How do you factor 8b ^ { 2} - 12b + 48b212b+4?

2 Answers
Jul 22, 2017

Use the factors of 8b^28b2 and 44 to factor out the polynomial.

Explanation:

8b^2-12b+48b212b+4

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

=4(2b^2-3b+1)=4(2b23b+1)

Now, just factor out the first and last term.

=(2b-1)(b-1)=(2b1)(b1)

Jul 22, 2017

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

Explanation:

First remove the common factor of 44

8b^2-12b+4 = 4(2b^2-3b+1)8b212b+4=4(2b23b+1)

Now we need to factor 2b^2-3b+12b23b+1

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

F = 2b^2F=2b2
O+I = -3bO+I=3b and
L = +1L=+1

So we start with

2b^2-3b+1 = (2b +- "something")(b +- "something")2b23b+1=(2b±something)(b±something)

Since L = +1L=+1 both "something"something's must be +1+1 or both are -11

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

Try both possibilities

(2b+1)(b+1) = 2b^2 "(of course") +2b+b " STOP!"(2b+1)(b+1)=2b2(of course)+2b+b STOP! that will give us +3b+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(2b1)(b1)=2b2(of course)2bb+1=2b23b+1 Good! That's it.

8b^2-12b+4 = 4(2b^2-3b+1) = 4(2b-1)(b-1)8b212b+4=4(2b23b+1)=4(2b1)(b1)