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.

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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.

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please help me!!
quadratic equations: there are 4 method to factor out the quadratic equations
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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
for the first one I would start it as (2a+ ?b) (a+?b)
so i would multiply the 2 and the 3 to get the middle term?
yes right: find two numbers when you multiply them you should get 6 but when you add them you should get 7ab
that would be 2a^2+8ab+6b^2 thats not right
hmm
okay im not following why do i add?
or would this one be prime?
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 ?
1 and 6
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
|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
\[\huge\rm \color{Red}{2a^2+6ab}+\color{blue}{1ab+3b^2}\] what is common in first two terms ?
so that makes (a+3b) (2a+b) right?
because you factor out the highest factor
of each group
yes that's correct!
okay cool
you can foil it to check your answer (2a+b)(a+3b)=2a^2+7ab+3b^2
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
do you want to try it first ? :)
im trying the other ab one first.
okay :=)
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
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
okay
\[ \huge\rm 3a^2 \color{Red}{-}8ab\color{Red}{-}28b^2 \] when you multiply two numbers you should get negative 84 bec `3 times -28=-84` and when you add them you should get `-8`
so 6 times 14 = 84 not negative but what if take -14 times 6 =
that equals the -8 that is needed
yes right thats what we need and thats two number you should put in the middle http://prntscr.com/880qre
\[\rm 3a^2+6ab-14ab+28b^2\]
okay like this 3a^2 -14ab+6ab-28b^2
then i separate them into groups and factor
yes right! :+)
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
yes right
okay now on to the last problem this one i have started it as |dw:1440355529657:dw|
wait hold on i think i just figured it out
how do you know it would be one minus and one plus parentheses ? i mean \[\huge\rm (6x-?)(x+??)\]
oh wow i feel dumb hahha i didnt realize that the number was supposed to be 1 so i did this |dw:1440355938178:dw|
well thank you natasha
|dw:1440356037300:dw| that's right !
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
|dw:1440356266572:dw|