Question

What is the complete factorization of the polynomial below?

x3 + 8x2 + 19x + 12

Answer 1

The correct answer is A

Answer 2

Option A.
`(x+1)(x+3)(x+4)`

Explanation:

Polynomial is an equation of the form
`p(x)=a_nx^n+a_(n-1)x^(n-1)+a_(n-2)x^(n-2)+...+a_1x+a_0` where
`a_n\,,\,a_(n-1)\,,\,a_(n-2)\,,\,...\,,a_1\,,\,a_0` are the coefficients such that
`a_n\\eq 0`

Let
`p(x)=x^3+8x^2+19x+12`

For
`x = -4` ,

`\displaystyle p(-4) &=(-4)^3+8(-4)^2+19(-4)+12\\\displaystyle =-64+128-76+12\\\displaystyle =0\\`

So, x+4 is a factor of p(x) .

{we know that x-a is a factor of p(x) if and only if p(a)=0}

Consider the following:

`p(x)=x^3+8x^2+19x+12\\=x^2(x)+8x(x)+19x+12\\=x^2(x+4)+8x(x+4)+19(x+4)-4x^2-32x-76+12\\=(x+4)(x^2+8x+19)-4x^2-32x-64\\=(x+4)(x^2+8x+19)-4x(x+4)-32(x+4)+16x+128-64\\=(x+4)(x^2+8x+19-4x-32)+16(x+4)+64-64\\=(x+4)(x^2+4x-13+16)\\=(x+4)(x^2+4x+3)\\=(x+4)(x^2+3x+x+3)\\=(x+4)\left [ x(x+3)+1(x+3) \right ]\\=(x+1)(x+3)(x+4)`

So, complete factorisation of p(x) is
`(x+1)(x+3)(x+4)`