Using factor theo

Let, f (x) = x4-2x3-7x2+8x+12

The constant term in f (x) is equal to +12 and factors of +12 are ,

Putting x = - 1 in f (x), we have

f (-1) = (-1)4 – 2 (-1)3 – 7 (-1)2 + 8 (-1) + 12

= 1 + 2 – 7 – 8 + 12

= 0

Therefore, (x + 1) is a factor of f (x).

Similarly, (x + 2), (x – 2) and (x - 3) are the factors of f (x).

Since, f (x) is a polynomial of degree 4. So, it cannot have more than four linear factors.

Therefore, f (x) = k (x + 1) (x + 2) (x - 2) (x - 3)

x4-2x3-7x2+8x+12 = k (x + 1) (x + 2) (x - 2) (x - 3)

Putting x = 0 on both sides, we get

0 - 0 – 0 + 0 + 12 = k (0 + 1) (0 + 2) (0 - 2) (0 - 3)

12 = 12k

k = 1

Putting k = 1 in f (x) = k (x + 1) (x + 2) (x - 2) (x - 3), we get

f (x) = (x + 1) (x + 2) (x - 2) (x - 3)

Hence,

x4-2x3-7x2+8x+12 = (x + 1) (x + 2) (x - 2) (x - 3)

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