# Finding a Quartic Polynomial Equation

1. Aug 22, 2007

### omg precal

1. The problem statement, all variables and given/known data

A quartic polynomial P(x) with real coefficients has zeros 2 + i and 3 - 2i. The other zeros are obviously 2 - i and 3 + 2i. If P(0) = 13, find a rule for P(x).

2. Relevant equations

-b/a = sum of all roots ; e/a = product of all roots

ax^4 + bx^3 + cx^2 + dx + e = 0

3. The attempt at a solution

I'm pretty sure e has to equal -13 because that would be the only way P(0) = 13, right?
If someone could kindly tell me how to solve this problem, it would be great.

2. Aug 22, 2007

### omg precal

3. Aug 22, 2007

### Dick

Write the polynomial as a*(x-(2+i))*(x-(2-i))*(x-(3+2i))*(x-(3-2i)). You could just multiply it out and find a.

4. Aug 22, 2007

### genneth

Simply multiply together the linear factors, and scale the coefficients to fit:

$$(x-(2+i))(x-(2-i))(x-(3-2i))(x-(3+2i)) = x^4 - 10x^3 + 42x^2 - 82 x + 65$$

Then scale by 5 to get the constant coefficient to be 13.

5. Aug 22, 2007

### omg precal

1/5x^4 - 2x^3 + 42/5x^2 - 82/5x + 13? that wouldnt work because the constant coefficient has to be negative 13 in order for P(0) to equal 13... would i then make everything negative? it already looks messy enough.

6. Aug 22, 2007

### Dick

P(0) IS 13. What are you talking about?

7. Aug 22, 2007

### omg precal

oops, my bad. thanks, Dick and genneth.

One more thing I'm having trouble with...

From a platform 35m above the ground, a ball is thrown upward with an initial speed of 30 m/s. The approximate height of the ball above the ground t seconds later is given by h(t) = 35 + 30t - 5t^2. After how many seconds does the ball hit the ground? What is the domain and range of h? After how many seconds does the ball reach its maximum height above the ground?

Thanks again!!!

8. Aug 22, 2007

### omg precal

jeez, you ppl on this forum are smart as hell. ^_^

9. Aug 22, 2007

### omg precal

Also need this one ASAP.

10. Aug 22, 2007

### omg precal

11. Aug 22, 2007

### Dick

Need your input first. It's not that hard. Start the problem out, how do you find when it hits the ground?

12. Aug 22, 2007

### omg precal

35 + 30t - 5t^2...

divide everything by 5?

7 + 6t - t^2?

rewrite it = t^2 - 6t - 7?

factor? (x - 7) (x + 1) ?

cant be negative, so 7 is the answer? after 7 seconds, the height is 0? i did that, but it seems waaaay too simple.

13. Aug 22, 2007

### omg precal

and also how would u find its maximum height in air?

>_<

how do u graph a quintic? should i just plug it in my calculator and copy it or is there a better way?

14. Aug 22, 2007

### HallsofIvy

Staff Emeritus
The problem was to use that height formula to determine when it hit the ground: h= 0. You want to solve the equation 35+ 30t- 5t^2= 0.

Now that it is an equation you can simplify the equation by dividing both sides by 5: 7+ 6t- t^2= 0.

Well, multiply the entire equation by -1: t^2- 6t- 7= 0.

Yes, you factor the right side of the equation and have (x-7)(x-1)= 0. Would you prefer it were harder?
Please remember that you have to deal with equations not just expressions.

h(t)= 35+ 6t- t^2 is a quadratic. Complete the square to get h(t)= H- (t-a)^2 (you find H and a). When t= a, h(t)= H. For t any other value, h(t)= H minus something and is lower. the maximum height is H and that happens when t= a.

Quintic? A fifth degree polynomial? Is this yet a third question in the same thread? I would recommend using a calculator.

15. Aug 22, 2007

### omg precal

Look at Bold.

Last edited: Aug 22, 2007