Using factor theorem, factorize each of the following polynomial:
x3-6x2+3x+10
Let, f (x) = x3-6x2+3x+10
The constant term in f (x) is equal to 10 and factors of 10 are ,
Putting x = - 1 in f (x), we have
f (-1) = (-1)3 – 6 (-1)2 + 3 (-1) + 10
= -1 – 6 – 3 + 10
= 0
Therefore, (x + 1) is a factor of f (x).
Similarly, (x - 2) and (x - 5) are the factors of f (x).
Since, f (x) is a polynomial of degree 3. So, it cannot have more than three linear factors.
Therefore, f (x) = k (x + 1) (x - 2) (x - 5)
x3-6x2+3x+10 = k (x + 1) (x - 2) (x - 5)
Putting x = 0 on both sides, we get
0 + 0 – 0 + 10 = k (0 + 1) (0 - 2) (0 - 5)
10 = 10k
k = 1
Putting k = 1 in f (x) = k (x + 1) (x - 2) (x - 5), we get
f (x) = (x + 1) (x - 2) (x - 5)
Hence,
x3-6x2+3x+10 = (x + 1) (x - 2) (x - 5)