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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.1

Question 1.

Which of the following expressions are polynomials. If not give reason:

(i) \(\frac{1}{x^{2}}\) + 3x – 4

(ii) x^{2} (x – 1)

(iii) \(\frac{1}{x}\) (x + 5)

(iv) \(\frac{1}{x^{-2}}+\frac{1}{x^{-1}}\) + 7

(v) \(\sqrt{5} x^{2}+\sqrt{3} x+\sqrt{2}\)

(vi) \(m^{2}-\sqrt[3]{m}+7 m-10\)

Solution:

Question 2.

Write the coefficient of x^{2} and x in each of the following polynomials,

(i) 4 + \(\frac{2}{5} x^{2}\) – 3x

(ii) 6 – 2x^{2} + 3x^{3} – \(\sqrt{7} x\)

(iii) πx^{2} – x + 2

(iv) \(\sqrt{3} x^{2}+\sqrt{2} x\) + 0.5

(v) x^{2} – \(\frac{7}{2} x\) + 8

Solution:

Question 3.

Find the degree of the following polynomials.

Solution:

Question 4.

Rewrite the following polynomial in standard form.

Solution:

Question 5.

Add the following polynomials and find the degree of the resultant polynomial.

(i) p(x) = 6x^{2} – 7x+ 2 q(x) = 6x^{3} – 7x + 15

(ii) h(x) = 7x^{3} – 6x + 1 f(x) = 7x^{2 }+ 17x – 9

(iii) f(x) = 16x^{4} – 5x^{2} + 9 g(x) = -6x^{3} + 7x – 15

Solution:

Question 6.

Subtract the second polynomial from the first polynomial and find the degree of the resultant polynomial.

(i) p(x) = 7x^{2} + 6x – 1 q(x) = 6x – 9

(ii) f(y) = 6y^{2} – 7y + 2 g(y) = 7y + y^{3}

(iii) h(z) = z^{5} – 6z^{4} + z f(z) = 6z^{2} + 10z – 7

Solution:

Question 7.

What should be added to 2x^{3} + 6x^{2} – 5x + 8 to get 3x^{3} – 2x^{2} + 6x + 15?

Solution:

(2x^{3} + 6x^{2} – 5x + 8) + Q(x) = 3x^{3} – 2x^{2} + 6x + 15

∴ Q(x) = (3x^{3} – 2x^{2} + 6x + 15) – (2x^{3} + 6x^{2} – 5x + 8)

The required polynomial is x^{3} – 8x^{2} + 11x + 7

Question 8.

What must be subtracted from 2x^{24} + 4x^{2} – 3x + 7 to get 3x^{3} – x^{2} + 2x + 1 ?

Solution:

(2x^{4} + 4x^{2} – 3x + 7) – Q(x) = 3x^{3} – x^{2} + 2x + 1

Q(x) = (2x^{4} + 4x^{2} – 3x + 7) – 3x^{3} – x^{2} + 2x + 1

The required polynomial = 2x^{4} + 4x^{2} – 3x + 7 – 3x^{3} + x^{2} – 2x – 1

= 2x^{4} – 3x^{3} + 5x^{2} – 5x + 6

Question 9.

Multiply the following polynomials and find the degree of the resultant polynomial:

(i) p(x) = x^{2} – 9 q(x) = 6x^{2} + 7x – 2

(ii) f(x) = 7x + 2 g(x) = 15x – 9

(iii) h(x) = 6x^{2} – 7x + 1 f(x) = 5x – 7

Solution:

(i) p(x) = x^{2} – 9 q(x) = 6x^{2} + 7x – 2

p(x) × q(x) = (x^{2} – 9) (6x^{2} + 7x – 2)

The required polynomial is 6x^{4} + 7x^{3} – 56x^{2} – 63x + 18, degree 4.

(ii) f(x) = 7x + 2 g(x) = 15x – 9

f(x) × g(x) = (7x + 2) (15x – 9)

The required polynomial is 105x^{2} – 33x – 18, degree 2.

(iii) h(x) = 6x^{2} – 7x + 1 f(x) = 5x – 7

h(x) × f(x) = (6x^{2} – 7x + 1) (5x – 7)

The required polynomial is 30x^{3} – 77x^{2} + 54x – 7, degree 3.

Question 10.

The cost of chocolate is Rs. (x + y) and Amir bought (x + y) chocolates. Find the total amount paid by him in terms of x andy. If x = 10, y = 5 find the amount paid by him.

Solution:

Amount paid = Number of chocolates × Cost of a chocolate

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

If x = 10, y = 5

The total amount paid by him

= 10^{2} + 2 × 10 × 5 + 5^{2} = 100 + 100 + 25 = Rs. 225

Question 11.

The length of a rectangle is (3x + 2) units and it’s breadth is (3x – 2) units. Find its area in terms of x. What will be the area if x = 20 units.

Solution:

Area of a rectangle = length × breadth

= (3x + 2) × (3x – 2) = (3x)^{2} – 2^{2} = [9x^{2} – 4] Sq. units

If x = 20, Area = 9 × 20^{2} – 4 = 9 × 400 – 4

= 3600 – 4 = 3596 Sq. units

Question 12.

p(x) is a polynomial of degree 1 and q(x) is a polynomial of degree 2. What kind of the polynomial p(x) × q(x) is ?

Solution:

p(x) is a polynomial of degree 1. q(x) is a polynomial of degree 2.

Then the p(x) × q(x) will be the polynomial of degree (1 + 2) = 3 (or)

Cubic polynomial