Question

Using the factor theorem, show that (x - 2) is a factor of x3 + x2 - 4x - 4.
Hence factorise the polynomial completely.​

Answer 1

Step-by-step explanation:

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Answer 2

x - 2 is a factor of $x^3 + x^2 - 4x - 4$

$x^3 + x^2 - 4x - 4 = \left(x+1\right)\left(x+2\right)\left(x-2\right)$

Solution:

Given that,

$p(x) = x^3+x^2 - 4x - 4$

By factor theorem,

If x - a is factor of p(x), then p(a) = 0

Therefore,

Given that, x - 2 is a factor

x - 2 = 0

x = 2

Substitute x = 2 in p(x)

$p(2) = 2^3 + 2^2 - 4(2) - 4\\\\p(2) = 8 + 4 - 8 - 4\\\\p(2) = 0$

Thus, x - 2 is a factor of x^3 + x^2 - 4x - 4

Factorize the polynomial completely.​

$x^3+x^2 - 4x - 4\\\\\left(x^3+x^2\right)+\left(-4x-4\right)\\\\\mathrm{Factor\:out\:}-4\mathrm{\:from\:}-4x-4\mathrm{:\quad }\\\\\left(x^3+x^2\right)-4\left(x+1\right)\\\\\mathrm{Factor\:out\:}x^2\mathrm{\:from\:}x^3+x^2\\\\x^2(x+1)-4(x+1)\\\\\mathrm{Factor\:out\:common\:term\:}x+1\\\\\left(x+1\right)\left(x^2-4\right)\\\\\mathrm{Factor}\:x^2-4:\quad \left(x+2\right)\left(x-2\right)\\\\\left(x+1\right)\left(x+2\right)\left(x-2\right)$

Thus the given polynomial is factored

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