Question

factories:x^4+2x^3-7x^2-8x+12

Answer 1

Hey friend, Harish here.

Here is your answer:

Solution:

⇒ $x^{4}+2x^{3}-7x^{2}-8x+12 = (x^{4}-8x)+(2x^{3}-7x^{2}+12)$

⇒ $x(x^{3}-8)+(2x^{3}-7x^{2}+12)=x(x^{3}-2^{3})+(2x^{3}-7x^{2}+12)$

We know that, a³ - b³ =  (a-b)(a² + ab + b²)

⇒ $x(x^{3}-2^{3})+(2x^{3}-7x^{2}+12)$

⇒ $x(x-2)(x^{2}+2x+4) +(2x^{3}-7x^{2}+12)$

⇒ $(x-2)(x)(x^{2}+2x+4) + ((x-2)(2x^{2}-3x-6))$

⇒ $(x-2)(x^{3} +2x^{2}+4x +2x^{2}-3x-6)$

⇒ $(x-2)(x^{3}+4x^{2}+x-6)$

⇒ $(x-2)(x-1)(x^{2}+5x+6)$

⇒  $(x-1)(x-2)(x^{2}+2x+3x+6)$

⇒ $(x-1)(x-2)(x(x+2)+3(x+2))$

⇒  $\bold{(x-1)(x-2)(x+2)(x+3)}$
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Hope my answer is helpful to you.  

Answer 2

Answer:

Answer:

x^4-2x^3-7x^2+8x+12

=x^4-3x^3+x^3-3x^2-4x^2+12x-4x+12

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

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

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

=(x-3)[x^2 (x+2)-x (x+2)-2 (x+2)]

=(x-3)(x+2)(x^2-x-2)

=(x-3)(x+2)(x^2+x-2x-2)

=(x-3)(x+2)[x (x+1)-2 (x+1)]

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

This is your required answer

Step-by-step explanation: