Answer:
Option A. [tex](x+1)(x+3)(x+4)[/tex]
Step-by-step explanation:
Polynomial is an equation of the form [tex]p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0[/tex] where [tex]a_n\,,\,a_{n-1}\,,\,a_{n-2}\,,\,...\,,a_1\,,\,a_0[/tex] are the coefficients such that [tex]a_n\neq 0[/tex]
Let [tex]p(x)=x^3+8x^2+19x+12[/tex]
For [tex]x = -4[/tex] ,
[tex]\displaystyle p(-4) &=(-4)^3+8(-4)^2+19(-4)+12\\\displaystyle =-64+128-76+12\\\displaystyle =0\\[/tex]
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:
[tex]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)[/tex]
So, complete factorisation of p(x) is [tex](x+1)(x+3)(x+4)[/tex]
