You could expand the right side of the given equation and equate like coefficients to find B.
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)
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
And someone please, please answer my first question... I'm really confused.....
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).3x^3 + 4x^2 - 13x - 12 = (Ax + 1)(x + B)(x -3) + C
hmm, i am intrigued.
suppose: 3x^{3} + 4x^{2} - 13x - 12 = (Ax + 1)(x + B)(x -3) + C
this means: 3x^{3} + 4x^{2} - 13x - 12 = Ax^{3} + (AB - 3A + 1)x^{2} + (-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.
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 into: , and you get
, so , so , so .
So you're done (um... not really): .
What about your concern over problem #1? That would worry me too. So let's check the answer:
.
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 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:
So if that's going to equal , then A, B, and C must satisfy all 4 of these equations:
, and .
Plugging in A = 3 and other simplifying gives 3 equations in the 2 unknonws B and C:
.
Just the first two become: , and .
Thus , and , and... well, that's enough to show that the problem has no solution.