# P3 partial fractionsWatch

This discussion is closed.
#1
express as a partial fraction:

3/(x^3+1)
0
14 years ago
#2
x^3+1= x^3 * x^1

then solve by factorising!!
0
14 years ago
#3
(Original post by SAM*)
x^3+1= x^3 * x^1

then solve by factorising!!
I'm pretty sure that x^3*x^1 = x^4
0
14 years ago
#4
(x+1) is a factor of (x^3+1); therefore, use polynomial division to find the full factorization.
0
14 years ago
#5
Express as a partial fraction:

(2x - 7)/(x-5)^2

I wrote:

(2x - 7)/(x-5)^2 = A/(x-5) + B/(x-5)^2

A(x-5)^2 + B(x-5) = 2x - 7

Now if I put x = 5, I will get 0 = -7, so what have I done wrong already? This is what the books tells me to do...

Also, why do you write:

a/(x+2)^2 = A/(x+2) + B/(x+2)^2

and not just:

a/(x+2)^2 = a/(x+2)(x+2) = A/(x+2) + B/(x+2)

Edit - ?!wtf.. sorry I meant to make a new thread, I guess I clicked the wrong button
0
#6
(Original post by mikesgt2)
(x+1) is a factor of (x^3+1); therefore, use polynomial division to find the full factorization.
sorry, i've only just started p3- i dont think i understand!
0
14 years ago
#7
Express as a partial fraction:

(2x - 7)/(x-5)^2

I wrote:

(2x - 7)/(x-5)^2 = A/(x-5) + B/(x-5)^2 correct

A(x-5)^2 + B(x-5) = 2x - 7 here is your mistake - when multiplying through by (x-5)^2 you should get A(x-5) + B = 2x - 7

Now if I put x = 5, I will get 0 = -7, so what have I done wrong already? This is what the books tells me to do...

Also, why do you write:

a/(x+2)^2 = A/(x+2) + B/(x+2)^2

and not just:

a/(x+2)^2 = a/(x+2)(x+2) = A/(x+2) + B/(x+2) because that wouldnt work! its easiest to just accept it! the pattern is always the same for partial fractions with squared/cubed etc. denominators
0
14 years ago
#8
(Original post by hihihihi)
express as a partial fraction:

3/(x^3+1)
x³+1 = (x+1)(x²-x+1)

3/(x³+1) = 3/(x+1)(x²-x+1) = A/(x+1) + (Bx + C)/(x²-x+1)

3 = A(x²-x+1) + (Bx + C)(x+1)

Put x = -1:

3 = A(1+1+1)
A = 1

Equate constants:

3 = A + C
C = 3 - 1 = 2

Equate coefficients of x²:

0 = A + B
B = -1

I hope!
0
14 years ago
#9
(Original post by hihihihi)
sorry, i've only just started p3- i dont think i understand!
what he means is (x+1) multiplied by something gives u (x^3+1),so:
(x+1)(something)=(x^3+1),so :something=(x^3+1)/(x+1),use long division for that.
0
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