How do you factor #8b ^ { 2} - 12b + 4#?

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Answer 1 · 2017-07-22T02:22:41

Use the factors of #8b^2# and #4# to factor out the polynomial.

#8b^2-12b+4#

Since they are all factors of 4, you can easily simplify them into simpler numbers.

#=4(2b^2-3b+1)#

Now, just factor out the first and last term.

#=(2b-1)(b-1)#

Answer 2 · 2017-07-22T02:33:04

Here's how I factor it. (Other people use different detaiols.)

First remove the common factor of #4#

#8b^2-12b+4 = 4(2b^2-3b+1)#

Now we need to factor #2b^2-3b+1#

Think about FOIL. If this can be factored using whole numbers we must have

#F = 2b^2#
#O+I = -3b# and
#L = +1#

So we start with

#2b^2-3b+1 = (2b +- "something")(b +- "something")#

Since #L = +1# both #"something"#'s must be #+1# or both are #-1#

(Or the expression cannot be factored using whole numbers.)

Try both possibilities

#(2b+1)(b+1) = 2b^2 "(of course") +2b+b " STOP!"# that will give us #+3b# -- not what we want.

Check the other possibility maybe wither one works and the expression cannot be factored using whole numbers.

#(2b-1)(b-1) = 2b^2 "(of course") -2b-b +1 = 2b^2-3b+1# Good! That's it.

#8b^2-12b+4 = 4(2b^2-3b+1) = 4(2b-1)(b-1)#