# Thread: Finding Unknown Coefficients of X (Another Problem)

1. ## Finding Unknown Coefficients of X (Another Problem)

For all x, find the values of the constants a and b in:

$\displaystyle (x+a)(x^2+bx+2)=x^3+2x^2-5x-6$

Multiplied out i get:

$\displaystyle x^3+bx^2+2x+ax^2+abx+2a=x^3+2x^2-5x-6$

I then move the 2x over to the other side of the equation and cancel the cubed x's giving me:

$\displaystyle bx^2+ax^2+2a=-4x^2-x-6$

Then i'm stuck as i have two seperate variables attatched to a like power of x :?

Any help would be great!

2. Originally Posted by BIOS
For all x, find the values of the constants a and b in:
$\displaystyle (x+a)(x^2+bx+2)=x^3+2x^2-5x-6$
Multiplied out i get:
$\displaystyle x^3+bx^2+2x+ax^2+abx+2a=x^3+2x^2-5x-6$
From that point you have:
$(a+b)x^2+(ab)x+2a=2x^2-7x-6$

But there must be something wrong with the question. It does not workout.

My Bad. Was doing the above while watching Dexter :P Okay so the format is still different from the problems i've done before. Does it matter what variable i equate in terms of order?

Can i say:

$\displaystyle 2a=-6$

$\displaystyle a=-3$

And then go from there? I tried but it doesn't work out :?

Sorry that 5x in the original question should read x. Apologies.....

4. With that correction then $b=1$.

5. Okay that works fine with original question. Sorry for the typo thanks for the help! So basically you re-factor when you have an instance of more than one variable attached to a like power of x independently (to make it clearer) and it doesn't matter what order you equate i.e in the above i equated:

$\displaystyle 2a = -6$

$\displaystyle a=-3$

Previously i could always equate the first term on either side of the equal sign whereas this time i obviously couldn't.