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)


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