Question

select the correct answer.
what is the completely factored form of this expression?
$3x^2 - 17x - 28$

a. $(3x + 4)(x + 7)$
b. $(3x + 4)(x - 7)$
c. $3x^2 - 17x - 28$
d. $(3x^2 + 4)(x - 7)$

Explanation:

Step1: Recall factoring quadratic formula

For a quadratic \(ax^2 + bx + c\), we find two numbers that multiply to \(a\times c\) and add to \(b\). Here, \(a = 3\), \(b=-17\), \(c = -28\). So \(a\times c=3\times(-28)=-84\). We need two numbers that multiply to \(-84\) and add to \(-17\). Those numbers are \(-21\) and \(4\) (since \(-21\times4=-84\) and \(-21 + 4=-17\)).

Step2: Split the middle term

Rewrite the quadratic as \(3x^2-21x + 4x-28\).

Step3: Factor by grouping

Group the first two and last two terms: \((3x^2 - 21x)+(4x - 28)\). Factor out common factors: \(3x(x - 7)+4(x - 7)\). Now factor out \((x - 7)\): \((3x + 4)(x - 7)\).

Answer:

B. \((3x + 4)(x - 7)\)