# Finding Unknown Coefficients of X (Another Problem)

• Sep 24th 2010, 01:15 PM
BIOS
Finding Unknown Coefficients of X (Another Problem)
For all x, find the values of the constants a and b in:

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

Multiplied out i get:

\$\displaystyle \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 \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!
• Sep 24th 2010, 01:25 PM
Plato
Quote:

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

From that point you have:
\$\displaystyle (a+b)x^2+(ab)x+2a=2x^2-7x-6\$

But there must be something wrong with the question. It does not workout.
• Sep 24th 2010, 01:38 PM
BIOS

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 \displaystyle 2a=-6\$

\$\displaystyle \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.....
• Sep 24th 2010, 01:43 PM
Plato
With that correction then \$\displaystyle b=1\$.
• Sep 24th 2010, 01:51 PM
BIOS
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 \displaystyle 2a = -6\$

\$\displaystyle \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.