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) 4x^{2} + 5x – 2 is a polynomial.

ii) y^{2} – 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) √3x^{2} + 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.

**Do These**

Question

Write two polynomials with variable ‘x’. [Page No. 29]

Solution:

5x^{2} + 2x – 8 and 3x^{2} – 2x + 6.

Question

Write three polynomials with variable ‘y’.

Solution:

y^{3} – y^{2} + y ; 2y^{2} + 7y – 9 + 3y^{3}; y^{4} – y + 6 + 2y^{2}.

Question

Is the polynomial 2x^{2} + 3xy + 5y^{2} 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) 7x^{3} + 5x^{2} + 2x – 6 – degree 3

ii) 7 – x + 3x^{2 }– degree 2

iii) 5p – √3 – degree 1

iv) 2 – degree 0

v) – 5 xy^{2} – degree 3

Question 2.

Write the co-efficient of x^{2} in each of the following. [Page No. 30]

Solution:

i) 15 – 3x + 2x^{2} : co-efficient of x^{2} is 2

ii) 1 -x^{2} : co-efficient of x^{2} is -1

iii) πx^{2} – 3x + 5 : co-efficient of x^{2} is π

iv) √2x^{2} + 5x – 1 : co-efficient of x^{2} 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.: 5x^{3} + 3x^{2} – 8x + 4; x^{3} – 8

**Try These**

Question 1.

Write a polynomial with 2 terms in variable x. [Page No. 31]

Solution:

2x + 3x^{2}

Question 2.

How can you write a polynomial with 15 terms in variable ‘x’. [Page No. 31]

Solution:

a_{14}p^{14} + a_{13}p^{13} + a_{12}p^{12}+ …………….+ a_{1}p + a_{0}

**Do This**

Question

Find the value of each of the follow ing polynomials for the indicated value of variables. [Page No. 33]

(i) p(x) = 4x^{2} – 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) = 2y^{3} – 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) = 4t^{4} + 3t^{3} – t^{2} + 6 at t = p, t ∈ R.

Solution:

The value of r(t) at t = p is

4p^{4} + 3p^{3} – p^{2} + 6

iv) s(z) = z^{3} – 1 at z – 1.

Solution:

The value of s(z) at z = 1 is 1^{3} – 1 = 0

v) p(x) = 3x^{2} + 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) = 5z^{3} – 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

**Try These**

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. x^{2} – 5x + 6

Solution:

x^{2} – 5x + 6 = 0

⇒ x^{2} – 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 x^{2} – 5x + 6.

3. x + 5

Solution:

x + 5 = 0

x = – 5

∴ x = – 5

**Do This**

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.

x^{2} + 1 has no zeroes. Why ? [Page No. 36]

Solution:

x^{2} + 1 = 0 ⇒ x^{2} = -1

No real number exists such that whose root is – 1.

∴ x^{2} + 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.

**Do These**

Question 1.

Divide 3y^{3} + 2y^{2} + y by ‘y’ and write division fact. [Page No. 38]

Solution:

(3y^{3} + 2y^{2} + y) ÷ y = \(\frac{3 y^{3}}{y}+\frac{2 y^{2}}{y}+\frac{y}{y}\)

= 3y^{2} + 2y + 1

Division fact = (3y^{2} + 2y + 1) y

= 3y^{3} + 2y^{2} + y

Question 2.

Divide 4p^{2} + 2p + 2 by ‘2p’ and write division fact.

Solution:

4p^{2} + 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 = 4p^{2} + 2p + 2

**Try These**

Show that (x – 1) is a factor of x^{n} – 1. [Page No. 45]

Solution:

Let p(x) = x^{n} – 1

Then p(1) = 1^{n} – 1 = 1 – 1 = 0

As p(1) = 0, (x – 1) is a factor of p(x).

**Do These**

Question

Factorise the following. [Page No. 46]

1. 6x^{2} + 19x + 15

Solution:

6x^{2} + 19x + 15 = 6x^{2} + 10x + 9x + 15

= 2x (3x + 5) + 3 (3x + 5)

= (3x + 5) (2x + 3)

2. 10m^{2} – 31m – 132

Solution:

10m^{2} – 31m – 132

= 10m^{2} – 55m + 24m – 132

= 5m (2m- 11) + 12 (2m- 11)

= (2m – 11) (5m + 12)

3. 12x^{2} + 11x + 2

Solution:

12x^{2} + 11x + 2

= 12x^{2} + 8x + 3x + 2

= 4x (3x + 2) + 1 (3x + 2)

= (3x + 2) (4x + 1)

**Try This**

Question

Try to draw the geometrical figures for other identities. [Page No. 49]

i) (x + y)^{2} ≡ x^{2} + 2xy + y^{2}

Step – 1 : Area of fig. I = x x = x^{2}

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 = y^{2}

Area of big square = sum of the areas of figures I, II, III and IV

∴ (x + y) (x + y) = x^{2} + xy + xy + y^{2}

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

ii) (x + y) (x – y) ≡ x^{2} – y^{2}

Step -1: Area of fig! I = x (x – y) = x^{2} – xy

Step – 2: Area of fig. II = (x – y) y = xy – y^{2}

Area of big rectangle = sum of areas of figures I & II

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

= x^{2} – y^{2}

∴ (x + y) (x-y) = x2-y2

iii) (x + a) (x + b) ≡ x^{2} + (a + b) x + ab

Step – 1 : Area of fig. I = x^{2}

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) = x^{2} + ax + bx + ab

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

**Do These**

Question

Find the following product using appropriate identities. [Page No. 49]

i) (x + 5) (x + 5)

Solution:

(x + 5) (X + 5) = (x + 5)^{2}

= x^{2} + 2(x) (5) + 5^{2}

= x^{2} + 10x + 25

ii) (p – 3) (p + 3)

Solution:

(p – 3) (p + 3)

= p^{2} – 3^{2}

= p^{2} – 9

iii) (y – 1) (y – 1)

Solution:

(y – 1) (y – 1)

= (y – 1)^{2}

= y^{2} – 2y + 1

iv) (t + 2) (t + 4)

Solution:

(t + 2) (t + 4)

= t^{2} + t(2 + 4) + 2 x 4

= t^{2} + 6t + 8

v) 102 x 98

Solution:

102 x 98 = (100 + 2) (100 -2)

= 100^{2} – 2^{2}

= 10000 – 4

= 9996

**Do These**

Question

Factorise the following using appro-priate identities. [Page No. 50]

i) 49a^{2} + 70ab + 25b^{2}

Solution:

49a^{2} + 70ab + 25b^{2}

= (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) t^{2} – 2t + 1

= (t)^{2} – 2(t) (1) + (1)^{2}

= (t – 1)^{2} = (t – 1) (t – 1)

iv) x^{2} + 3x + 2

Solution:

x^{2} + 3x + 2 = x2 + (2 + 1) x + (2 x 1)

(x + 2) (x + 1)

**Do These**

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)

= p^{2} + 4q^{2} + r^{2} + 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)

= 16x^{2} + 4y^{2} + 9z^{2} – 16xy + 12yz – 24zx.

Question iii)

Factorise 4a^{2} + b^{2} + c^{2} – 4ab + 2bc – 4ca

using identity. [Page No. 52]

Solution:

4a^{2} + b^{2} + c^{2} – 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)

**Try These**

Question

How can you find (x – y)^{3} without actual multiplication ? Verify with actual multiplication. [Page No. 52]

Solution:

(x – y)^{3} = x^{3} – 3x^{2}y + 3xy^{2} – 3y^{3} from identity.

By actual multiplication

(x – y)^{3} = (x – y)^{2} (x – y)

= (x^{2} – 2xy + y^{2}) (x – y)

= x^{3} – 2x^{2}y + xy^{2} – x^{2}y + 2xy^{2} – y^{3}

= x^{3} – 3x^{2} y + 3xy^{2} – y^{3}

Both are equal.

**Do These**

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)

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

= x^{3} + 1 + 3x^{2} + 3x = x^{3} + 3x^{2} + 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}

= 27m^{3} – 54m^{2}n + 36mn^{2} – 8n^{3}

Question 3.

Factorise a^{3} – 3a^{2}b + 3ab^{2} – b^{3}. [Page No. 54]

Solution:

a^{3} – 3a^{2}b + 3ab^{2} – b^{3}

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

= (a – b)^{3}

= (a – b) (a – b) (a – b)

**Do These**

Question 1.

Find the product (a – b – c) (a^{2} + b^{2} + c^{2} – ab + be – ca) without actual multi-plication. [Page No. 55]

Solution:

The problem is incorrect.

Question 2.

Factorise 27a^{3} + b^{3} + 8c^{3} – 18abc using identity. [Page No. 55]

Solution:

27a^{3} + b^{3} + 8c^{3} – 18abc

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

= (3a + b + 2c) (9a^{2} + b^{2} + 4c^{2} – 3ab – 2be – 6ca)