AP 9th Class Maths Chapter 2 Polynomials Important Questions

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) = 2x² – x + a, find the value of ‘a’.

Solution:
p(x) = 2x² – x + a. As the zero the polynomial is – 2, we know that p(-2) = 0.
2x² – 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(a² + b²) = (a + b)², then show that a = b.
Solution:
2(a² + b²) = (a + b)², (given)
⇒ 2a² + 2b² = a² + b² + 2ab
⇒ a² + b² – 2ab = 0
⇒ (a – b)² = 0
⇒ a – b = 0
⇒ a = b

Question 4.
If a + b + c = 12 and ab + be + ca = 47, find a² + b² + c².
Solution:
We know that
(a + b+ c)² = a² + b² + c² + 2ab + 2bc + 2ca
⇒ a² + b² + c² = (a + b + c)² – 2(ab + bc + ca)
= (12)² – 2(47)
= 144 – 94 = 50

Question 5.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Solution:
Let p(x) = x³ – ax² + 6x – a
⇒ x – a = 0
⇒ x = a
∴ Remainder = p(a)
= (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a
= 5a

Question 6.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Solution:
7 + 3x will be a factor of 3x³ + 7x only if
7 + 3x divides 3x³ + 7x leaving no remainder.
Let p(x) = 3x³ + 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 3x³ + 7x.

Question 7.
Using suitable identity, find the value of \frac{87^3+13^3}{87^2-87 \times 13+13^2}\(\)
Solution:
\(\frac{87^3+13^3}{87^2-87 \times 13+13^2}\) = \(\) (∵ a³ + b³ = (a + b)(a² – ab + b²)
= 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
= (x² + y² + z² + 2xy + 2yz + 2xz) – [x² + (-y)² + (-z)² + 2x(-y) + 2(-y) (- z) + 2(- z)x]
= (x² + y² + z² + 2xy + 2yz + 2zx) – (x² + y² + z² – 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)]
= (8x³ + y³ + 12x²y + 6xy²) + (8x³ – y³ – 12x²y + 6xy²)
= 16x³ + 12xy²
= 4x(4x² + 3y²)

Question 10.
If x + a is a factor of ma – nx – 3x², then prove that a = \(\frac{m+n}{3}\).
Solution:
Let p(x) = ma – nx – 3x²
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 – 3a² = 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) = x³ – 3x² + 4x – 9 is divided by (x – 1) and (x + 2).
Solution:
p(x) = x³ – 3x² + 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) 343a³ – 729b³
Solution:
343a³ – 729b³
= (7a)³ – (9b)³
= (7a – 9b) [(7a)² + (9b)² + (7a)(9b)]
[∵ x³ – y³ = (x – y) (x² + y² + xy)]
= (7a – 9b) (49a² + 81b² + 63ab)

ii) 25x³y – 121xy³
Solution:
25x³y – 121xy³
= xy (25x² – 121y²)
= 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)]]
[∵ x³ + y³ = (x + y)(x² + y² – xy)]
= (5x – y) x (4x² + 4y² + 8xy + 9x² + 9y² – 18xy – 6x² + 6y²]
= (5x – y)(7x² + 19y² – 10xy)

Question 13.
If x + \(\frac{1}{x}\) = 4, find the value of x³ + \(\frac{1}{x^3}\).
Solution:
We know that,
(x + \(\frac{1}{x}\))³ = x³ + \(\frac{1}{x^3}\) + 3(x)(\(\frac{1}{x}\))(x + \(\frac{1}{x}\))
(x + \(\frac{1}{x}\))³ = x³ + \(\frac{1}{x^3}\) + 3(x + \(\frac{1}{x}\))
(4)³ = x³ + \(\frac{1}{x^3}\) + 3(4)
64 = x³ + \(\frac{1}{x^3}\) + 12
x³ + \(\frac{1}{x^3}\) = 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) = 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 2.
Verify that
p³ + q³ + r³ – 3pqr = (p + q + r) (p² + q² + r² – pq – qr – rp)
Solution:
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 + r³ – pqr – pqr – pq
= p³ + q³ + r³ – 3pqr = LHS
∴ LHS = RHS.
Hence proved.

Question 3.
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
⇒ 2 a = 2 b
∴ a = b
if a = b, from equation (1)
c = 0

Question 4.
Divide the polynomial 3y⁴ – 4y³ – 3y + 4 by (y – 1)
Solution:

∴ Quotient = 3y³ – y² – y – 4 and Remainder = 0

Question 5.
Factorise : x³ – 3x² – 10x + 24
Solution:
Let p(x) = x³ – 3x² – 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) = x²(x – 2) – x(x – 2) – 12(x – 2)
= (x – 2) (x² – x – 12)
= (x – 2) (x² – 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 8x³ + 27y³.
Solution:
2x + 3y = 13
Cubing on both sides,
⇒ (2x + 3y)³ = (13)³
⇒ (2x)³ + (3y)³ + 3(2x)(3y)(2x + 3y) = 2197
⇒ 8x³ + 27y³ + 18xy (2x + 3y) = 2197
⇒ 8x³ + 27y³ + 18(6)(13) = 2197 (∵ xy = 6, 2x + 3y = 13 given)
⇒ 8x³ + 27y³ + 1404 = 2197
⇒ 8x³ + 27y³ = 2197 – 1404
⇒ 8x³ + 27y³ = 793

Question 7.
Find the value of ab + bc + ca if a + b + c = 9 and a² + b² + c² = 35.
Solution:
We know that,
(a + b + c)² = a² + b² + c² + 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) = a² – b²
(100 + 2) (100 – 2) = (100)² – 2²
= 10000 – 4
= 9996
∴ 102 × 98 = 9996

ii) 104 × 106 = (100 + 4) (100 + 6)
(x + a)(x + b) = x² + 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 : x¹⁵ – 2, t¹⁵ + 3
ii) Ex: x³, -3x³, t³
iii) Ex : x + 5, 3x – 2, -4x + 6
iv) Ex : 2x² – 3x + 7, 4x² – 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) x³ + 1
ii) x³
iii) x + y
iv) y² + 2y + 1

Question 12.
If \(\frac{1}{2}\) [(x – y)² + (y – z)² + (z – x)²] = 0,
show that x² + y² + z² – xy – yz – zx = 0.
Solution:
\(\frac{1}{2}\) [(x – y)² + (y – z)² + (z – x)²] = 0 (Given)
⇒ \(\frac{1}{2}\) [x² + y² – 2xy + y² + z² – 2yz + z² + x² – 2zx] = 0
⇒ \(\frac{1}{2}\) [2x² + 2y² + 2z² – 2xy – 2yz – 2zx] = 0
⇒ \(\frac{1}{2}\) × 2[x² + y² + z² – xy – yz – zx] = 0
⇒ x² + y² + z² – xy – yz – zx = 0

Question 13.
If a + b + c = 6 and a² + b² + c² = 14, find ab + bc + ca.
Solution:
a + b + c = 6 (Given)
a² + b² + c²2 = 14 (Given)
We know that
(a + b + c)² ≡ a² + b² + c² + 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
4x² + 8 – \(\sqrt{128}\)x.
Solution:
Let the edge of the cube by y.
Then lateral surface area of the cube = 4y²
According to the question,
4y² = 4x² + 8 – \(\sqrt{128}\)x
= 4x² + 8 – 8√2x = 4[x² – 2√2x + 2]
= 4[x² – 2√2x + (√2)²] = 4[x – √2]²
⇒ y² = [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 3kx² + 2kx – 5k ? What value of Pooja is depicted here ?
Solution:
Volume = 3kx² + 2kx – 5k
= k(3x² + 2x – 5)
= k(3x² + 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 x³ + 2x² – 5ax – 7 and x³ + ax² – 12x + 6 are divided by x + 1 and x – 2 respectively and 2A + B = 6. Find the value of ‘a’.
Solution:
Let f(x) = x³ + 2x² – 5ax – 7 and
p(x) = x³ + ax² – 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 ax³ + bx² + 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) = ax³ + bx² + 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 2x⁴ + 5x³ – 2x² + 2x – 4 by 2x + 1 and verify the remainder using remainder theorem.
Solution:

∴ Remainder = -6
and Quotient = x³ + 2x² – 2x + 2
Let p(x) = 2x⁴ + 5x³ – 2x² + 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 y³ + 3y² – 3py + q, find p and q.
Solution:
Let f(y) = y³ + 3y² – 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 + 11x³ – x² + 7x + 27

∴ Quotient = 2x³ + x² – \(\frac{5}{3}\)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 2x³ – 3x² – 11x + 6. If yes, factorise the polynomial.
Solution:
Let f(x) = 2x³ – 3x² – 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 x² – 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) [a² + 4b² + 4ab + 4a² + c²
– 4ac + a² + 4c² + 4ac – 2a² + ac – 4ab + 2bc + 2a² + 4ac – ca – 2c² + a² + 2ab + 2ac + 4bc]
= (2a + 2b – 3c)
(7a² + 4b² + 3c² + 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:
(a² – b²)³ + (b² – c³)² + (c² – a²)³
= 3(a² – b²)(b² – c²)(c² – a²)
(∵ If x + y + z = 0, then x³ + y³ + z³ = 3xyz)
(Here, a² – b² + b² – c² + c² – a²= 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 x³ + y³ + z³ = 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 x² + \(\frac{1}{x^2}\).
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 = x² + \(\frac{1}{x^2}\) + 2
⇒ (2√5)² = x² + \(\frac{1}{x^2}\) + 2
⇒ 20 = x² + \(\frac{1}{x^2}\) + 2
⇒ x² + \(\frac{1}{x^2}\) = 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) = 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.
Read the following table and answer the following questions given below.

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 + 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
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