1}{x^3}\) = 52

Question 14.

Find the zero of the polynomial.

i) p(x) = 3x + 5

ii) P(y) = 3y

Solution:

i) p(x) = 3x + 5

3x + 5 = 0

3x = -5

x = –\(\frac{5}{3}\)

ii) p(y) = 3y

3y = 0

y = \(\frac{0}{3}\)

y = 0

Question 15.

Find the sum and product of the co-efficient of x and y of y – 2x – 1 = 0.

Solution:

Given equation y – 2x – 1 = 0

Co-efficient of y = 1

Co-efficient of x = -2 .

Sum of the co-efficients of x and y

= 1 + (-2)

= -1

Product of the co-efficient of x and y

= 1 × (-2)

= -2

### 9th Class Maths Polynomials 4 Marks Important Questions

Question 1.

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 2.

Verify that

p^{3} + q^{3} + r^{3} – 3pqr = (p + q + r) (p^{2} + q^{2} + r^{2} – pq – qr – rp)

Solution:

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} + r^{3} – pqr – pqr – pq

= p^{3} + q^{3} + r^{3} – 3pqr = LHS

∴ LHS = RHS.

Hence proved.

Question 3.

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

⇒ 2 a = 2 b

∴ a = b

if a = b, from equation (1)

c = 0

Question 4.

Divide the polynomial 3y^{4} – 4y^{3} – 3y + 4 by (y – 1)

Solution:

∴ Quotient = 3y^{3} – y^{2} – y – 4 and Remainder = 0

Question 5.

Factorise : x^{3} – 3x^{2} – 10x + 24

Solution:

Let p(x) = x^{3} – 3x^{2} – 10x + 24

By trial, we find that

p(2) = (2)^{3} – 3(2)^{2} – 10(2) + 24

= 8 – 12 – 20 + 24 = 0

∴ By factor theorem, (x – 2) is a factor of p(x).

Now,

p(x) = x^{2}(x – 2) – x(x – 2) – 12(x – 2)

= (x – 2) (x^{2} – x – 12)

= (x – 2) (x^{2} – 4x + 3x – 12)

= (x – 2) [x(x – 4) + 3(x – 4)]

= (x – 2) (x – 4) (x + 3)

Question 6.

If 2x + 3y= 13 and xy = 6, find the value of 8x^{3} + 27y^{3}.

Solution:

2x + 3y = 13

Cubing on both sides,

⇒ (2x + 3y)^{3} = (13)^{3}

⇒ (2x)^{3} + (3y)^{3} + 3(2x)(3y)(2x + 3y) = 2197

⇒ 8x^{3} + 27y^{3} + 18xy (2x + 3y) = 2197

⇒ 8x^{3} + 27y^{3} + 18(6)(13) = 2197 (∵ xy = 6, 2x + 3y = 13 given)

⇒ 8x^{3} + 27y^{3} + 1404 = 2197

⇒ 8x^{3} + 27y^{3} = 2197 – 1404

⇒ 8x^{3} + 27y^{3} = 793

Question 7.

Find the value of ab + bc + ca if a + b + c = 9 and a^{2} + b^{2} + c^{2} = 35.

Solution:

We know that,

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) (9)^{2} = 35 + 2(ab + bc + ca)

81 = 35 + 2(ab + bc + ca)

2(ab + bc + ca) = 81 – 35

2(ab + bc + ca) = 46

∴ ab + bc + ca = 23

Question 8.

Evaluate the following product, without actual multiplication.

i) 102 × 98

ii) 104 × 106

Solution:

i) 102 × 98 = (100 + 2) (100 – 2)

(a + b)(a – b) = a^{2} – b^{2}

(100 + 2) (100 – 2) = (100)^{2} – 2^{2}

= 10000 – 4

= 9996

∴ 102 × 98 = 9996

ii) 104 × 106 = (100 + 4) (100 + 6)

(x + a)(x + b) = x^{2} + x(a + b) + ab

(100 + 4) (100 + 6)

= (100)^{2} + 100(4 + 6) + 4 × 6

= 10000 + 100 × 10 + 24

= 10000 + 1000 + 24

= 11,024

∴ 104 × 106 = 11,024

Question 9.

Give one example for each polynomial of a

i) binomial of degree 15

ii) monomial of degree 3

iii) linear polynomial in x

iv) quadratic polynomial in one variable

Solution:

i) Ex : x^{15} – 2, t^{15} + 3

ii) Ex: x^{3}, -3x^{3}, t^{3}

iii) Ex : x + 5, 3x – 2, -4x + 6

iv) Ex : 2x^{2} – 3x + 7, 4x^{2} – x – 3

Question 10.

Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Solution:

2(2) + 3(1) = k

∴ k = 7

Question 11.

Give one example of each of the following.

i) binomial of degree 3

ii) monomial of degree 3

iii) linear polynomial in x and y

iv) quadratic polynomial in one vari¬able ‘y’

Solution:

i) x^{3} + 1

ii) x^{3}

iii) x + y

iv) y^{2} + 2y + 1

Question 12.

If \(\frac{1}{2}\) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}] = 0,

show that x^{2} + y^{2} + z^{2} – xy – yz – zx = 0.

Solution:

\(\frac{1}{2}\) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}] = 0 (Given)

⇒ \(\frac{1}{2}\) [x^{2} + y^{2} – 2xy + y^{2} + z^{2} – 2yz + z^{2} + x^{2} – 2zx] = 0

⇒ \(\frac{1}{2}\) [2x^{2} + 2y^{2} + 2z^{2} – 2xy – 2yz – 2zx] = 0

⇒ \(\frac{1}{2}\) × 2[x^{2} + y^{2} + z^{2} – xy – yz – zx] = 0

⇒ x^{2} + y^{2} + z^{2} – xy – yz – zx = 0

Question 13.

If a + b + c = 6 and a^{2} + b^{2} + c^{2} = 14, find ab + bc + ca.

Solution:

a + b + c = 6 (Given)

a^{2} + b^{2} + c^{2}2 = 14 (Given)

We know that

(a + b + c)^{2} ≡ a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca …………… (1)

Substitute the values in (1), we have

(6)^{2} = 14 + 2ab + 2bc + 2ca

36 = 14 + 2(ab + bc + ca)

36 – 14 = 2(ab + bc + ca)

⇒ 22 = 2(ab + bc + ca)

⇒ ab + bc + ca = \(\frac{22}{2}\) = 11.

Question 14.

The lateral surface area of a cube is 4 times the square of its edge. Find the edge of a cube whose lateral surface area is given by

4x^{2} + 8 – \(\sqrt{128}\)x.

Solution:

Let the edge of the cube by y.

Then lateral surface area of the cube = 4y^{2}

According to the question,

4y^{2} = 4x^{2} + 8 – \(\sqrt{128}\)x

= 4x^{2} + 8 – 8√2x = 4[x^{2} – 2√2x + 2]

= 4[x^{2} – 2√2x + (√2)^{2}] = 4[x – √2]^{2}

⇒ y^{2} = [x – √2]^{2} ⇒ y = x – √2,

Hence the edge of the cube is [x – √2].

Question 15.

Pooja distributed cuboidal gifts to the children in an orphanage on her birthday. What are the possible expressions for the dimensions of the cuboid of the volume is 3kx^{2} + 2kx – 5k ? What value of Pooja is depicted here ?

Solution:

Volume = 3kx^{2} + 2kx – 5k

= k(3x^{2} + 2x – 5)

= k(3x^{2} + 5x – 3x – 5)

= k[x(3x + 5) – 1(3x + 5)]

= k(3x + 5)(x – 1)

∴ The possible expressions for the dimensions of the cuboid are k, 3k + 5 and x – 1.

The sympathetic nature of Pooja for the children in an orphanage is depicted here.

### 9th Class Maths Polynomials 8 Marks Important Questions

Question 1.

Let A and B be the remainders when polynomials x^{3} + 2x^{2} – 5ax – 7 and x^{3} + ax^{2} – 12x + 6 are divided by x + 1 and x – 2 respectively and 2A + B = 6. Find the value of ‘a’.

Solution:

Let f(x) = x^{3} + 2x^{2} – 5ax – 7 and

p(x) = x^{3} + ax^{2} – 12x + 6.

Then,

A = Remainder when f(x) is divided by

(x + 1) = f(- 1) [∵ x + 1 = 0 ⇒ x = – 1 by Remainder theorem]

= (-1)^{3} + 2(-1)^{2} – 5a(- 1) – 7

= – 1 + 2 + 5a – 7

= 5a – 6

B = Remainder when p(x) is divided by

x – 2 = p(2) [∵ x – 2 = 0 ⇒ x = 2 by Remainder theorem]

= (2)^{3} + a(2)^{2} – 12(2) + 6

= 8 + 4a – 24 + 6 = 4a – 10

Now, 2A + B = 6 (given)

⇒ 2(5a – 6) + 4a – 10 = 6

⇒ 10a – 12 + 4a – 10 = 6

⇒ 14a – 22 = 6

14a = 28

∴ a = 2

Question 2.

If ax^{3} + bx^{2} + x – 6 has x + 2 as a factor and leaves remainder 4 when divided by x – 2, find ‘a’ and ‘b’

Solution:

Let f(x) = ax^{3} + bx^{2} + x – 6

If (x + 2) is a factor of f(x), then by factor theorem f(-2) = 0 [∵ x + 2 = 0 ⇒ x = -2]

⇒ a(- 2)^{3} + b(- 2)^{2} + (- 2) – 6 = 0

⇒ – 8a + 4b – 2 – 6 = 0

⇒ 8a – 4b = – 8

⇒ 2a – b = – 2 _______ (1)

(dividing throughout by 4)

If f(x) leaves remainder 4 when divided by x – 2, then by remainder theorem,

f(2) = 4 [∵ x – 2 = 0 ⇒ x = 2]

⇒ a(2)^{3} + b(2)^{2} + 2 – 6 = 4

⇒ 8a + 4b – 4 = 4

⇒ 8a + 4b = 8

⇒ 2a + b = 2 _________ (2)

(dividing throughout by 4)

Solving (1) and (2) for ‘a’ and ‘b’

⇒ a = \(\frac{0}{4}\)

⇒ a = 0

(1) ⇒ 2(0) – b = -2

⇒ -b = -2

⇒ b = 2

∴ a = 0, b = 2

Question 3.

Divide the polynomial 2x^{4} + 5x^{3} – 2x^{2} + 2x – 4 by 2x + 1 and verify the remainder using remainder theorem.

Solution:

∴ Remainder = -6

and Quotient = x^{3} + 2x^{2} – 2x + 2

Let p(x) = 2x^{4} + 5x^{3} – 2x^{2} + 2x – 4

∴ Remainder = p(\(\frac{-1}{2}\)) (By remainder theorem 2x + 1 = 0 ⇒ x = \(\frac{-1}{2}\))

Hence, the remainder is verified.

Question 4.

If (y + 1) and (y + 2) are factors of the polynomial y^{3} + 3y^{2} – 3py + q, find p and q.

Solution:

Let f(y) = y^{3} + 3y^{2} – 3py + q

If (y + 1) is a factor of f(y), then by factor theorem,

f(-1) = 0 [∵ y + 1 = 0 ⇒ y = -1]

⇒ (-1)^{3} + 3(-1)^{2} – 3p(- 1) + q = 0

⇒ – 1 + 3 + 3p + q = 0

⇒ 3p + q = -2 _______ (1)

If (y + 2) is a factor of f(y), then by factor theorem.

f(- 2) = 0 [∵ y + 2 = 0 ⇒ y = -.2]

⇒ (- 2)^{3} + 3(- 2)^{3} – 3p(- 2) + q = 0

⇒ – 8 + 12 + 6p + q = 0

⇒ 6p + q = – 4 _________ (2)

Solving (1) and (2), we get

(1) ⇒ 3(\(\frac{-2}{3}\)) + q = – 2

⇒ – 2 + q = -2 ⇒ q = 0

∴ P = \(\frac{-2}{3}\) q = 0

Question 5.

Find the quotient and remainder when 6x + 11x – x + 7x + 27 is divided by (3x + 4). Also check the remainder obtained by using remainder theorem.

Solution:

Let f(x) = 6x + 11x^{3} – x^{2} + 7x + 27

∴ Quotient = 2x^{3} + x^{2} – \(\frac{5}{3}\)x + \(\frac{41}{9}\)

and Remainder = \(\frac{79}{9}\)

Also, remainder obtained by using remainder theorem when f(x) is divided by 3x + 4.

Which is the same as obtained above by actual division.

Question 6.

Verify – 2 and 3 are zeroes of the polynomial 2x^{3} – 3x^{2} – 11x + 6. If yes, factorise the polynomial.

Solution:

Let f(x) = 2x^{3} – 3x^{2} – 11x + 6

f(- 2) = 2(- 2)^{3} – 3(- 2)^{2} – 11(- 2) + 6

= – 16 – 12 + 22 + 6 = 0

Hence, – 2 is a zero of f(x).

f(3) = 2(3)^{3} – 3(3)^{2} – 11(3) + 6

= 54 – 27 – 33 + 6 = 0

Hence, 3 is a zero of f(x).

∵ -2 and 3 are zeores of f(x).

∴ f(x) is divisible by (x + 2) and (x – 3) both.

⇒ f(x) is divisible by (x + 2) (x – 3)

⇒ f(x) is divisible by x^{2} – x – 6

∴ f(x) = (x + 2)(x – 3) (2x – 1)

Question 7.

Factorise : (a + 2b)^{3} + (2a – c)^{3} – (a + 2c)^{3} + 3(a + 2b) (2a – c) (a + 2c)

Solution:

(a + 2b)^{3} + (2a – c)^{3} – (a + 2c)^{3} + 3(a + 2b) (2a – c) (a + 2c)

= (a + 2b)^{3} + (2a – c)^{3} + [- (a + 2c)]^{3} – 3(a + 2b)(2a – c) [- (a + 2c)]

= [(a + 2b) + (2a – c)] + [- (a + 2c)]] × [(a + 2b)^{2} + (2a – c)^{2} + [- (a + 2c)]^{2}

– (a + 2b) (2a – c) – (2a – c) [- (a + 2c)] – [- (a + 2c)] (a + 2b)]

= (2a + 2b – 3c) [a^{2} + 4b^{2} + 4ab + 4a^{2} + c^{2}

– 4ac + a^{2} + 4c^{2} + 4ac – 2a^{2} + ac – 4ab + 2bc + 2a^{2} + 4ac – ca – 2c^{2} + a^{2} + 2ab + 2ac + 4bc]

= (2a + 2b – 3c)

(7a^{2} + 4b^{2} + 3c^{2} + 2ab + 6bc + 6ca)

Question 8.

Simplify:

\(\frac{\left(a^2-b^2\right)^3+\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3}{(a-b)^3+(b-c)^3+(c-a)^3}\)

Solution:

(a^{2} – b^{2})^{3} + (b^{2} – c^{3})^{2} + (c^{2} – a^{2})^{3}

= 3(a^{2} – b^{2})(b^{2} – c^{2})(c^{2} – a^{2})

(∵ If x + y + z = 0, then x^{3} + y^{3} + z^{3} = 3xyz)

(Here, a^{2} – b^{2} + b^{2} – c^{2} + c^{2} – a^{2}= 0)

= 3(a – b)(a + b) (b – c) (b + c) (c – a)(c + a) …………… (1)

(a – b)^{3} + (b – c)^{3} + (c – a)^{3}

= 3(a – b)(b – c) (c – a) …………… (2)

(∵ If x + y + z = 0, then x^{3} + y^{3} + z^{3} = 0) (Here, a – b + b – c + c – a = 0)

∴ \(\frac{\left(a^2-b^2\right)^3+\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3}{(a-b)^3+(b-c)^3+(c-a)^3}\)

= \(\frac{3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{3(a-b)(b-c)(c-a)}\) [from (1) and (2)]

= (a + b)(b + c) (c + a)

Question 9.

If ab + bc + ca = 0, find the value of \(\frac{1}{\mathbf{a}^2-\mathbf{b c}}\) + \(\frac{1}{\mathbf{b}^2-\mathbf{c a}}\) + \(\frac{1}{\mathbf{c}^2-\mathbf{a b}}\).

Solution:

We have, ab + be + ca = 0 ……… (1)

∴ \(\frac{1}{\mathbf{a}^2-\mathbf{b c}}\) + \(\frac{1}{\mathbf{b}^2-\mathbf{c a}}\) + \(\frac{1}{\mathbf{c}^2-\mathbf{a b}}\)

Question 10.

If x = 2 + √5, find the value of x^{2} + \(\frac{1}{x^2}\).

Solution:

x = 2 + √5 …………. (1)

∴ \(\frac{1}{x}\) = \(\frac{1}{2+\sqrt{5}}\) = \(\frac{1}{2+\sqrt{5}}\) \(\frac{2-\sqrt{5}}{2-\sqrt{5}}\)

= \(\frac{2-\sqrt{5}}{(2)^2-(\sqrt{5})^2}\) = \(\frac{2-\sqrt{5}}{4-5}\) = \(\frac{2-\sqrt{5}}{-1}\)

= √5 – 2 ………….. (2)

Adding (1) and (2), we get

x + \(\frac{1}{x}\) = (2 + √5) + (√5 – 2) = 2√5

We know that

(x + \(\frac{1}{x}\))^{2} = x^{2} + \(\frac{1}{x^2}\) + 2

⇒ (2√5)^{2} = x^{2} + \(\frac{1}{x^2}\) + 2

⇒ 20 = x^{2} + \(\frac{1}{x^2}\) + 2

⇒ x^{2} + \(\frac{1}{x^2}\) = 18

**AP 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