# Polynomial with two unknowns

1. Jan 26, 2008

### AFG34

[SOLVED] polynomial with two unknowns

1. The problem statement, all variables and given/known data
The graph of f(x)= 3x^4 + 14x^3 + px^2 + qx + 24 has x-intercepts -4 and 2. Determine the function.

3. The attempt at a solution

I could solve it if there were only one unknown but i don't know how to do it if there are two unknowns.

What i did so far is plug -4 for x and 0 for the output, got an expression = 0
did the same thing for 2
since both equal 0, i set them equal to each other, simplified and got: 2p-q=13.3
Don't know what to do next.

2. Jan 26, 2008

### rocomath

Rather then setting them equal to each other. Solve for p with one of your x-intercepts then plug it in your other set with the other x-intercept.

3. Jan 26, 2008

### AFG34

but there are two unknowns in each expression

4. Jan 26, 2008

### Hootenanny

Staff Emeritus
What roco is getting at is that you can create a system of two simultaneous equations thus;

$$f(-4) = 0$$

$$f(2) = 0$$

5. Jan 26, 2008

### AFG34

yes i know that, that is what i initially did. So you get 2 equations, both equal to 0, both have q and p in them. But i don't know what to do next.

6. Jan 26, 2008

### Hootenanny

Staff Emeritus
Have you never solved simultaneous equations before?

7. Jan 26, 2008

### AFG34

no, i haven't.

8. Jan 26, 2008

### Hootenanny

Staff Emeritus
Okay, in that case if you post the two equations you obtain I shall walk you through the process.

9. Jan 26, 2008

### AFG34

1) 0 = 16p - 4q - 104
2) 0 = 4p +2q + 184

then i divided both by 2:
1) 0 = 8p - 2q - 52
2) 0 = 2p + q + 92

10. Jan 26, 2008

### Hootenanny

Staff Emeritus
Good, so now multiply (2) by 2 and then add the two equations.

11. Jan 26, 2008

### rocomath

Did you ever learn to solve matrices in Algebra class?

12. Jan 26, 2008

### AFG34

8p + 132

ok so p = -11, q = -70

thnx

Last edited: Jan 26, 2008
13. Jan 26, 2008

### AFG34

nope

14. Jan 26, 2008

### rocomath

I'll go step by step. Let me type this up.

15. Jan 26, 2008

### Hootenanny

Staff Emeritus
If the OP hasn't met simultaneous equations it's pretty safe to say that they haven't been introduced to linear algebra (which IMHO is over-kill for a question such as this).
Firstly, what you have written is an expression, not an equation. Secondly, you might want to check your coefficent of p.

16. Jan 26, 2008

### AFG34

0 = 12p + 132 is an equation
ya i typed it wrong
checked the back of the book, got the right answer (f(x)= 3x^4 + 14x^3 - 11x^2 - 70x + 24). ok i know how to do them now, thnx

hootenanny, funny name

Last edited: Jan 26, 2008
17. Jan 26, 2008

### rocomath

Want to learn the fast way?

18. Jan 26, 2008

### AFG34

sure

got a quick question:
is the graph of y = log3^(x+4) the same as the graph of y = log3^x+4?
*the base is 3 not 10

I think the second one is y=log3^x moved up by 4 units.

Last edited: Jan 26, 2008
19. Jan 26, 2008

### rocomath

This will only apply to linear equations.

$$a_1 x+b_1 y=c_1$$
$$a_2 x+b_2 y=c_2$$

$$x=\frac{\left|\begin{array}{cc}c_1 & b_1 \\ c_2 & b_2\end{array}\right|}{\left|\begin{array}{cc}a_1 & b_1 \\ a_2 & b_2\end{array}\right|}=\frac{c_1 b_2 - b_1 c_2}{a_1 b_2 - b_1 a_2}$$

$$y=\frac{\left|\begin{array}{cc}a_1 & c_1 \\ a_2 & c_2\end{array}\right|}{\left|\begin{array}{cc}a_1 & b_1 \\ a_2 & b_2\end{array}\right|}=\frac{a_1 c_2 - c_1 a_2}{a_1 b_2 - b_1 a_2}$$

Notice that the denominator for both x & y are the same (determinant). While the numerators vary.

You can also solve for 3 unknowns but that takes too long to type.

Last edited: Jan 26, 2008
20. Jan 26, 2008

### rocomath

I'm confused to as what your problem is. Is it ...

$$y=\log_3{(x+4)}$$

?