FACTORING!
please help me factor x^2+3x+9/4

- anonymous

FACTORING!
please help me factor x^2+3x+9/4

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- anonymous

@jagr2713

- myininaya

hint: this may help"
\[x^2+bx+\frac{c}{a} \\ \frac{1}{a}(ax^2+bax+\frac{c(a)}{a}) \\ \frac{1}{a}(ax^2+bax+c)\]

- anonymous

ok so my equation should look like \[1/4(4x^2+3(4)x+9)\]

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## More answers

- myininaya

yep it is like dividing everything by 4 at the same time as multiplying everything by 4
and we know 4/4=1
so yep

- myininaya

and 3(4) can be written as 12 of course

- myininaya

you can also see if you need to factor anymore

- anonymous

right, so now my equation should be \[x^2+12+9\]

- myininaya

well you are missing 1/4( )
and a 4 in front of x^2
and a x next to the 12

- myininaya

\[\frac{1}{4}(4x^2+12x+9)\]

- myininaya

anyways let's look at the thing in ( )

- myininaya

can you find two numbers that multiply to be 4(9)=36
and add up to be 12

- anonymous

something that are multiples to 36 and add to 12?

- myininaya

yep

- anonymous

6 and 6

- myininaya

yes yes! :)

- myininaya

\[\frac{1}{4}(4x^2+6x+6x+9)\]
now we can ignore the 1/4 and bring it down later
but we need to factor the thing inside the ( )by grouping

- myininaya

we will bring the 1/4 down later
\[4x^2+6x+6x+9 \\ (4x^2+6x)+(6x+9) \\ 2x(2x+3)+(6x+9)\]
you try factoring the 6x+9

- anonymous

(6x+9) will equal 3x(2x+3)

- myininaya

well 9 doesn't have a factor of x
so you cannot factor an x out of it unless you divide by x
but that is making is nastier than it should be
6x+9=3(2x+3)

- myininaya

we will bring down 1/4 later
\[2x(2x+3)+3(2x+3)\]
now one last step

- anonymous

oh ok, that makes sense!

- anonymous

I'm not sure how to do the last step

- myininaya

example to factor
a(b+c)+m(b+c)
notice both terms both a(b+c) and m(b+c) have the common factor (b+c)
so I can factor (b+c) out of both terms in the sum
(b+c)(a+m)

- myininaya

so you see (2x+3) in both terms
the terms being 2x(2x+3) and 3(2x+3)

- myininaya

what is left of the first term when factoring out the (2x+3)
is 2x
and of the second term is +3

- myininaya

2x(2x+3)+3(2x+3)
(2x+3)(2x+3)

- myininaya

don't forget to bring down the 1/4

- anonymous

so its just 1/4(2x+3)(2x+3)

- myininaya

or 1/4(2x+3)^2

- myininaya

or...

- myininaya

\[\frac{1}{4}(2x+3)^2 \\ \frac{1}{2^2}(2x+3)^2 \\ \frac{1^2}{2^2}(2x+3)^2 \\ (\frac{1}{2}(2x+3))^2 \\ (\frac{2x+3}{2})^2 \]
there is another or :p

- myininaya

you could separate the fraction inside the ( )^2

- myininaya

but anyways yeah you are done

- anonymous

so its just \[(x+\frac{ 3 }{ 2 })^2\]

- myininaya

yep

- anonymous

Thank you so much, I really appreciate your help!

- myininaya

if we would have noticed the following to begin with:
\[x^2+3x+(\frac{3}{2})^2=(x+\frac{3}{2})^2 \]
we wouldn't had so much work

- myininaya

\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \]

- anonymous

oh yes! That's very true, we missed that. Oh well :)

- myininaya

for example:
\[x^2+5x+(\frac{5}{2})^2=(x+\frac{5}{2})^2 \\ x^2-5x+(\frac{5}{2})^2=(x-\frac{5}{2})^2 \\ x^2-3x+(\frac{-3}{2})^2=(x-\frac{3}{2})^2 \\ x^2-2x+(\frac{-2}{2})^2=(x-1)^2 \]

- anonymous

that actually makes perfect sense, thank you so much for showing me that formula! It's a lot easier than all the other ways I have been taught.

- myininaya

no problem :)

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