anonymous
  • anonymous
Partial fractions help.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
.......yes??
anonymous
  • anonymous
will someone please set up (4 5 or 6) and do one of 7 and 8 please?
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amistre64
  • amistre64
lets do 4 :)

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amistre64
  • amistre64
3 ----------- (2x+1)(3x-2)
amistre64
  • amistre64
We need on "fraction" for each factor in the bottom. A B 3 ------ + ------ = ----------- (2x+1) (3x-2) (2x+1)(3x-2) is thsi correct?
anonymous
  • anonymous
That looks good
amistre64
  • amistre64
After we split it up into its "baser" units, we can try to "re-solve" it.
amistre64
  • amistre64
A(3x-2) B(2x+1) ----------- + ----------- correct? (2x+1)(3x-2) (3x-2)(2x+1)
amistre64
  • amistre64
the bottoms become redundant and we want the tops to equal out.
anonymous
  • anonymous
Are you asking me whether you're right or are you just putting the questions there?
anonymous
  • anonymous
guys no 4 is just setting up
amistre64
  • amistre64
A(3x-2) + B(2x+1) = 3 lol.... they are really rhetorical, but I do tend to mess up alot ;)
amistre64
  • amistre64
3x(A) -2(A) +2x(B) +1(B) = 3 combine like terms...
amistre64
  • amistre64
3x(A) + 2x(B) -2(A) +1(B) ; factor out the variables now
amistre64
  • amistre64
x(3A +2B) -1(2A +B) = 0x + 3
amistre64
  • amistre64
3A +2B = 0 -2A -B = 3
amistre64
  • amistre64
we can solve for A and B thru a system of equations now
amistre64
  • amistre64
B = -3-2A 3A +2(-3-2A) = 0 3A -6 -4A = 0 -A = +6 A = -6
amistre64
  • amistre64
B = -3-2(-6) B = -3+12 B = 9
amistre64
  • amistre64
A B 3 ------ + ------ = ----------- (2x+1) (3x-2) (2x+1)(3x-2) -3 9 3 ------ + ------ = ----------- (2x+1) (3x-2) (2x+1)(3x-2)
amistre64
  • amistre64
and thats our decomposition; any questions?
anonymous
  • anonymous
for the set up only where should i stop?
amistre64
  • amistre64
you should stop when all the possible "denominators" are accounted for: for example, if the denominator is (x+2)^3 we need to account for.. (x+2) and (x+2)^2 and (x+2)^3 as possible denominators in our setup
anonymous
  • anonymous
what do we do for 3 terms in numerator?
amistre64
  • amistre64
the 3 in the numberator is what we use to set up the condition for our "unknowns". A + B have to end up equaling 3 in the end.... its our goal, our guide, our calibration.
amistre64
  • amistre64
If we get stuff like x^2(A+B) and there is no value for x^2 in our numberator, we set it up by 0x^2
amistre64
  • amistre64
I might be missing a "+C" in there but I cant remember :)
anonymous
  • anonymous
can you do 5 or 6 please? i am confused
amistre64
  • amistre64
I can...... :)
amistre64
  • amistre64
3x^2 +x -1 -------------- (x^2+2) (x^2+1) can we factor the bottom any more? I dont see that we can....
anonymous
  • anonymous
no
amistre64
  • amistre64
A B 3x^2 +x -1 ------- + ------- = ---------- (x^2+2) (x^2+1) yadayada Try to "solve" by getting like denominators now :) like normal...
amistre64
  • amistre64
A(x^2+1) +B(x^2+2) = 3x^2 +x -1 ------------------ ----------- yadayada yadayada makes sense?
anonymous
  • anonymous
yeah. thank you. I just needed the concept.
amistre64
  • amistre64
now we can "turn" that left side into the right side like this :)
amistre64
  • amistre64
x^2(A)+1(A) +x^2(B)+2(B) = 3x^2 +x -1 x^2(A) + x^2(B) +(A+2B) = 3x^2 +x -1 x^2(A+B) + (A+2B) = 3x^2 +x -1 ------------------ x^2(A+B) = 3x^2 A + B = 3 -------------------
amistre64
  • amistre64
A + 2B = -1 A = -1 -2B ------------------
amistre64
  • amistre64
-1 -2B +B = 3 -B = 4 B = -4
anonymous
  • anonymous
i get it now.
anonymous
  • anonymous
thank you very much
anonymous
  • anonymous
gtg now
anonymous
  • anonymous
seeya
amistre64
  • amistre64
youre welcome :) bye..

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