11. Given a polynomial with the degree 3. If it is divided by $\displaystyle x^2+2x-3$, the remainder is 2x + 1. If it is divided by $\displaystyle x^2+2x$, the remainder is 3x - 2. The polynomial is ....
A. $\displaystyle \frac23x^3+\frac43x^2+3x-2$
B. $\displaystyle \frac23x^3+\frac43x^2+3x+2$
C. $\displaystyle \frac23x^3+\frac43x^2-3x+2$
D. $\displaystyle x^3+2x^2+3x-2$
E. $\displaystyle 2x^3+4x^2+3x+2$

The book says that the answer is A, but I don't understand the part when they suddenly substitute f(-2) = -8, where does that come from? I tried doing it myself and got a (the coefficient of $\displaystyle x^3$) as $\displaystyle -\frac13$. Can you tell me what's wrong?

plug $-2$ into polynomial A

\begin{align*} &\dfrac 2 3 (-2)^3 + \dfrac 4 3 (-2)^2 +3(-2) - 2 = \\ \\ &-\dfrac{16}{3} + \dfrac{16}{3} -6 -2 = -8 \end{align*}

I get

$\displaystyle -\frac{1}{3}x^3+\frac{1}{3}x^2+5x-2$

Let's say the original polynomial is $a(x)$. Then you have:

$$a(x) = p(x)(x^2+2x-3)+2x+1$$ Equation 1

$$a(x) = q(x)(x^2+2x)+3x-2$$ Equation 2

$x^2+2x=0$ when $x=0$ or $x=-2$.

So, from the second equation, we have $a(0) = -2$ and $a(-2) = -8$. Plugging in zero to each of the given polynomials, you limit the possibilities to A or D. Plugging in $x=-2$ to only those two, you get -8 for both A and D still. So, let's find where $x^2+2x-3=0$.

$$(x+3)(x-1) = 0$$

$$x=-3,1$$

Using the first equation, we plug in and find $a(-3) = p(-3)(0)+2(-3)+1 = -5$ and $a(1) = p(1)(0)+2(1)+1 = 3$.

Plugging in $x=1$ to A and D gives 3 for A but it gives 4 for D. So, A is the only possibility among the choices. Let's try plugging in -3:
$$\dfrac{2}{3}(-3)^3+\dfrac{4}{3}(-3)^2+3(-3)-2 = -17 \neq -5$$

So, A is not the correct answer either. In fact, there are no correct answers among those given.

Let's check the solution that Idea came up with.
$$f(x) = -\dfrac{x^3}{3}+\dfrac{x^2}{3}+5x-2$$

$$f(0) = -2$$
That checks out.

$$f(-2) = \dfrac{8}{3}+\dfrac{4}{3}+5(-2)-2 = -8$$
That checks out.

$$f(1) = -\dfrac{1}{3}+\dfrac{1}{3}+5-2 = 3$$
That checks out.

$$f(-3) = 9+3-15-2 = -5$$
That checks out.

So, Idea's solution meets all of the criteria we are given and is likely the correct result.

Originally Posted by romsek
plug $-2$ into polynomial A

\begin{align*} &\dfrac 2 3 (-2)^3 + \dfrac 4 3 (-2)^2 +3(-2) - 2 = \\ \\ &-\dfrac{16}{3} + \dfrac{16}{3} -6 -2 = -8 \end{align*}
Turns out A isn't the answer, though.
Thanks for everyone, anyway.