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------------------------------------------------ Prove that $(x + 2)$ is a factor of $6x^3 + 19x^2 + 16x + 4$ by using the factor theorem. Also, factorize $6x^3 + 19x^2 + 16x + 4$.

A. Proof: Divide $6x^3 + 19x^2 + 16x + 4$ by $(x + 2)$ and show that the remainder is zero.
Factorization: $(x + 2)(3x^2 + 13x + 2)$

B. Proof: Divide $6x^3 + 19x^2 + 16x + 4$ by $(x + 2)$ and show that the remainder is zero.
Factorization: $(x + 2)(3x^2 + 10x + 2)$

C. Proof: Divide $6x^3 + 19x^2 + 16x + 4$ by $(x + 2)$ and show that the remainder is 6.
Factorization: $(x + 2)(3x^2 + 13x + 2)$

D. Proof: Divide $6x^3 + 19x^2 + 16x + 4$ by $(x + 2)$ and show that the remainder is 4.
Factorization: $(x + 2)(3x^2 + 10x + 2)$

To proof that '(x + 2)' is a factor of '6x^3 + 19x^2 + 16x + 4', substitute -2 (the value to make x + 2 equal to zero) into the polynomial equation. If resultant is zero, then '(x + 2)' is a factor. For the factorization, divide the given polynomial by '(x + 2)', and the quotient is '(3x^2 + 7x + 2)'.

Explanation:

The subject of this question is algebra, specifically polynomial division and the factor theorem. To prove that '(x + 2)' is a factor of the polynomial '6x^3 + 19x^2 + 16x + 4', you need to use the Factor Theorem which states that a polynomial f(x) has a factor '(x - p)' if and only if f(p) = 0. In this scenario, p = -2 because '(x + 2)' is equivalent to '(x - -2)'.

Substitute -2 into 6x^3 + 19x^2 + 16x + 4 and if the value obtained is zero, then '(x + 2)' is a factor. Therefore, 6*(-2)^3 + 19*(-2)^2 + 16*(-2) + 4 = 0 which confirms that '(x + 2)' is a factor. For the factorization, divide the given polynomial by (x+2) using long division or synthetic division. The quotient you obtain is the other factor, which should be '(3x^2 + 7x + 2)'. Hence, option B is correct.

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