Find Polynomial Given Remainder After Division

[Monoxdifly]

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?

 

[MarkFL]

I would begin by writing:

\(\displaystyle \frac{ax^3+bx^2+cx+d}{(x+3)(x-1)}=Q(x)+\frac{2x+1}{(x+3)(x-1)}\)

Or:

\(\displaystyle f(x)=ax^3+bx^2+cx+d=Q_1(x)(x+3)(x-1)+2x+1\)

Now, we may state, by looking at the roots of the divisor:

\(\displaystyle f(-3)=-27a+9b-3c+d=-5\)

\(\displaystyle f(1)=a+b+c+d=3\)

Next, we may write:

\(\displaystyle f(x)=ax^3+bx^2+cx+d=Q_2(x)x(x+2)+3x-2\)

Hence:

\(\displaystyle f(0)=d=-2\)

\(\displaystyle f(-2)=-8a+4b-2c+d=-8\)

Now, with \(d=-2\), we obtain the following 3 X 3 system of equations:

\(\displaystyle -27a+9b-3c=-3\implies -9a+3b-c=-1\)

\(\displaystyle a+b+c=5\)

\(\displaystyle -8a+4b-2c=-6\implies -4a+2b-c=-3\)

Solving this system, there results:

\(\displaystyle (a,b,c,d)=\left(-\frac{1}{3},\frac{1}{3},5,-2\right)\)

Thus:

\(\displaystyle f(x)=-\frac{1}{3}x^3+\frac{1}{3}x^2+5x-2\)

I have verified, using W|A, that the above cubic results in the correct remainders in both cases. Your book is wrong about A being the correct choice, I would say.

 

[Monoxdifly]

Glad to see I was right, though I still can't comprehend why you got f(-2) as -8.

 

[MarkFL]

Glad to see I was right, though I still can't comprehend why you got f(-2) as -8.

Suppose \(f(x)\) is divided by the divisor \(D(x)\). This will result in a quotient \(Q(x)\) and a remainder \(R(x)\):

\(\displaystyle \frac{f(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}\)

or:

\(\displaystyle f(x)=Q(x)D(x)+R(x)\)

Now, if \(x=r\) is a root of the divisor such that \(D(r)=0\), then we will have:

\(\displaystyle f(r)=R(r)\)

Does that make sense?

 

[Monoxdifly]

I see, so you substituted it to 3x - 2.

 

[MarkFL]

I see, so you substituted it to 3x - 2.

Yes, using the reasoning in my previous post, we can write:

\(\displaystyle f(-2)=3(-2)-2=-8\)

 

[Monoxdifly]

Okay, it is clear now. Thanks Mark!