Question

Using factor theorem, factorize each of the following polynomial:
x⁴-2x³-7x²+8x+12

Answer 1

$\large{\tt{Answer:}}$

$\sf{Factors \: of \:12 = \pm1, \pm 2, \pm3, \pm4, \pm6, \pm12}$

When x = 2:

(2)^4 - 2(2)^3 - 7(2)^2 + 8(2) + 12

=> 16 - 16 - 28 + 16 + 12

=> 0

Thus, x = 2

=> x - 2 = 0 is a factor of p(x)

When x = -2:

(-2)^4 - 2(-2)^3 - 7(-2)^2 + 8(-2) + 12

=> 16 + 16 - 28 - 16 + 12

=> 0

Thus, x = -2

=> x + 2 = 0 is a factor of p(x)

When x = 3:

(3)^4 - 2(3)^3 - 7(3)^2 + 8(3) + 12

=> 81 - 54 - 63 + 24 + 12

=> 0

Thus, x = 3

=> x - 3 is a factor of p(x)

When x = -1:

(-1)^4 - 2(-1)^3 - 7(-1)^2 - 8(-1) + 12

=> 1 + 2 - 7 + 8 + 12

=> 0

Thus, x = -1

=> x + 1 = 0 is a factor of p(x)

Therefore, x^4 - 2x^3 - 7x^2 + 8x + 12 = (x - 2)(x + 2)(x - 3)(x + 1)

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Answer 2

Factors Of 12=±1,±2,±3,±4,±6,±12

When x = 2:

(2)^4 - 2(2)^3 - 7(2)^2 + 8(2) + 12

=> 16 - 16 - 28 + 16 + 12

=> 0

Thus, x = 2

=> x - 2 = 0 is a factor of p(x)

When x = -2:

(-2)^4 - 2(-2)^3 - 7(-2)^2 + 8(-2) + 12

=> 16 + 16 - 28 - 16 + 12

=> 0

Thus, x = -2

=> x + 2 = 0 is a factor of p(x)

When x = 3:

(3)^4 - 2(3)^3 - 7(3)^2 + 8(3) + 12

=> 81 - 54 - 63 + 24 + 12

=> 0

Thus, x = 3

=> x - 3 is a factor of p(x)

When x = -1:

(-1)^4 - 2(-1)^3 - 7(-1)^2 - 8(-1) + 12

=> 1 + 2 - 7 + 8 + 12

=> 0

Thus, x = -1

=> x + 1 = 0 is a factor of p(x)

Therefore, x^4 - 2x^3 - 7x^2 + 8x + 12 = (x - 2)(x + 2)(x - 3)(x + 1)

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