Polynomials Comparing coefficients: (x^2-a)(3x-2) = 3x^3+bx^2+cx+10

[sznaiden]

Hi I’m stuck on this question comparing coefficients

 

[Otis]

Hi I’m stuck

Hi sznaiden. Thanks for showing your work, so far.

Each side of your second equation contains an x³-term, an x²-term, an x-term and a constant term. The coefficients on like-terms must be the same because the two sides are equal. We see that's true for the cubed term; the coefficient on each side is 3.

Also, when we write a term containing both variable(s) and parameter(s) like -3xa, we generally position the variable(s) last as shown here:

3x^3 – 2x^2 – 3ax + 2a

just to make clear that -3a is the coefficient.

Here's a hint: Start comparing coefficients on each side beginning with the constant terms (2a and 10). They must be equal, right? After you find the value of a, replace all the a-symbols in your equation with that value, and continue.
[imath]\;[/imath]

 

[stapel]

Hi I’m stuck on this question comparing coefficients:
given \(\displaystyle (x^2-a)(3x-2) = 3x^3+bx^2+cx+10\), find \(\displaystyle a\), \(\displaystyle b\), and \(\displaystyle c\)
\(\displaystyle 3x^3-2x^2-3ax+2a = 3x^3+bx^2+cx+10\)

I'm sorry, but I'm not seeing your work so far (other than the expanded product)...?

You know that you are equating coefficients, so you should have included that part in your work shown:

1. \(\displaystyle 3 = 3\) from the \(\displaystyle x^3\) term
2. \(\displaystyle -2 = +b\) from the \(\displaystyle x^2\) term
3. \(\displaystyle -3a = +c\) from the \(\displaystyle x\) term
4. \(\displaystyle +2a=+10\) from the constant term

Where are you stuck in solving the simple linear equation in (2) (where no work is needed)?
Where are you stuck in (4) (where you need only divide through once)?

Once you have figured out (2) and (4), please reply with your work so far on (3). Thank you!

Eliz.