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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.7

Question 1.

Factorize: x^{4} + 1. (Hint: Try completing the square.)

Solution:

Question 2.

If x^{2} + x + 1 is a factor of the polynomial 3x^{3} + 8x^{2} + 8x + a, then find the value of a.

Solution:

Let 3x^{3} + 8x^{2} + 8x + a = (x^{2} + x + 1) (3x + a) .

Equating coefficient of x

8 = a + 3

8 – 3 = a

a = 5

### Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.7 Additional Questions Solved

Question 1.

Solve for x^{2} – 7x^{3} + 8x^{2} + 8x – 8 = 0. given 3 – \(\sqrt{5}\) is a root

Solution:

when 3 – \(\sqrt{5}\) is a root, 3 + \(\sqrt{5}\) is the other root.

S.o.r. = (3 – \(\sqrt{5}\)) + (3 + \(\sqrt{5}\)) = 6

The equation is x^{2} – 6x + 4 = 0

Now x^{4} – 7x^{3} + 8x^{2} + 8x – 8 = (x^{2} – 6x + 4) (x^{2} + px – 2)

Equating co-eff of x

12 + 4p = 8

4p = 8 – 12 = -4

So the other factor is x^{2} – x – 2

Now solving x^{2} – x – 2 = 0

Question 2.

Solve the equation x^{3} + 5x^{2} – 16x – 14 = 0. given x + 7 is a root

Solution:

x^{3} + 5x^{2} – 16x – 14 = (x + 7) (x^{2} + px – 2)

Equating co-eff of x

7p – 2 = -16

7p = -16 + 2 = -14

⇒ p = -2

So the other factor is x^{2} – 2x – 2