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factor completely. 6f² - 22f + 12

Question

factor completely.
6f² - 22f + 12

Explanation:

Step1: Factor out the GCF

First, find the greatest common factor (GCF) of \(6f^2\), \(-22f\), and \(12\). The GCF of 6, 22, and 12 is 2. So we factor out 2:
\(6f^2 - 22f + 12 = 2(3f^2 - 11f + 6)\)

Step2: Factor the quadratic

Now factor the quadratic \(3f^2 - 11f + 6\). We need two numbers that multiply to \(3\times6 = 18\) and add up to -11. The numbers are -9 and -2. Rewrite the middle term:
\(3f^2 - 9f - 2f + 6\)

Step3: Group and factor

Group the terms:
\((3f^2 - 9f) + (-2f + 6)\)
Factor out the GCF from each group:
\(3f(f - 3) - 2(f - 3)\)

Step4: Factor out the common binomial

Now factor out \((f - 3)\):
\((3f - 2)(f - 3)\)

Step5: Combine with the GCF

Remember we factored out a 2 earlier, so the completely factored form is:
\(2(3f - 2)(f - 3)\)

Answer:

\(2(3f - 2)(f - 3)\)