anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
please help me!!
Nnesha
  • Nnesha
quadratic equations: there are 4 method to factor out the quadratic equations
Nnesha
  • Nnesha
s*

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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