Hey guys i have three questions can you help me? Ive just started college and I need help with this basic algebra.
1. 2a^2+7ab+3b^2
2. 3a^2-8ab-28b^2
3. 36x^2-12x+1
if you could also explain to me how to do this that would be even better.

- anonymous

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- schrodinger

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- anonymous

please help me!!

- Nnesha

quadratic equations:
there are 4 method to factor out the quadratic equations

- Nnesha

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## More answers

- Nnesha

i'll go with the easy one
Ac method also known as headphone method
where you have to find two numbers when you multiply them you should get product of AC and when you add them yo should get middle term

- anonymous

for the first one I would start it as
(2a+ ?b) (a+?b)

- anonymous

so i would multiply the 2 and the 3 to get the middle term?

- Nnesha

yes right:
find two numbers when you multiply them you should get 6 but when you add them you should get 7ab

- anonymous

that would be 2a^2+8ab+6b^2 thats not right

- Nnesha

hmm

- anonymous

okay im not following why do i add?

- anonymous

or would this one be prime?

- Nnesha

you should find `two` numbers when you `add` them you should get `middle` term
but when you multiply same numbers you should get product of ac
what two numbers would you multiply to get 6 but when you add them you get 7 ?

- anonymous

1 and 6

- Nnesha

yes right
leading coefficient isn't so we have to use factor by groping method |dw:1440352936888:dw| carry down the first and last term
and write the numbers u just found in the middle

- Nnesha

|dw:1440353069776:dw|
make a group of `2` terms
find the common factor of `1st two terms
and then common factor of last 2 terms

- Nnesha

\[\huge\rm \color{Red}{2a^2+6ab}+\color{blue}{1ab+3b^2}\]
what is common in first two terms ?

- anonymous

so that makes (a+3b) (2a+b) right?

- anonymous

because you factor out the highest factor

- anonymous

of each group

- Nnesha

yes that's correct!

- anonymous

okay cool

- Nnesha

you can foil it to check your answer
(2a+b)(a+3b)=2a^2+7ab+3b^2

- anonymous

okay i foiled it and i got the same answer so its correct let me try the other ab problem the problem 36x^2-12x+1 I am also still having troubles with

- Nnesha

do you want to try it first ? :)

- anonymous

im trying the other ab one first.

- Nnesha

okay :=)

- anonymous

okay im stuck
i have 3a^2 -8ab-28b^2
I multiplied the 3 by the 28 and got 84 then used the factors 6 and 14
then I got
3a^2-6ab-42ab-28b^2
and then i put them in groups like you said
(3a^2-6ab) (-42ab -28b^2)
then i took 3 and a out of the first group and negative 14b out of the second group
then I was left with (3a-14b) (a-2b) (3a+2b) im confused on what i did wrong

- Nnesha

oky so by just looking at two numbers i can say thats wrong
6 times 28 =64 okay
you have to be careful about signs

- anonymous

okay

- Nnesha

\[ \huge\rm 3a^2 \color{Red}{-}8ab\color{Red}{-}28b^2 \]
when you multiply two numbers you should get negative 84
bec `3 times -28=-84`
and when you add them you should get `-8`

- Nnesha

so 6 times 14 = 84 not negative
but what if take -14 times 6 =

- anonymous

that equals the -8 that is needed

- Nnesha

yes right thats what we need
and thats two number you should put in the middle http://prntscr.com/880qre

- Nnesha

\[\rm 3a^2+6ab-14ab+28b^2\]

- anonymous

okay like this 3a^2 -14ab+6ab-28b^2

- anonymous

then i separate them into groups and factor

- Nnesha

yes right! :+)

- anonymous

okay i got the correct answer so then it is (3a-14b)(a+2b) and i used foil to make sure it was correct and got 3a^2-8ab-28b^2

- Nnesha

yes right

- anonymous

okay now on to the last problem this one i have started it as |dw:1440355529657:dw|

- anonymous

wait hold on i think i just figured it out

- Nnesha

how do you know it would be one minus and one plus parentheses ?
i mean \[\huge\rm (6x-?)(x+??)\]

- anonymous

oh wow i feel dumb hahha i didnt realize that the number was supposed to be 1
so i did this |dw:1440355938178:dw|

- anonymous

well thank you natasha

- Nnesha

|dw:1440356037300:dw|
that's right !

- Nnesha

there is ano|dw:1440356070415:dw| 0ther way of factoring quadratic i would like to show that one too
~find factor of first and last term
then cross multiply if you get middle term then the numbers you multiplid are factors of the quadratic equation

- Nnesha

|dw:1440356266572:dw|