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 ...

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

?