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Ask by George Alexander. in the United States

Mar 25,2025

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Question Watch Video Show fxamplea Factor completely. $$ x^{3}+6 x^{2}-14 x-84 $$ Answer Attompt in ons of o $$ \square $$ Andmit Answer

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1. We are given the polynomial  
   $$
   x^3 + 6x^2 - 14x - 84
   $$
   and we want to factor it completely.

2. First, we look for a rational root. By testing values, we can try $$x = -6$$. Substitute $$x = -6$$ into the polynomial:
   $$
   (-6)^3 + 6(-6)^2 - 14(-6) - 84 = -216 + 216 + 84 - 84 = 0.
   $$
   Since the result is zero, $$x = -6$$ is a root, which means $$(x + 6)$$ is a factor.

3. Next, we divide the polynomial by $$(x + 6)$$ using synthetic division with the value $$-6$$. The coefficients of the polynomial are $$1$$ (for $$x^3$$), $$6$$ (for $$x^2$$), $$-14$$ (for $$x$$), and $$-84$$.

$$
   \begin{array}{r|rrrr}
   -6 & 1 & 6 & -14 & -84 \
      &   & -6 & 0  & 84  \
   \hline
      & 1 & 0 & -14 & 0 \
   \end{array}
   $$
   
   The synthetic division gives a quotient of $$x^2 - 14$$.

4. Therefore, we have factored the polynomial as:
   $$
   x^3 + 6x^2 - 14x - 84 = (x + 6)(x^2 - 14).
   $$

5. The quadratic factor $$x^2 - 14$$ does not factor further over the integers. If desired, it can be factored over the real numbers as:
   $$
   x^2 - 14 = (x - \sqrt{14})(x + \sqrt{14}).
   $$

6. The completely factored form over the integers is:
   $$
   (x + 6)(x^2 - 14).
   $$

$$
\square
$$
The polynomial $$ x^3 + 6x^2 - 14x - 84 $$ factors completely as $$ (x + 6)(x^2 - 14) $$.

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Question Watch Video Show fxamplea Factor completely. Answer Attompt in ons of o Andmit Answer