Question

(i)Express
2 − x + 8x2
(1 − x)(1 + 2x)(2 + x)
in partial fractions.

Answer 1

Is that:\[\bf \large \frac{ 2-x+8x^2 }{ (1-x)(1+2x)(2+x) }\]
@eyna_nuzrey

Answer 2

yes! @genius12

Answer 3

Notice that we have 3 linear factors in the denominator, i.e. \(\bf 1-x , 1+2x, 2+x\).
So we "decompose" this fraction in to 3 partial fractions with each linear factor as the denominator:\[\bf \frac{2-x+8x^2}{(1-x)(1+2x)(2+x)}=\frac{ A }{ 1-x }+\frac{ B }{ 1+2x }+\frac{ C }{ 2+x }\]

Answer 4

Because each factor is linear, we can assume that the numerator of each fraction is a "constant".

Answer 5

@eyna_nuzrey By any chance, are you trying to integrate this rational function through the method of partial fraction decomposition?

Answer 6

nope @genius12

Answer 7

so you've just been asked to find the partial fractions I see. Well, do you understand up to point I've showed you so far? Like how I broke up that 1 fraction in to 3 fractions each with a linear factor and the numerator of each is assumed to be some constant? @eyna_nuzrey

Answer 8

yes @genius12

Answer 9

Ok now to solve for each constant A, B, C, we know that if we were to combine the 3 fractions in to 1 single fraction, we would have to equalise their denominators. When we do that and put our 3 partial fractions all under 1 fraction we see the following:\[\bf \small \frac{ A }{ 1-x }+\frac{ B }{ 1+2x }+\frac{ C }{ 2+x }=\frac{ A(1+2x)(2+x)+B(1-x)(2+x)+C(1-x)(1+2x) }{ (1-x)(1+2x)(2+x) }\]

Answer 10

@eyna_nuzrey Do you see what I just did and do you understand how I got there?

Answer 11

@eyna_nuzrey You know the drill: You have to reply.

Answer 12

yes i understand it.. i also got that but i don't know what to do for the next step after i expanding it.. @genius12

Answer 13

Now we know from original rational function that our numerator is \(\bf 2-x+8x^2\).
This means that we can equate the numerator of our original rational function and the numerator that we just got after combining the 3 partial fractions because we know they must be equal:\[\bf A(1+2x)(2+x)+B(1-x)(2+x)+C(1-x)(1+2x)=2-x+8x^2\]

Answer 14

Do you understand this step? @eyna_nuzrey

Answer 15

yes @genius12

Answer 16

Ok good. Now we set 'x' to a certain value so that 2 of the 3 constants A,B,C cancel out so that we can solve for 1. First I will solve for A by setting x = 1. By doing that, the 'B' and 'C' will cancel out since they are multiplied by (x - 1) and setting x = 1 will make them 0. This will allow me to solve for 'A':\[\bf A(1+2(1))(2+(1))+B(0)(2+1)+C(0)(1+2(1))=2-(1)+8(1)^2\]\[\bf 9A=9 \implies A = 1\]Similarly, we solve for 'B' and 'C' by setting x to x = -1/2 and x = -2 respectively. As you can see, in each case, the key is to set 'x' such that the other two constants get cancelled out so that you can solve for one of them. Now I will solve for B by setting x = -1/2:\[\bf B(1-(-1/2))(2+(-1/2))=2-(-1/2)+8(-1/2)^2\]\[\bf \frac{ 9 }{ 4 }B=\frac{9}{2} \implies B = 2\]Now I will solve for C by setting x = -2:\[\bf C(1-(-2))(1+2(-2))=2-(-2)+8(-2)^2\]\[\bf -9C=36 \implies C = -4\]Now that we know A, B, C, we can plug these in values back in to our partial fraction decomposition: \[\bf \frac{ 2-x+8x^2 }{ (1-x)(1+2x)(2+x) }=\frac{ A }{ 1-x }+\frac{ B }{ 1+2x }+\frac{ C }{ 2+x }\]\[\bf =\frac{ 1 }{ 1-x }+\frac{ 2}{ 1+2x }+\frac{ -4 }{ 2+x }=\frac{ 1 }{ 1-x }+\frac{ 2}{ 1+2x }-\frac{ 4 }{ 2+x }\]

Answer 17

@eyna_nuzrey And we are done. I hope you understood the process.

Answer 18

ok thank you @genius12

Answer 19

@eyna_nuzrey Make sure you read through what I did and understand it fully.

Answer 20

@genius12 ok done!

Answer 21

@genius12 wait for a minute.. i'm trying to understand the answer...

Answer 22

lol ok

Answer 23

Ask me any questions you have. Is there still something you don't understand?

Answer 24

@genius12 nope.. thanks for helping me..