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factor completely. $-7u^2 - 12u + 4$

Question

factor completely.
$-7u^2 - 12u + 4$

Explanation:

Step1: Factor out -1

First, we factor out -1 from the quadratic expression to make the coefficient of \(u^2\) positive. This gives us:
\(-(7u^2 + 12u - 4)\)

Step2: Factor the quadratic inside the parentheses

We need to factor \(7u^2 + 12u - 4\). We look for two numbers that multiply to \(7\times(-4)= -28\) and add up to 12. The numbers are 14 and -2.
We rewrite the middle term using these numbers:
\(7u^2 + 14u - 2u - 4\)

Step3: Group and factor

Group the first two terms and the last two terms:
\((7u^2 + 14u) + (-2u - 4)\)
Factor out the greatest common factor from each group:
\(7u(u + 2) - 2(u + 2)\)

Step4: Factor out the common binomial factor

Now, we can factor out \((u + 2)\) from both terms:
\((7u - 2)(u + 2)\)

Step5: Include the -1 we factored out earlier

Putting it all together with the -1, we get:
\(-(7u - 2)(u + 2)\) or we can also write it as \((-7u + 2)(u + 2)\) (by distributing the -1 to \(7u - 2\))

Answer:

\(-(7u - 2)(u + 2)\) (or \((-7u + 2)(u + 2)\))