AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation InText Questions

Think, Discuss and Write

Question
Which of the following expressions are polynomials ? Which are not ? Give reasons. [Page No. 28]
Solution:
i) 4x² + 5x – 2 is a polynomial.
ii) y² – 8 is a polynomial.
iii) 5 is a constant polynomial.
iv) \(2 x^{2}+\frac{3}{x}-5\) is not a polynomial as x is in denominator.
v) √3x² + 5y is a polynomial.
vi) \(\frac{1}{x+1}\) is not a polynomial as the variable x is in denominator.
vii) √x is not a polynomial as its exponent is not an integer.
viii) 3xyz is a polynomial.

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Question
Write two polynomials with variable ‘x’. [Page No. 29]
Solution:
5x² + 2x – 8 and 3x² – 2x + 6.

Question
Write three polynomials with variable ‘y’.
Solution:
y³ – y² + y ; 2y² + 7y – 9 + 3y³; y⁴ – y + 6 + 2y².

Question
Is the polynomial 2x² + 3xy + 5y² in one variable ?
Solution:
No. It is in two variables x and y.

Question
Write the formulae of area and volume of different solid shapes. Find out the variables and constants in them. [Page No. 29]
Solution:

Question 1.
Write the degree of each of the following polynomials. [Page No. 30]
Solution:
i) 7x³ + 5x² + 2x – 6 – degree 3
ii) 7 – x + 3x² – degree 2
iii) 5p – √3 – degree 1
iv) 2 – degree 0
v) – 5 xy² – degree 3

Question 2.
Write the co-efficient of x² in each of the following. [Page No. 30]
Solution:
i) 15 – 3x + 2x² : co-efficient of x² is 2
ii) 1 -x² : co-efficient of x² is -1
iii) πx² – 3x + 5 : co-efficient of x² is π
iv) √2x² + 5x – 1 : co-efficient of x² is √2

Think, Discuss and Write

Question
How many terms a cubic (degree 3) polynomial with one variable can have? Give examples. [Page No. 31]
Solution:
A cubic polynomial can have atmost 4 terms.
E.g.: 5x³ + 3x² – 8x + 4; x³ – 8

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Question 1.
Write a polynomial with 2 terms in variable x. [Page No. 31]
Solution:
2x + 3x²

Question 2.
How can you write a polynomial with 15 terms in variable ‘x’. [Page No. 31]
Solution:
a₁₄p¹⁴ + a₁₃p¹³ + a₁₂p¹²+ …………….+ a₁p + a₀

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Question
Find the value of each of the follow ing polynomials for the indicated value of variables. [Page No. 33]
(i) p(x) = 4x² – 3x + 7 at x = 1.
Solution:
The value of p(x) at x = 1 is
4(1)² – 3(1) + 7 = 8

ii) q(y) = 2y³ – 4y + √11 at y = 1.
Solution:
The value of q(y) at y = 1 is
2(1)³ – 4(1) + √11 = -2 + √11

iii) r(t) = 4t⁴ + 3t³ – t² + 6 at t = p, t ∈ R.
Solution:
The value of r(t) at t = p is
4p⁴ + 3p³ – p² + 6

iv) s(z) = z³ – 1 at z – 1.
Solution:
The value of s(z) at z = 1 is 1³ – 1 = 0

v) p(x) = 3x² + 5x – 7 at x = 1.
Solution:
The value of p(x) at x = 1 is
3(1)² + 5(1) — 7 = 1.

vi) q(z) = 5z³ – 4z + √2 at 7. = 2.
Solution:
The value of q(z) at z = 2 is
5(2)³ – 4(2) + √2 = 40 – 8 + √2
= 32 + √2

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Question
Find zeroes of the following polyno¬mials. [Page No. 34]

1. 2x-3
Solution:
2x – 3 = 0
2x = 3
x = \(\frac{3}{2}\)
∴ x = \(\frac{3}{2}\) is the zero of 2x – 3.

2. x² – 5x + 6
Solution:
x² – 5x + 6 = 0
⇒ x² – 3x – 2x + 6 = 0
⇒ x (x – 3) – 2 (x – 3) = 0
⇒ (x – 2) (x – 3) = 0
⇒ x – 2 = 0 or x – 3 = 0
⇒ x = 2 or x = 3
∴ x = 2 or 3 are the zeroes of x² – 5x + 6.

3. x + 5
Solution:
x + 5 = 0
x = – 5
∴ x = – 5

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Fill in the bianks : [Page No. 35]

Linear polynomial	Zero of the polynomial
x + a	– a
x – a	a
ax + b	\( \frac{-b}{a} \)
ax – b	\( \frac{b}{a} \)

Solution:

Linear polynomial	Zero of the polynomial
x + a	– a
x – a	a
ax + b	\( \frac{-b}{a} \)
ax – b	\( \frac{b}{a} \)

Think, and Discuss

Question 1.
x² + 1 has no zeroes. Why ? [Page No. 36]
Solution:
x² + 1 = 0 ⇒ x² = -1
No real number exists such that whose root is – 1.
∴ x² + 1 has no zeroes.

Question 2.
Can you tell the number of zeroes of a polynomial of degree ‘n’ will have? [Page No. 36]
Solution:
A polynomial of degree n will have n- zeroes.

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Question 1.
Divide 3y³ + 2y² + y by ‘y’ and write division fact. [Page No. 38]
Solution:
(3y³ + 2y² + y) ÷ y = \(\frac{3 y^{3}}{y}+\frac{2 y^{2}}{y}+\frac{y}{y}\)
= 3y² + 2y + 1
Division fact = (3y² + 2y + 1) y
= 3y³ + 2y² + y

Question 2.
Divide 4p² + 2p + 2 by ‘2p’ and write division fact.
Solution:
4p² + 2p ÷ 2 = \(\frac{4 p^{2}}{2 p}+\frac{2 p}{2 p}+\frac{2}{2 p}\)
= 2p + 1 + \(\frac{1}{\mathrm{P}}\)
Division fact:
(2p + 1 + \(\frac{1}{\mathrm{P}}\)).2p = 4p² + 2p + 2

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Show that (x – 1) is a factor of xⁿ – 1. [Page No. 45]
Solution:
Let p(x) = xⁿ – 1
Then p(1) = 1ⁿ – 1 = 1 – 1 = 0
As p(1) = 0, (x – 1) is a factor of p(x).

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Question
Factorise the following. [Page No. 46]

1. 6x² + 19x + 15
Solution:
6x² + 19x + 15 = 6x² + 10x + 9x + 15
= 2x (3x + 5) + 3 (3x + 5)
= (3x + 5) (2x + 3)

2. 10m² – 31m – 132
Solution:
10m² – 31m – 132
= 10m² – 55m + 24m – 132
= 5m (2m- 11) + 12 (2m- 11)
= (2m – 11) (5m + 12)

3. 12x² + 11x + 2
Solution:
12x² + 11x + 2
= 12x² + 8x + 3x + 2
= 4x (3x + 2) + 1 (3x + 2)
= (3x + 2) (4x + 1)

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Question
Try to draw the geometrical figures for other identities. [Page No. 49]
i) (x + y)² ≡ x² + 2xy + y²

Step – 1 : Area of fig. I = x x = x²
Step – 2 : Area of fig. II = x y = xy
Step – 3 : Area of fig. III = x y = xy
Step – 4 : Area of fig. IV = y y = y²

Area of big square = sum of the areas of figures I, II, III and IV
∴ (x + y) (x + y) = x² + xy + xy + y²
(x + y)² = x² + 2xy + y²

ii) (x + y) (x – y) ≡ x² – y²

Step -1: Area of fig! I = x (x – y) = x² – xy
Step – 2: Area of fig. II = (x – y) y = xy – y²
Area of big rectangle = sum of areas of figures I & II
(x + y) (x – y) = x² – xy + xy – y²
= x² – y²
∴ (x + y) (x-y) = x2-y2

iii) (x + a) (x + b) ≡ x² + (a + b) x + ab

Step – 1 : Area of fig. I = x²
Step – 2 : Area of fig. II = ax
Step – 3 : Area of fig. Ill = bx
Step – 4 : Area of fig. IV = ab
∴ Area of big rectangle = Sum of areas of four small figures.
∴ (x + a) (x + b) = x² + ax + bx + ab
(x + a) (x + b) = x² + (a + b) x + ab

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Question
Find the following product using appropriate identities. [Page No. 49]
i) (x + 5) (x + 5)
Solution:
(x + 5) (X + 5) = (x + 5)²
= x² + 2(x) (5) + 5²
= x² + 10x + 25

ii) (p – 3) (p + 3)
Solution:
(p – 3) (p + 3)
= p² – 3²
= p² – 9

iii) (y – 1) (y – 1)
Solution:
(y – 1) (y – 1)
= (y – 1)²
= y² – 2y + 1

iv) (t + 2) (t + 4)
Solution:
(t + 2) (t + 4)
= t² + t(2 + 4) + 2 x 4
= t² + 6t + 8

v) 102 x 98
Solution:
102 x 98 = (100 + 2) (100 -2)
= 100² – 2²
= 10000 – 4
= 9996

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Question
Factorise the following using appro-priate identities. [Page No. 50]
i) 49a² + 70ab + 25b²
Solution:
49a² + 70ab + 25b²
= (7a)² + 2 (7a) (5b) + (5b)²
= (7a + 5b)²
= (7a + 5b)(7a + 5b)

ii) \(\frac{9}{16} x^{2}-\frac{y^{2}}{9}\)
Solution:

iii) t² – 2t + 1
= (t)² – 2(t) (1) + (1)²
= (t – 1)² = (t – 1) (t – 1)

iv) x² + 3x + 2
Solution:
x² + 3x + 2 = x2 + (2 + 1) x + (2 x 1)
(x + 2) (x + 1)

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Question i)
Write (p + 2q + r)² in expanded form. [Page No. 52]
Solution:
(p + 2q + r)² = (p)² + (2q)² + (r)²
+ 2 (P) (2q) + 2 (2q) (r) + 2(r) (p)
= p² + 4q² + r² + 4pq + 4qr + 2rp

Question ii)
Expand (4x – 2y – 3z)² using identity. [Page No. 52]
Solution:
(4x – 2y – 3z)² = (4x)² + (- 2y)² + (- 3z)² + 2 (4x) (- 2y) + 2 (- 2y) (- 3z) + 2 (- 3z) (4x)
= 16x² + 4y² + 9z² – 16xy + 12yz – 24zx.

Question iii)
Factorise 4a² + b² + c² – 4ab + 2bc – 4ca
using identity. [Page No. 52]
Solution:
4a² + b² + c² – 4ab + 2bc – 4ca
= (2a)² + (- b)² + (- c)² + 2(2a) (- b) + 2 (- b) (- c) + 2(- c) (2a)
= (2a – b – c)² = (2a – b – c) (2a – b – c)

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Question
How can you find (x – y)³ without actual multiplication ? Verify with actual multiplication. [Page No. 52]
Solution:
(x – y)³ = x³ – 3x²y + 3xy² – 3y³ from identity.
By actual multiplication
(x – y)³ = (x – y)² (x – y)
= (x² – 2xy + y²) (x – y)
= x³ – 2x²y + xy² – x²y + 2xy² – y³
= x³ – 3x² y + 3xy² – y³
Both are equal.

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Question 1.
Expand (x + 1)³ using an identity. [Page No. 54]
Solution:
(x + 1)³ = (x)³ + (1)³ + 3 (x) (1) (x + 1)
= x³ + 1 + 3x (x + 1)
= x³ + 1 + 3x² + 3x = x³ + 3x² + 3x + 1

Question 2.
Compute (3m – 2n)³. [Page No. 54]
Solution:
(3m-2n)³
=(3m)³ – 3 (3m)² (2n) + 3 (3m) (2n)² – (2n)³
= 27m³ – 54m²n + 36mn² – 8n³

Question 3.
Factorise a³ – 3a²b + 3ab² – b³. [Page No. 54]
Solution:
a³ – 3a²b + 3ab² – b³
= (a)³ – 3 (a)² (b) + 3 (a) (b)² – (b)³
= (a – b)³
= (a – b) (a – b) (a – b)

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Question 1.
Find the product (a – b – c) (a² + b² + c² – ab + be – ca) without actual multi-plication. [Page No. 55]
Solution:
The problem is incorrect.

Question 2.
Factorise 27a³ + b³ + 8c³ – 18abc using identity. [Page No. 55]
Solution:
27a³ + b³ + 8c³ – 18abc
= (3a)³ + (b)³ + (2c)³ – 3(3a) (b) (2c)
= (3a + b + 2c) (9a² + b² + 4c² – 3ab – 2be – 6ca)