MHB Find Polynomial Given Remainder After Division

[Monoxdifly]

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

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 [math]x^3[/math]) as [math]-\frac13[/math]. Can you tell me what's wrong?

 

[MarkFL]

I would begin by writing:

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

Or:

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

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

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

$$f(1)=a+b+c+d=3$$

Next, we may write:

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

Hence:

$$f(0)=d=-2$$

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

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

$$-27a+9b-3c=-3\implies -9a+3b-c=-1$$

$$a+b+c=5$$

$$-8a+4b-2c=-6\implies -4a+2b-c=-3$$

Solving this system, there results:

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

Thus:

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

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 \(R(x)\):

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

or:

$$f(x)=Q(x)D(x)+R(x)$$

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

$$f(r)=R(r)$$

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:

$$f(-2)=3(-2)-2=-8$$

 

[Monoxdifly]

Okay, it is clear now. Thanks Mark!