Question

factor the polynomial. $2x^2 + 13x - 7$
a $(x - 2)(x + 7)$
b $(x - 7)(2x + 1)$
c $(x + 7)(2x - 1)$
d $(2x + 7)(x - 1)$
e $(x - 7)(2x - 1)$

Explanation:

Step1: Recall factoring method

For a quadratic \(ax^2 + bx + c\), find two numbers \(m\) and \(n\) such that \(m \times n = a \times c\) and \(m + n = b\). Here, \(a = 2\), \(b = 13\), \(c = -7\). So \(m \times n = 2\times(-7)= -14\) and \(m + n = 13\). The numbers are \(14\) and \(-1\) (since \(14\times(-1)= -14\) and \(14 + (-1)=13\)).

Step2: Split the middle term

Rewrite \(13x\) as \(14x - x\):
\(2x^2 + 14x - x - 7\)

Step3: Group and factor

Group the first two and last two terms:
\((2x^2 + 14x) + (-x - 7)\)
Factor out common terms:
\(2x(x + 7) - 1(x + 7)\)

Step4: Factor out \((x + 7)\)

\((x + 7)(2x - 1)\)

Answer:

C. \((x + 7)(2x - 1)\)