I'm taking a 13 question test with Algebra and its module 8.01 - 8.04 . If you could help me with 2 questions, 1 from each so I understand each method, that's be great :)

[8.03]:Factor completely: x2 − 3x + 7
[8.04]: Factor completely: 2a2 − a − 10

Question

I'm taking a 13 question test with Algebra and its module 8.01 - 8.04 .  If you could help me with 2 questions, 1 from each so I understand each method, that's be great :)

[8.03]:Factor completely: x2 − 3x + 7
[8.04]: Factor completely: 2a2 − a − 10

anonymous · Answer

I was helped with 8.01 and I completely understand it now :)

ganeshie8 · Answer

good :) lets see 8.02

ganeshie8 · Answer

a2b + 3a2 + 2b + 6

ganeshie8 · Answer

it has 4 terms

ganeshie8 · Answer

factor first two separately, and last two separately

ganeshie8 · Answer

a2b + 3a2 + 2b + 6 
^        ^

amistre64 · Answer

does your schools "honor code" have any misgivings about posting these test questions ?

amistre64 · Answer

or perhaps im reading your post incorrectly, and these are examples of some problems that the test is about?

its important that we do not violate a schools honor code.  posting similar questions to get help with regards to a method is more suitable.  So its prolly best, to avoid any confusion, if you just leave out the whole "this is a test" information that may or maynot be getting misread.

anonymous · Answer

No, it shouldn't because they have the same thing on their page but I don't understand how they put it.

anonymous · Answer

I don't understand how to factor them

ganeshie8 · Answer

a2b + 3a2 + 2b + 6 
^        ^

ganeshie8 · Answer

find the gcf of those two terms

anonymous · Answer

a?

ganeshie8 · Answer

thats all ?

anonymous · Answer

a^2

ganeshie8 · Answer

yes ! so pull that out

ganeshie8 · Answer

a2b + 3a2 + 2b + 6 
^        ^

a2(b+3) + 2b + 6

anonymous · Answer

(b + 3)(a + 2)?

ganeshie8 · Answer

nope, lets go step by step ok

anonymous · Answer

okay..

ganeshie8 · Answer

a2(b+3) + 2b + 6
                 ^      ^

ganeshie8 · Answer

you can pull out 2 from those two terms

ganeshie8 · Answer

a2(b+3) + 2b + 6
                 ^      ^

a2(b+3) + 2(b+3)

anonymous · Answer

Okay, I see what you did.  After that would you distribute?

ganeshie8 · Answer

no, (b+3) is common in both terms... factor that out

anonymous · Answer

ganes is doing good job so ill leave this too him lol

ganeshie8 · Answer

ty :)

ganeshie8 · Answer

a2(b+3) + 2b + 6
                 ^      ^

a2(b+3) + 2(b+3)

(b+3)(a2+2)

ganeshie8 · Answer

we're done.

anonymous · Answer

np :P

anonymous · Answer

So you just took the ones before the parenthesis and put them in parenthesis last after all the other work?

ganeshie8 · Answer

not exactly

ganeshie8 · Answer

a2(b+3) + 2b + 6
                 ^      ^

a2(b+3) + 2(b+3)

ganeshie8 · Answer

see that, (b+3) is there in both terms

anonymous · Answer

right

ganeshie8 · Answer

since its there in both terms, you can pull that out

ganeshie8 · Answer

a2(b+3) + 2(b+3)
    -----      ----

ganeshie8 · Answer

that leaves a2 in first term, and 2 in second term.

anonymous · Answer

Ohhhh, okay

ganeshie8 · Answer

a2(b+3) + 2(b+3)
    -----      ----


(b+3)(a2+2)
_____

ganeshie8 · Answer

two down !! two more to go ha...  :)

ganeshie8 · Answer

8.03 - maybe post it as a new q, someone will explain...

anonymous · Answer

Okay, quick question.. Is 2x3 + 6x2 + 2x + 6 
 (2x + 6)(x2 + 1) ?

ganeshie8 · Answer

Yes ! you can factor out 2 also like below :-

(2x + 6)(x2 + 1)

2(x+3)(x2+1)

anonymous · Answer

Awesome ! thanks sooo much !!

ganeshie8 · Answer

you're wlcme :)