Using factor theorem, factorize each of the following polynomial:
x3-23x2+142x-120
Let, f (x) = x3-23x2+142x-120
The factors of the constant term – 120 are
Putting x = 1, we have
f (1) = (1)3 – 23 (1)2 + 142 (1) – 120
= 1 – 23 + 142 – 120
= 0
So, (x – 1) is a factor of f (x)
Let us now divide
f (x) = x3-23x2+142x-120 by (x - 1) to get the other factors of f (x)
Using long division method, we get
x3-23x2+142x-120 = (x – 1) (x2 – 22x + 120)
x2 – 22x + 120 = x2 – 10x – 12x + 120
= x (x – 10) – 12 (x – 10)
Hence, x3-23x2+142x-120 = (x – 1) (x - 10) (x - 12)