# Thread: Third Degree Polynomial Function question

1. ## Third Degree Polynomial Function question

I have been working on this for hours and I cannot seem to get all the numbers to work. please. How would I do this?

Find a third degree polynomial function such that
f(0)=3 and whose zeros are 1, 2, and 3.

I have been working on this for hours and I cannot seem to get all the numbers to work. please. How would I do this?

Find a third degree polynomial function such that
f(0)=3 and whose zeros are 1, 2, and 3.

Hi,

So firstly, how can you write a cubic in terms of its zeroes? Once you do that, now you have to adjust somehow for f(0)=3. Once you answer my first question, plug in x=0 and see what you get.

3. f(0) = 3 and whose zeros are 1, 2, and 3.

(x - a)(x - b)(x - c) = x^3 -(a + b + c)x^2 + (ab + bc + ac)x - (abc) = 0?

4. Am I supposed to randomly pic numbers out and put them in a 3rd degree polynomial and keep reworking the problem until I get answers 1, 2,3 ? I think this is where I am confused.

Am I supposed to randomly pic numbers out and put them in a 3rd degree polynomial and keep reworking the problem until I get answers 1, 2,3 ? I think this is where I am confused.
No, no. You're making this too hard.

Think of a quadratic. You factor these all the time into something like (x+a)(x-b). This means that -a and b are zeroes of the function. Why? Because plugging in either of those for x makes one of the parenthesis 0, which makes the whole thing 0 since everything is multiplied.

You can go backwards too. If I said a quadratic had zeroes of x=2,1 , what would this be? It is just y=(x-2)(x-1). If you expand that out, it will look like a normal quadratic. The same idea goes with a cubic and with all powers. You can write any polynomial as a product of it's zeroes.

6. x^3-6x^2+11x-6=0
?

Am I supposed to randomly pic numbers out and put them in a 3rd degree polynomial and keep reworking the problem until I get answers 1, 2,3 ? I think this is where I am confused.
Polynomials can always be factored into products of polynomials of lesser degree. Look at this example:

$(x+c)(x+d) = x^2 + (c+d)x + cd$

The two factors on the left are polynomials of degree 1, and their product is a polynomial of degree 2. Treat it like addition- 1 + 1 = 2.
Here's another example:

$(x+a)(x^2 + bx + c) = x^3 + (b+a)x^2 + (ab + c)x + ca$

Note that polynomial of degree 1 factored with polynomial of degree 2 is, by this addition rule, 1 + 2 = 3, so we've got a polynomial of degree 3 out of it.

But of course, there are other ways to make three. What about 1 + 1 + 1? If we multiply three polynomials of degree 1 together, we're going to get a cubic (degree 3).

Now recall that we know the three roots to our polynomial, which are 1, 2 and 3. Let's write out our polynomial factors:

$f(x) = (x + a)(x + b)(x + c)$

We know that all we have to do is to get one of those linear factors to be equal to 0 to find a root. If f(1) = 0, then one of those factors must be:

$(x+(-1))$

Going on, if f(2) = 0 as well, then one of the other factors must be $(x+(-2))$, and so on. In fact, the whole picture is:

$f(x) = (x-1)(x-2)(x-3)$

If you've followed me thus far, you should be able to expand those terms. That's not quite the whole story, though. Although it will work for f(1), f(2), f(3), it might not for f(0), so you'll have to adjust for a constant term.

x^3-6x^2+11x-6=0
?
y=(x-1)(x-2)(x-3)

That's how you write the cubic with x=1,2,3 being zeroes. x=0 gives a y-value of -6 though, so we have to add 9 to this whole thing so that x=0 gives a y-value of 3.

So I get y=(x-1)(x-2)(x-3)+9. That might be what you wrote, I just didn't expand it out.

9. Originally Posted by Jameson
y=(x-1)(x-2)(x-3)

That's how you write the cubic with x=1,2,3 being zeroes. x=0 gives a y-value of -6 though, so we have to add 9 to this whole thing so that x=0 gives a y-value of 3.

So I get y=(x-1)(x-2)(x-3)+9. That might be what you wrote, I just didn't expand it out.
I'll probably get shot for contradicting the Administrator, but when you add 9 you no longer have a polynomial that vanishes when x = 1, 2 or 3. What you need to do to the polynomial (x-1)(x-2)(x-3) is to multiply it by a suitable constant so that it takes the value 3 at x=0.

10. Originally Posted by Opalg
I'll probably get shot for contradicting the Administrator, but when you add 9 you no longer have a polynomial that vanishes when x = 1, 2 or 3. What you need to do to the polynomial (x-1)(x-2)(x-3) is to multiply it by a suitable constant so that it takes the value 3 at x=0.
Duh! Thank you for the catch. No infractions this time

My mistake. I'll finish the problem correctly now since I messed it up before. Like I said x=0 gives a y-value of -6, so we must now find a number that multiplied by -6 yields 9. So -6S=9 -> S=-3/2

Final solution should be $y=\left( -\frac{3}{2} \right) (x-1)(x-2)(x-3)$

Hopefully Opalg won't find any more mistakes...

11. Originally Posted by Jameson
Duh! Thank you for the catch. No infractions this time

My mistake. I'll finish the problem correctly now since I messed it up before. Like I said x=0 gives a y-value of -6, so we must now find a number that multiplied by -6 yields 9. 9? Shouldn't that be 3?? So -6S=9 -> S=-3/2

Final solution should be $y=\left( -\frac{\color{red}1}{2} \right) (x-1)(x-2)(x-3)$

Hopefully Opalg won't find any more mistakes...