1. ## (Simple?) Polynomial problem

Hi, I am from Poland so please forgive my english.

I seem to have problems solving the following assignment (translated)

"There are polynomials:

P(x) = (2x+1)^3
Q(x) = 8x+a
R(x) = 4x^2-1
S(x) = 8x^2 + cx

What
a and c is required for P(x) + Q(x) - 3R(x) = S(x)?"

After many conversions, I came up with this:

(2x+1)^3 +
a = 12x^2 - 3 + cx

But after that I have no idea what next, plus I may not even be heading the right direction... help?

2. Hello, Skydriver!

There must be a typo.
As stated, the problem has no solution.

$\begin{array}{ccc}P(x) &=& (2x+1)^3 \\ Q(x) &=& 8x+a \\ R(x) &=& 4x^2-1 \\ S(x) &=& 8x^{{\color{red}3}} + cx \end{array}$

What $a$ and $c$ is required for: . $P(x) + Q(x) - 3R(x) \:= \:S(x)$ ?
The equation is:

. . $P(x) + Q(x) - 3R(x) \;=\;S(x)$

. . $(2x+1)^3 + (8x+a) - 3(4x^2-1) \;=\;8x^3 + cx$

. . $8x^3 + 12x^2 + 6x + 1 + 8x + a - 12x^2 + 3 \;=\;8x^3 + cx$

. . $14x + a + 4 \;=\;cx$

We have two equal polynomials.
Two polynomials are equal if their corresponding coefficients are equal.

So we have: . $\begin{array}{ccccccc}14 \:=\: c & \Rightarrow & \boxed{c \:=\:14} \\ a+4 \:=\:0 & \Rightarrow &\boxed{ a \:=\:\text{-}4} \end{array}$