1. ## Polynomials

Given that 3x^3 + 4x^2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C for all values of x, find the values of A, B and C.

What I did:

f(x) = 3x^3 + 4x^2 - 13x - 12

A = 3 (by inspection)

C = f(3) = 66

My questions:
1. Why is it not possible to find C by subsituting -(1/3) into x, since (3x + 1) would = 0?
2. How do I solve for B? (Ans: 4)

2. ## Re: Polynomials

You could expand the right side of the given equation and equate like coefficients to find B.

3. ## Re: Polynomials

Originally Posted by MarkFL2
You could expand the right side of the given equation and equate like coefficients to find B.
I guess you mean this.

3x^3 + 4x^2 - 13x - 12 = (3x^2 - 8x - 3)(x + B) + C

4 = 3B - 8
3B = 12
B = 4

Why can't I seem to get the answer when I compare coefficients of x, though?
-13x = -8Bx - 3x
-13 = -8B - 3
8B = 16
B = 2

4. ## Re: Polynomials

Are you sure you aren't given:

$\displaystyle 3x^3+4x^2-13x-12=(Ax+1)(x+B)(x-3)+Cx$ ?

In this case, expanding the right side, we have:

$\displaystyle 3x^3+4x^2-13x-12=Ax^3+(AB-3A+1)x^2-(3AB-B-C+3)x-(3B)$

This implies:

$\displaystyle A=3$

$\displaystyle 3B-9+1=4\:\therefore\:B=4$

$\displaystyle 36-4-C+3=13\:\therefore\:C=22$

5. ## Re: Polynomials

Nope, there's no wrong with the question.

6. ## Re: Polynomials

3x^3 + 4x^2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C
There are no constants A,B,C that make this equation true for all x. This can be proven by contradiction from your own reasoning. A must be 3, so that means C must equal both f(3) and f(-1/3), however f(3) is not equal to f(-1/3).

7. ## Re: Polynomials

hmm, i am intrigued.

suppose: 3x3 + 4x2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C

this means: 3x3 + 4x2 - 13x - 12 = Ax3 + (AB - 3A + 1)x2 + (-3AB + B - 3)x - 3B + C

equating coefficients:

A = 3
AB - 3A + 1 = 4
-3AB + B - 3 = -13
-3B + C = -12

the last equation gives us: B = (C+12)/3. so we have 2 equations in C:

3((C+12)/3) - 9 + 1 = 4, which leads to: C = 0.

(-9)((C+12)/3) + (C+12)/3 - 3 = -13, which leads to: C = -33/4.

this makes no sense at all....i strongly suggest you check the problem again.

8. ## Re: Polynomials

I checked again... no problems. The answers I got/given might not be correct though. I think I may ask my teacher tomorrow...

9. ## Re: Polynomials

Originally Posted by fActor
Given that 3x^3 + 4x^2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C for all values of x, find the values of A, B and C.

What I did:

f(x) = 3x^3 + 4x^2 - 13x - 12

A = 3 (by inspection)

C = f(3) = 66

My questions:
1. Why is it not possible to find C by subsituting -(1/3) into x, since (3x + 1) would = 0?
2. How do I solve for B? (Ans: 4)
Your reasoning to get A and C are perfect. Your question #1 is very good - it shows you're thinking.

For reasons that will become apparent, let me do #2 first:
Since you've found C, and you're looking for B, notice that B shows up in the constant term. You get the constant term by setting x=0:

Plug in $\displaystyle x=0$ into: $\displaystyle 3x^3 + 4x^2 - 13x - 12 = (3x + 1)(x + B)(x - 3) + 66$, and you get

$\displaystyle -12 = (1)(B)(-3) + 66$, so $\displaystyle -12 = -3B +66$, so $\displaystyle -3B = -78$, so $\displaystyle B = 26$.

So you're done (um... not really): $\displaystyle 3x^3 + 4x^2 - 13x - 12 = (3x + 1)(x + 26)(x -3) + 66$.

What about your concern over problem #1? That would worry me too. So let's check the answer:

$\displaystyle (3x + 1)(x + 26)(x -3) + 66 = (3x + 1)(x^2 + 23x - 78)$

$\displaystyle = 3x^3 + (1 + 69)x^2 + (23 - 234)x - 78 = 3x^3 + 70x^2 - 211x - 78$.

Ooops. What did we do wrong?

Nothing.

The problem is itself wrong. There ARE no such numbers A, B, and C that make those polymonials equal. The setup declared otherwise, and we did everything right, GIVEN their false assertion that those two polynomials can be made equal.

And that gives the answer to your #1: "Yes, you're absolutely right. Plugging in $\displaystyle x = -1/3$ should've caused the polynomial to evaluate to 66. But since there was a false assumption to begin with, everything after that error is without valid mathematical justification."

If you let me begin by assuming 1 = 0, I can pretty much prove anything. (In fact, I can prove anything in one line. I think it's called "ex contradictione non sequitur quodlibet")

Here's what's going on. A 3rd degree polynomial has 4 coefficients. Thus when you equate those polynomials, you'll have to equate them coefficient by coefficient. That gives you 4 equations. How many unknowns? Just 3: A, B, and C. Whenever you have more equations than unknowns, you run the risk of having no solutions. And that's what's happened here.

Working it out:

$\displaystyle (Ax + 1)(x + B)(x -3) + C = (Ax + 1)(x^2 + (B-3)x -3B) + C$

$\displaystyle = A x^3 + ( A(B-3) + 1) x^2 + ( -3AB + (B-3) ) x + (-3B) + C$

$\displaystyle = A x^3 + ( AB -3A + 1) x^2 + ( -3AB + B -3 ) x + ( C - 3B )$

So if that's going to equal $\displaystyle 3x^3 + 4x^2 - 13x - 12$, then A, B, and C must satisfy all 4 of these equations:

$\displaystyle A = 3, AB -3A + 1 = 4, -3AB + B -3 = -13$, and $\displaystyle C - 3B = -12$.

Plugging in A = 3 and other simplifying gives 3 equations in the 2 unknonws B and C:

$\displaystyle 3B - 9 + 1 = 4, -9B + B - 3 = -13, C - 3B = -12$.

Just the first two become: $\displaystyle 3B = 12$, and $\displaystyle -8B = -10$.

Thus $\displaystyle B = 4$, and $\displaystyle B = 5/4$, and... well, that's enough to show that the problem has no solution.

10. ## Re: Polynomials

This is what I did. I divided $\displaystyle 3x^3+4x^2-13x-12$ by $\displaystyle x-3$ to get $\displaystyle 3x^2+13x+26+\frac{66}{x-3}$ the 66 is your C value because it was added on. So I solved for what the true equation is by solving for z. $\displaystyle -12 = 66 + z => z = -78 => 3x^2+13x+26+\frac{66}{x-3}$$\displaystyle = (3x^2+13x+26)(x-3)+66$ By using the quadratic equation on $\displaystyle 3x^2+13x+26$ you can solve for your other unknown values.