Question

show what you know : factor trinomials
3 of 4
factor : $3x^2 - 20x - 7$
$3x + 1$
$3x - 1$
$x + 7$
$x - 7$

Explanation:

Step1: Recall factoring trinomial \(ax^2 + bx + c\)

For \(3x^2 - 20x - 7\), \(a = 3\), \(b=-20\), \(c = -7\). We need two numbers \(m\) and \(n\) such that \(m\times n=3\times(-7)= - 21\) and \(m + n=-20\). The numbers are \(-21\) and \(1\) since \(-21\times1=-21\) and \(-21 + 1=-20\).

Step2: Rewrite the middle term

Rewrite \(-20x\) as \(-21x+x\): \(3x^2-21x + x-7\).

Step3: Group and factor

Group the first two and last two terms: \((3x^2 - 21x)+(x - 7)=3x(x - 7)+1(x - 7)\). Then factor out \((x - 7)\): \((3x + 1)(x - 7)\). So the factors are \(3x + 1\) and \(x - 7\).

Answer:

A. \(3x + 1\), D. \(x - 7\) (assuming the options are labeled A: \(3x + 1\), B: \(3x - 1\), C: \(x + 7\), D: \(x - 7\))