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OpenStudy (anonymous):

7X^2 +2 all over (2x+1)(x^2 +1)

OpenStudy (anonymous):

what about it?

OpenStudy (anonymous):

i got A/2x+1 + Bx+C/(X^2+1)

OpenStudy (anonymous):

is that right? because of the x^2

OpenStudy (anonymous):

this one is easier

OpenStudy (anonymous):

i figure out A = 3

OpenStudy (anonymous):

no. factors in denominator are only \[(2x+1)(x^2+1)\]

OpenStudy (anonymous):

x is not a factor of the denominator

OpenStudy (anonymous):

oh no that is not right sorry

OpenStudy (anonymous):

B = 2?

OpenStudy (anonymous):

harder, not easier. because \[x^2+1\] has degree = 2, you need \[\frac{Ax+B}{x^2+1}+\frac{C}{2x+1}\]

OpenStudy (anonymous):

i led you astray. because the denominator has degree 2, the numerator can have degree 1

OpenStudy (anonymous):

so this one is harder

OpenStudy (anonymous):

you have \[(Ax+B)(2x+1)+c(x^2+1)=7x^2+2\]

OpenStudy (anonymous):

now you have to multiply out and equate like coefficients. you get \[2Ax^2+Ax+2Bx+B+Cx^2+C=7x^2+2\]

OpenStudy (anonymous):

oh ok thats the step i miss

OpenStudy (anonymous):

now collect terms on the left. i hope i didn't screw this up. you get \[(2A+C)x^2+(A+2B)x+(B+C)=7x^2+2\]

OpenStudy (anonymous):

giving the following equations by "equating like coefficients" \[2A+C=7\] \[A+2B=0\]\[B+C=2\]

OpenStudy (anonymous):

hope it is clear where i got those from

OpenStudy (anonymous):

what did you do?

OpenStudy (anonymous):

first step was to collect terms. that is for powers of x

OpenStudy (anonymous):

\[2Ax^2+Cx^2=(2A+C)x^2\] for example that is on the left. on the right you have \[7x^2\] so you know that \[2A+C=7\]

OpenStudy (anonymous):

likewise \[Ax+2Bx=(A+2B)x\] on the left, and you have no x term on the right, so you know \[A+2B=0\]

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

and finally the "constant" on the left is \[B+C\] and on the right it is 2 giving \[B+C=2\]

OpenStudy (anonymous):

now your job is to solve this system of equations. 3 equations, 3 unknowns

OpenStudy (anonymous):

may ways but i will say this: second equation gives \[A=-2B\]put that in first equation you get \[-4B+C=7\] \[B+C=2\] subtract to get \[-5B=5\] so \[B=-1\]

OpenStudy (anonymous):

now it is easy, since \[B+C=2\] \[-1+C=2\] \[C=3\]

OpenStudy (anonymous):

\[2A+C=7\]\[2A+3=7\] \[2A=4\] \[A=2\]

OpenStudy (anonymous):

problem cooked up with nice integer solutions. as you can imagine it is usually not so simple at the last step

OpenStudy (anonymous):

hahhaa thank you soo much!

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