Question

Simplify the following (6b-2)(6b+2)+4(b^2-3b)

Find the factors of the following
e^2+8e+15
5f^2-15f+10

Does g^2-9 have factors? If so, how there is no "middle" term?

Answer 1

expand it first

(6b-2)(6b+2)+4(b^2-3b)
= 36b^2 - 4 + 4b^2 - 12b
= 40b^2 - 12b - 4

e^2 + 8e + 15
-- can you think of two numbers that add up to 8, and multiply to get 15 ?

5f^2 - 15f + 10
= 5(f^2 - 3f + 2)
= 5(f-1)(f-2)

g^2 - 9
is the difference of squares, you should know that one.

look at the two factors at the front of your first question and what happened when I expanded it.
Can you reverse the process?

Answer 2

(6b-2)(6b+2) +4(b^2-3b)
= 36b^2 -4 +4b^2 -12b
= 40b^2 -12b -4
= 4(10b^2 -3b -1)
= 4(5b -1)(2b + 1)
I don't what would be considered the simplets form.

e^2 +8e +15 = (e+3)(e+5)

5f^2 -15f +10 = 5(f^2 -3f +2)
= 5(f-1)(f-2)

Does g^2-9 have factors?
yes. (g-3)(g+3)
The middle terms can cancel whenever you have the difference of two squares.