Using factor theorem, factorize each of the following polynomial:
3x3-x2-3x+1
Let, f (x) = 3x3-x2-3x+1
The factors of the constant term
The factor of the coefficient of x3 is 3. Hence, possible rational roots of f (x) are:
We have,
f (1) = 3 (1)3 – (1)2 – 3 (1) + 1
= 3 – 1 – 3 + 1
= 0
So, (x – 1) is a factor of f (x)
Let us now divide
f (x) = 3x3-x2-3x+1 by (x - 1) to get the other factors of f (x)
Using long division method, we get
3x3-x2-3x+1 = (x – 1) (3x2 + 2x – 1)
Now,
3x2 + 2x - 1 = 3x2 + 3x – x – 1
= 3x (x + 1) – 1 (x + 1)
= (3x – 1) (x + 1)
Hence, 3x3-x2-3x+1 = (x – 1) (x + 1) (3x – 1)