By using the factor theorem,show that $(x+2)$ is a factor of the polynomial $6x^3 + 19x^2 + 16x + 4$ and then factorise the polynomial completely.
(A) Let $p(x) = 6x^3 + 19x^2 + 16x + 4$. According to the factor theorem,$(x+2)$ is a factor if $p(-2) = 0$.
$p(-2) = 6(-2)^3 + 19(-2)^2 + 16(-2) + 4 = 6(-8) + 19(4) - 32 + 4 = -48 + 76 - 32 + 4 = 0$.
Since $p(-2) = 0$,$(x+2)$ is a factor.
Now,divide $6x^3 + 19x^2 + 16x + 4$ by $(x+2)$ to get the quadratic $6x^2 + 7x + 2$.
Factorising $6x^2 + 7x + 2 = 6x^2 + 4x + 3x + 2 = 2x(3x+2) + 1(3x+2) = (2x+1)(3x+2)$.
Thus,the complete factorisation is $(x+2)(2x+1)(3x+2)$.
$p(-2) = 6(-2)^3 + 19(-2)^2 + 16(-2) + 4 = 6(-8) + 19(4) - 32 + 4 = -48 + 76 - 32 + 4 = 0$.
Since $p(-2) = 0$,$(x+2)$ is a factor.
Now,divide $6x^3 + 19x^2 + 16x + 4$ by $(x+2)$ to get the quadratic $6x^2 + 7x + 2$.
Factorising $6x^2 + 7x + 2 = 6x^2 + 4x + 3x + 2 = 2x(3x+2) + 1(3x+2) = (2x+1)(3x+2)$.
Thus,the complete factorisation is $(x+2)(2x+1)(3x+2)$.
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