Q. 3

# Using factor theo

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)

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