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QUESTION IMAGE

factor the polynomial completely using the x method. \\(x^2 + 16x + 48\…

Question

factor the polynomial completely using the x method.

\\(x^2 + 16x + 48\\)

which equivalent four-term polynomial can be created using the x method?

\\(x^2 + 8x + 8x + 48\\)
\\(x^2 - 12x - 4x + 48\\)
\\(x^2 + 12x + 4x + 48\\)
\\(x^2 - 8x - 8x + 48\\)

Explanation:

Response

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"new_concepts": [
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<reasoning>

Identify coefficients for the X method

For the trinomial \(x^2 + 16x + 48\), the standard form is \(ax^2 + bx + c\):
\[
a = 1, \quad b = 16, \quad c = 48
\]
The top of the X contains \(ac = 1 \cdot 48 = 48\), and the bottom contains \(b = 16\).

Find two factors

We need two numbers that multiply to \(ac = 48\) and add to \(b = 16\):
\[
12 \cdot 4 = 48 \quad \text{and} \quad 12 + 4 = 16
\]

Rewrite the middle term

Using the factors \(12\) and \(4\), split the middle term \(16x\) into \(12x + 4x\):
\[
x^2 + 12x + 4x + 48
\]
</reasoning>

<answer>
<mcq-option>(A) \(x^2 + 8x + 8x + 48\)</mcq-option>
<mcq-option>(B) \(x^2 - 12x - 4x + 48\)</mcq-option>
<mcq-correct>(C) \(x^2 + 12x + 4x + 48\)</mcq-correct>
<mcq-option>(D) \(x^2 - 8x - 8x + 48\)</mcq-option>
</answer>

<post_analysis>
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Answer:

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<reasoning>

Identify coefficients for the X method

For the trinomial \(x^2 + 16x + 48\), the standard form is \(ax^2 + bx + c\):
\[
a = 1, \quad b = 16, \quad c = 48
\]
The top of the X contains \(ac = 1 \cdot 48 = 48\), and the bottom contains \(b = 16\).

Find two factors

We need two numbers that multiply to \(ac = 48\) and add to \(b = 16\):
\[
12 \cdot 4 = 48 \quad \text{and} \quad 12 + 4 = 16
\]

Rewrite the middle term

Using the factors \(12\) and \(4\), split the middle term \(16x\) into \(12x + 4x\):
\[
x^2 + 12x + 4x + 48
\]
</reasoning>

<answer>
<mcq-option>(A) \(x^2 + 8x + 8x + 48\)</mcq-option>
<mcq-option>(B) \(x^2 - 12x - 4x + 48\)</mcq-option>
<mcq-correct>(C) \(x^2 + 12x + 4x + 48\)</mcq-correct>
<mcq-option>(D) \(x^2 - 8x - 8x + 48\)</mcq-option>
</answer>

<post_analysis>
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