How Do You Solve a Polynomial with Two Unknowns and Given Roots?

[AFG34]

[SOLVED] polynomial with two unknowns

Homework Statement

The graph of f(x)= 3x^4 + 14x^3 + px^2 + qx + 24 has x-intercepts -4 and 2. Determine the function.

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.

 

[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.

 

[AFG34]

but there are two unknowns in each expression

 

[Hootenanny]

but there are two unknowns in each expression

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

$$f(-4) = 0$$

$$f(2) = 0$$

 

[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.

 

[Hootenanny]

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.

Have you never solved simultaneous equations before?

 

[AFG34]

no, i haven't.

 

[Hootenanny]

i don't think so.

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

 

[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

 

[Hootenanny]

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

 

[rocomath]

Did you ever learn to solve matrices in Algebra class?

 

[AFG34]

8p + 132

ok so p = -11, q = -70

thnx

 

[AFG34]

Did you ever learn to solve matrices in Algebra class?

nope

 

[rocomath]

nope

Well either way, your answer isn't right b/c your answer should have ended up being in terms of a solution.

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

 

[Hootenanny]

Did you ever learn to solve matrices in Algebra class?

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).

8p + 132

ok so p = -11

Firstly, what you have written is an expression, not an equation. Secondly, you might want to check your coefficient of p.

 

[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

 

[rocomath]

0 = 12p + 132 is an equation
ya i typed it wrong
checked the back of the book, got the right answer. ok i know how to do them now, thnx

Want to learn the fast way?

 

[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.

 

[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.

 

[rocomath]

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.

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

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

?

 

[AFG34]

ya that and the second one is the same except there is no bracket around x+4

 

[rocomath]

ya that and the second one is the same except there is no bracket around x+4

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

$$y=\log_3 x+4$$

No they are not the same

2nd one, as you said it translated up 4 units. While the 1st one is translated to the left 4 units.

$$y=f(x+c)$$ translation to the left c units.

$$y=f(x)+c$$ translation up c units.

 

[AFG34]

thnx very much :D