using the x method to factor

$ax^2 + bx + c$

which statement is true of $x^2 + 8x - 6$?

(x - 2) is a factor of the polynomial.
(x + 2) is a factor of the polynomial.
(x - 4) is a factor of the polynomial.
the polynomial is prime.

Question

using the x method to factor

$ax^2 + bx + c$

which statement is true of $x^2 + 8x - 6$?

(x - 2) is a factor of the polynomial.
(x + 2) is a factor of the polynomial.
(x - 4) is a factor of the polynomial.
the polynomial is prime.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "X Method Factoring", "Factoring Polynomials" ], "new_concepts": [], "current_concepts": [ "X Method Factoring", "Factoring Polynomials", "Prime Polynomials" ] } ## Identify the coefficients of the quadratic polynomial $$ \begin{aligned} &x^2 + 8x - 6\ &a = 1,\quad b = 8,\quad c = -6\ &ac = 1 \cdot (-6) = -6 \end{aligned} $$ ## Find factors of ac that sum to b $$ \begin{aligned} &\text{Factors of } -6: \ &1 \text{ and } -6 \implies \text{sum} = -5 \ &-1 \text{ and } 6 \implies \text{sum} = 5 \ &2 \text{ and } -3 \implies \text{sum} = -1 \ &-2 \text{ and } 3 \implies \text{sum} = 1 \end{aligned} $$ ## Determine if the polynomial is factorable $$ \begin{aligned} &\text{No integer factors of } -6 \text{ sum to } 8.\ &\text{Therefore, the polynomial cannot be factored over the integers.}\ &\text{The polynomial is prime.} \end{aligned} $$

(x - 2) is a factor of the polynomial.

(x + 2) is a factor of the polynomial.

(x - 4) is a factor of the polynomial.

The polynomial is prime.

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Factoring Polynomials" ] }

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QUESTION IMAGE

Question

Explanation:

Identify the coefficients of the quadratic polynomial

Find factors of ac that sum to b

Determine if the polynomial is factorable

Answer: