Question

1. Determine which of the following polynomials has (x + 1) a factor :
(1) x3 + x2 + x + 1
(ii) x4 + x + x2 + x + 1
(ii) x4 + 3x + 3x2 + x + 1
(iv) x-x- (2 + V2)x+ 12​

Answer 1

Answer:

(i) 6 ans (ii) 9 (iii) 15 (iv) v=15 ans

Answer 2

Answer :-

The zero of x + 1 is -1.

(i) Let p (x) = x3 + x2 + x + 1

∴ p (-1) = (-1)3 + (-1)2 + (-1) + 1 .

= -1 + 1 – 1 + 1

⇒ p (- 1) = 0

So, (x+ 1) is a factor of x3 + x2 + x + 1.

(ii) Let p (x) = x4 + x3 + x2 + x + 1

∴ P(-1) = (-1)4 + (-1)3 + (-1)2 + (-1)+1

= 1 – 1 + 1 – 1 + 1

⇒ P (-1) ≠ 1

So, (x + 1) is not a factor of x4 + x3 + x2 + x+ 1.

(iii) Let p (x) = x4 + 3x3 + 3x2 + x + 1 .

∴ p (-1)= (-1)4 + 3 (-1)3 + 3 (-1)2 + (- 1) + 1

= 1 – 3 + 3 – 1 + 1 = 1

⇒ p (-1) ≠ 0

So, (x + 1) is not a factor of x4 + 3x3 + 3x2 + x+ 1.

(iv) Let p (x) = x3 – x2 – (2 + √2) x + √2

∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2

= -1 – 1 + 2 + √2 + √2

= 2√2

⇒ p (-1) ≠ 0

So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2.