The Student Room Group

Help please... partial fractions!!!!!!

Given that

(x^2-36)/(x^2-11x+30) * (25-x^2)/(Ax^2+Bx+C) * (6x^2+7x-3)/(3x^2+17x-6) (x+5)/(6-x)

Find the values of the constants A, B and C where A, B and C are integers.


thanks

Scroll to see replies

Original post by Kercmaher
Given that

(x^2-36)/(x^2-11x+30) * (25-x^2)/(Ax^2+Bx+C) * (6x^2+7x-3)/(3x^2+17x-6) (x+5)/(6-x)

Find the values of the constants A, B and C where A, B and C are integers.


thanks


What have you tried?
Complete the squares and cancel down some terms in the numerator and denominator.

I did this in my recent mock exam O_o

Make sure you notice that there will be 2 terms which are similar BUT do not "cancel down", make sure you look carefully.
(edited 5 years ago)
Reply 3
Original post by RDKGames
What have you tried?

Ive tried timesing the numerators all together and the denomenators all together... and got a big mess
The question is a lot, if you get a piece of paper and write it landscape, it will be a lot easier to see.
Reply 5
Original post by Guarddyyy
The question is a lot, if you get a piece of paper and write it landscape, it will be a lot easier to see.

Okay just a moment
Original post by Kercmaher
Ive tried timesing the numerators all together and the denomenators all together... and got a big mess


It would be good if you can simplify x236x211x+30\dfrac{x^2 -36}{x^2-11x+30} and 6x2+7x33x2+17x6\dfrac{6x^2+7x-3}{3x^2+17x-6} first.
These quadratics all factorise.
Original post by Kercmaher
Okay just a moment


I misled you, it's a difference of 2 squares - aka (x+y)(x-y)
Reply 8
Original post by RDKGames
It would be good if you can simplify x236x211x+30\dfrac{x^2 -36}{x^2-11x+30} and 6x2+7x33x2+17x6\dfrac{6x^2+7x-3}{3x^2+17x-6} first.
These quadratics all factorise.

Okay so I have factorised the first fraction to (x+6)/(x-5)
Original post by Kercmaher
Okay so I have factorised the first fraction to (x+6)/(x-5)


And the second fraction?
Reply 10
Original post by RDKGames
And the second fraction?


(2x+3)/(x+6)

thank you so much for the help btw haha
Original post by RDKGames
What have you tried?


Hi,

Kindly DM me.
Original post by Kercmaher
(2x+3)/(x+6)

thank you so much for the help btw haha


Yep. So now you have that:

x+6x5(5+x)(5x)Ax2+Bx+C2x+3x+6x+56x\dfrac{x+6}{x-5} \cdot \dfrac{(5+x)(5-x)}{Ax^2+Bx+C} \cdot \dfrac{2x+3}{x+6} \equiv \dfrac{x+5}{6-x}

You are able to cancel further on the LHS. Can you do that? Notice that 5x=(x5)5-x = -(x-5)
Reply 13
Original post by RDKGames
Yep. So now you have that:

x+6x5(5+x)(5x)Ax2+Bx+C2x+3x+6x+56x\dfrac{x+6}{x-5} \cdot \dfrac{(5+x)(5-x)}{Ax^2+Bx+C} \cdot \dfrac{2x+3}{x+6} \equiv \dfrac{x+5}{6-x}

You are able to cancel further on the LHS. Can you do that? Notice that 5x=(x5)5-x = -(x-5)

So the (x-5) and the -(x-5) cancel and the x+6? To give (x+5)(2x+3) on the top and (Ax^2+Bx+C) on the bottom? not sure where the - from the front of -(x-5) goes at the top
Original post by Kercmaher
So the (x-5) and the -(x-5) cancel and the x+6? To give (x+5)(2x+3) on the top and (Ax^2+Bx+C) on the bottom? not sure where the - from the front of -(x-5) goes at the top


The - can stay in front of the fraction, so that you now have (x+5)(2x+3)Ax2+Bx+Cx+56x-\dfrac{(x+5)(2x+3)}{Ax^2+Bx+C} \equiv \dfrac{x+5}{6-x}.

Distributing the -ve into the denominator of the RHS gives an equivalent relation of (x+5)(2x+3)Ax2+Bx+Cx+5x6\dfrac{(x+5)(2x+3)}{Ax^2+Bx+C} \equiv \dfrac{x+5}{x-6}.

Now you can denote y=Ax2+Bx+Cy=Ax^2+Bx+C (for the sake of simplicity). Can you rearrange this identity for yy now?
The (5-x) and (x-5) cancel to give (-1), which is the key to getting the right solutions

Because... if x = 10, (5-10)/(10-5) = -5/5 = -1
(edited 5 years ago)
Reply 16
Original post by RDKGames
The - can stay in front of the fraction, so that you now have (x+5)(2x+3)Ax2+Bx+Cx+56x-\dfrac{(x+5)(2x+3)}{Ax^2+Bx+C} \equiv \dfrac{x+5}{6-x}.

Distributing the -ve into the denominator of the RHS gives an equivalent relation of (x+5)(2x+3)Ax2+Bx+Cx+5x6\dfrac{(x+5)(2x+3)}{Ax^2+Bx+C} \equiv \dfrac{x+5}{x-6}.

Now you can denote y=Ax2+Bx+Cy=Ax^2+Bx+C (for the sake of simplicity). Can you rearrange this identity for yy now?

Sorry to be annoying and rubbish at this but I dont understand how you moved the -1 to the denominator of the RHS?
Original post by Kercmaher
Sorry to be annoying and rubbish at this but I dont understand how you moved the -1 to the denominator of the RHS?


Multiply both sides by -1. The -1 on the LHS gets cancelled but now you gained it on the RHS. It doesn't matter whether you distribute it into the numerator or denominator (*), so give it to the denomonator so that your xx coefficients are positive on that RHS fraction.

(*) note that ba=ba=ba-\dfrac{b}{a} = \dfrac{-b}{a} = \dfrac{b}{-a} are all the same thing.
(edited 5 years ago)
Reply 18
Original post by RDKGames
Multiply both sides by -1. The -1 on the LHS gets cancelled but now you gained it on the RHS. It doesn't matter whether you distribute it into the numerator or denominator (*), so give it to the denomonator so that your xx coefficients are equal positive on that RHS fraction.

(*) note that ba=ba=ba-\dfrac{b}{a} = \dfrac{-b}{a} = \dfrac{b}{-a} are all the same thing.

Oh my goodness Ive got it!! SO is the answer A=2 B=-.9 C=-18 ???
Original post by Kercmaher
Oh my goodness Ive got it!! SO is the answer A=2 B=-.9 C=-18 ???


Yes. :smile:

Quick Reply

Latest