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

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

Question 1.
Factorize: x⁴ + 1. (Hint: Try completing the square.)
Solution:

Question 2.
If x² + x + 1 is a factor of the polynomial 3x³ + 8x² + 8x + a, then find the value of a.
Solution:
Let 3x³ + 8x² + 8x + a = (x² + 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 x² – 7x³ + 8x² + 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 x² – 6x + 4 = 0
Now x⁴ – 7x³ + 8x² + 8x – 8 = (x² – 6x + 4) (x² + px – 2)
Equating co-eff of x
12 + 4p = 8
4p = 8 – 12 = -4
So the other factor is x² – x – 2
Now solving x² – x – 2 = 0

Question 2.
Solve the equation x³ + 5x² – 16x – 14 = 0. given x + 7 is a root
Solution:
x³ + 5x² – 16x – 14 = (x + 7) (x² + px – 2)
Equating co-eff of x
7p – 2 = -16
7p = -16 + 2 = -14
⇒ p = -2
So the other factor is x² – 2x – 2