Can you guys help me factor trinomials?

Question

I really, really, really, don't understand this stuff. If you could that would be great. Okay...here's the problem:

8a3(the a is to the third just remeber that in all the other ones too.)

8a3 -12a2+6a-9=(I have to group them)

8a3-12a2+6a-9=(8a3-12a2)+(6a-9)
=4a2(2a-3)+3(2a-3)

I thought the answer was (2a-3)(2a-3), but the book says it's (4a2+3)(2a-3). I don't get that. Please explain it to me. Thanks!!

Answer 1

in this type of problem, we'll use "factoring by grouping" because it has 4 terms in it. Which means, we have to group the first two terms (8a3 - 12a) and the last two terms (6a-9)... and just combined them later...

8a3-12a2+6a-9

= (8a3-12a2)+(6a-9) i seperated the first two terms from the last two...

factor the first two... (8a3-12a2)

Look for something that they both have in common... i suggest that u write them in prime factorization so that u'll get the hang of it..

2.2.2.a.a.a = 8a3

2.2.3.a.a= 12a2

what do they have in common? they both have 2.2.a.a... which is equal to 4a2... so that's what u factor out...

4a2( )

if u take out 4a2 from (8a3-12a2), what will be left inside the parenthesis? the answer is (2a-3)

So 4a2(2a-3) is the answer when u try to factor the first two terms....

now let's do the last two (6a-9)

2.3.a =6a

3.3 = 9

they both have 3 in common, just one 3... even though the first one has two, the second one only has one....

so that means 3( )

what should u place in the parenthesis? answer is 2a-3

3(2a-3) u can check it to make sure

now let's combined our two answers

4a2(2a-3) +3(2a-3)

look closely... they have (2a-3) so factor it out....

you'll get (2a-3)(4a2+3) final anawer (don't forget the parenthesis)

Answer 2

The reason you were wrong is because in order for the leading term to be third degree, you needed one of the a's to be raised to the second power: that way the new term would be to the third power.