Question

which shows the four-term polynomial and factored form of \\(x^2 + 6x - 27\\)?

\\(x^2 + 3x - 9x - 27 = (x + 3)(x - 9)\\)
\\(x^2 + 6x - 3x - 27 = (x + 6)(x - 3)\\)
\\(x^2 + 9x - 3x - 27 = (x + 9)(x - 3)\\)
\\(x^2 + 3x - 6x - 27 = (x + 3)(x - 6)\\)

Explanation:

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<reasoning>

Find two numbers that multiply to -27 and add to 6

\[

$$\begin{aligned} &p \cdot q = -27 \\ &p + q = 6 \\ &\implies p = 9,\quad q = -3 \end{aligned}$$

\]

Rewrite the middle term using these two numbers

\[
x^2 + 6x - 27 = x^2 + 9x - 3x - 27
\]

Factor by grouping

\[

$$\begin{aligned} x^2 + 9x - 3x - 27 &= x(x + 9) - 3(x + 9) \\ &= (x + 9)(x - 3) \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \(x^2 + 3x - 9x - 27 = (x + 3)(x - 9)\)</mcq-option>
<mcq-option>(B) \(x^2 + 6x - 3x - 27 = (x + 6)(x - 3)\)</mcq-option>
<mcq-correct>(C) \(x^2 + 9x - 3x - 27 = (x + 9)(x - 3)\)</mcq-correct>
<mcq-option>(D) \(x^2 + 3x - 6x - 27 = (x + 3)(x - 6)\)</mcq-option>
</answer>

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Answer:

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<reasoning>

Find two numbers that multiply to -27 and add to 6

\[

$$\begin{aligned} &p \cdot q = -27 \\ &p + q = 6 \\ &\implies p = 9,\quad q = -3 \end{aligned}$$

\]

Rewrite the middle term using these two numbers

\[
x^2 + 6x - 27 = x^2 + 9x - 3x - 27
\]

Factor by grouping

\[

$$\begin{aligned} x^2 + 9x - 3x - 27 &= x(x + 9) - 3(x + 9) \\ &= (x + 9)(x - 3) \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \(x^2 + 3x - 9x - 27 = (x + 3)(x - 9)\)</mcq-option>
<mcq-option>(B) \(x^2 + 6x - 3x - 27 = (x + 6)(x - 3)\)</mcq-option>
<mcq-correct>(C) \(x^2 + 9x - 3x - 27 = (x + 9)(x - 3)\)</mcq-correct>
<mcq-option>(D) \(x^2 + 3x - 6x - 27 = (x + 3)(x - 6)\)</mcq-option>
</answer>

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