Question

factor completely.
-2w² - w + 21

Explanation:

Step1: Factor out -1

First, we factor out -1 from the quadratic expression to make the coefficient of \(w^2\) positive. So, \(-2w^2 - w + 21=- (2w^2 + w - 21)\)

Step2: Factor the quadratic inside the parentheses

Now we need to factor \(2w^2 + w - 21\). We look for two numbers that multiply to \(2\times(-21)=- 42\) and add up to 1. The numbers are 7 and - 6.
We rewrite the middle term using these numbers: \(2w^2+7w - 6w - 21\)
Then we group the terms: \((2w^2 + 7w)+(-6w - 21)\)
Factor out the common factors from each group: \(w(2w + 7)-3(2w + 7)\)
Now we can factor out \((2w + 7)\): \((2w + 7)(w - 3)\)

Step3: Combine with the -1 factored out earlier

Since we had factored out -1 in Step 1, the completely factored form of the original expression is \(-(2w + 7)(w - 3)\) or we can also write it as \((-2w - 7)(w - 3)\) (by distributing the -1 to one of the factors) or \((2w + 7)(-w + 3)\)

Answer:

\(-(2w + 7)(w - 3)\) (or equivalent forms like \((-2w - 7)(w - 3)\) or \((2w + 7)(-w + 3)\))