Using factor theorem, factorize each of the following polynomial:
x3-3x2-9x-5
Let, f (x) = x3-3x2-9x-5
The factors of the constant term - 5 are
Putting x = -1, we have
f (-1) = (-1)3 – 3 (-1)2 – 9 (-1) - 5
= -1 – 3 + 9 - 5
= 0
So, (x + 1) is a factor of f (x)
Let us now divide
f (x) = x3-3x2-9x-5 by (x + 1) to get the other factors of f (x)
Using long division method, we get
x3-3x2-9x-5 = (x + 1) (x2 - 4x 5)
x2 - 4x - 5 = x2 - 5x + x - 5
= x (x - 5) + 1 (x - 5)
= (x + 1) (x - 5)
Hence, x3+13x2+32x+20 = (x + 1) (x + 1) (x - 5)
= (x + 1)2 (x – 5)