anonymous
  • anonymous
FACTORING! please help me factor x^2+3x+9/4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@jagr2713
myininaya
  • 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
  • anonymous
ok so my equation should look like \[1/4(4x^2+3(4)x+9)\]

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myininaya
  • 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
  • myininaya
and 3(4) can be written as 12 of course
myininaya
  • myininaya
you can also see if you need to factor anymore
anonymous
  • anonymous
right, so now my equation should be \[x^2+12+9\]
myininaya
  • myininaya
well you are missing 1/4( ) and a 4 in front of x^2 and a x next to the 12
myininaya
  • myininaya
\[\frac{1}{4}(4x^2+12x+9)\]
myininaya
  • myininaya
anyways let's look at the thing in ( )
myininaya
  • myininaya
can you find two numbers that multiply to be 4(9)=36 and add up to be 12
anonymous
  • anonymous
something that are multiples to 36 and add to 12?
myininaya
  • myininaya
yep
anonymous
  • anonymous
6 and 6
myininaya
  • myininaya
yes yes! :)
myininaya
  • 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
  • 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
  • anonymous
(6x+9) will equal 3x(2x+3)
myininaya
  • 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
  • myininaya
we will bring down 1/4 later \[2x(2x+3)+3(2x+3)\] now one last step
anonymous
  • anonymous
oh ok, that makes sense!
anonymous
  • anonymous
I'm not sure how to do the last step
myininaya
  • 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
  • myininaya
so you see (2x+3) in both terms the terms being 2x(2x+3) and 3(2x+3)
myininaya
  • myininaya
what is left of the first term when factoring out the (2x+3) is 2x and of the second term is +3
myininaya
  • myininaya
2x(2x+3)+3(2x+3) (2x+3)(2x+3)
myininaya
  • myininaya
don't forget to bring down the 1/4
anonymous
  • anonymous
so its just 1/4(2x+3)(2x+3)
myininaya
  • myininaya
or 1/4(2x+3)^2
myininaya
  • myininaya
or...
myininaya
  • 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
  • myininaya
you could separate the fraction inside the ( )^2
myininaya
  • myininaya
but anyways yeah you are done
anonymous
  • anonymous
so its just \[(x+\frac{ 3 }{ 2 })^2\]
myininaya
  • myininaya
yep
anonymous
  • anonymous
Thank you so much, I really appreciate your help!
myininaya
  • 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
  • myininaya
\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \]
anonymous
  • anonymous
oh yes! That's very true, we missed that. Oh well :)
myininaya
  • 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
  • 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
  • myininaya
no problem :)

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