Question

factor completely.
$2r^3 - 4r^2 + 7r - 14$

Explanation:

Step1: Group the terms

We group the first two terms and the last two terms:
$$(2r^3 - 4r^2)+(7r - 14)$$

Step2: Factor out the GCF from each group

From the first group, the GCF of \(2r^3\) and \(-4r^2\) is \(2r^2\), so we factor that out: \(2r^2(r - 2)\). From the second group, the GCF of \(7r\) and \(-14\) is \(7\), so we factor that out: \(7(r - 2)\). Now the expression becomes:
$$2r^2(r - 2)+7(r - 2)$$

Step3: Factor out the common binomial factor

We can see that \((r - 2)\) is a common factor in both terms, so we factor that out:
$$(r - 2)(2r^2 + 7)$$

Answer:

\((r - 2)(2r^2 + 7)\)