You are able to cancel further on the LHS. Can you do that? Notice that 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
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 −Ax2+Bx+C(x+5)(2x+3)≡6−xx+5.
Distributing the -ve into the denominator of the RHS gives an equivalent relation of Ax2+Bx+C(x+5)(2x+3)≡x−6x+5.
Now you can denote y=Ax2+Bx+C (for the sake of simplicity). Can you rearrange this identity for y now?
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 x coefficients are positive on that RHS fraction.
(*) note that −ab=a−b=−ab are all the same thing.
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 x coefficients are equal positive on that RHS fraction.
(*) note that −ab=a−b=−ab are all the same thing.
Oh my goodness Ive got it!! SO is the answer A=2 B=-.9 C=-18 ???