If -2 is a zero of the cubic polynomial p(x) | vedclass

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(D) Given that $-2$ is a zero of $p(x) = x^3 + 6x^2 + 11x + 6$,it implies that $(x + 2)$ is a factor of $p(x)$.To find the remaining zeros,divide $p(x

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$-1, -3$

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(D) Given that $-2$ is a zero of $p(x) = x^3 + 6x^2 + 11x + 6$,it implies that $(x + 2)$ is a factor of $p(x)$.To find the remaining zeros,divide $p(x)$ by $(x + 2)$ using polynomial long division or synthetic division:$(x^3 + 6x^2 + 11x + 6) \div (x + 2) = x^2 + 4x + 3$.Now,factorize the quadratic polynomial $x^2 + 4x + 3$:$x^2 + 3x + x + 3 = x(x + 3) + 1(x + 3) = (x + 1)(x + 3)$.Setting these factors to zero gives $x + 1 = 0$ and $x + 3 = 0$,which results in $x = -1$ and $x = -3$.Thus,the remaining zeros are $-1$ and $-3$.

If $-2$ is a zero of the cubic polynomial $p(x) = x^3 + 6x^2 + 11x + 6$,then find the remaining zeros of $p(x)$.

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