# Cubic function factored with zeroes

1. Apr 5, 2006

### endeavor

Let f(x) = ax3+bx2+cx+d
and let f have x-intercepts x1, x2, and x3.

How can I show that the expression for f(x) becomes
f(x) = a(x-x1)(x-x2)(x-x3) ?

2. Apr 5, 2006

### Gale

you need to show some work or effort. we don't just give out answers. do you know what an intercept is? when the line crosses the x-axis, y equals? you could also just expand your second equation and see what you get, and then work backwards.

3. Apr 5, 2006

### endeavor

I know what an intercept is. I just don't know how to figure the problem out.
I thought about using synthetic division, but after using x1 as the zero, it seemed like there had to be an easier and faster way...
I got the following after using synthetic division only one time:
f(x) = (x - x1)(ax2 + 2ax12 + bx + bx1 + c)

I know f(x1)=f(x2)=f(x3)=0, which would mean ax13 + bx12 + cx1 = ax23 + bx22 + cx2 = ax33 + bx32 + cx3
Is there a way to solve these 3 equations for a, b, and c, in terms of x1, x2, and x3?

In case you're wondering, this is part of a bigger problem which could be solved by finding only a and b in terms of x1, x2, and x3. The solutions manual uses f(x) = a(x-x1)(x-x2)(x-x3) to find a and b, not solving for the above 3 equations.

By working backwards, I can't arrive at the original equation; at least I don't think so, because there are a, b, c, and d in the original compared to only a in the new equation...

Isn't there any easy way to figure this out??

4. Apr 5, 2006

### endeavor

I think I don't really need to show this... I only need to give a reason for this. Is there a theorem or postulate that states this?

5. Apr 5, 2006

### Gale

a, b, c and x1, x2, x3 are all just arbitrary constants. any combination of them are also constants.

anyway, an intercept by definition is a point on the x axis right? so y=0. if you have any polynomial function with any x intercepts then each intercept is a solution to the equation when y=0.

so you have f(x) = a(x-x1)(x-x2)(x-x3), this would imply that x1 x2 and x3 are intercepts because when f(x)=0 they are solutions to the equation. in other words, when y=0, that's what x equals. there's really no theorems for it, its just the definition of what we call intercepts. you have to understand what they really are. if you were given a 4th degree polynomial, it could be reduced to a similar equation, except with another factor of (x-x4).

6. Apr 5, 2006

### endeavor

I know that f(x) = (x-x1)(x-x2)(x-x3) * p(x)

where p is another polynomial?

but f is cubic, so there can't be another polynomial... right?

and since there is a coefficient for the first factor, namely a, a has to be put in the final equation. therefore,
f(x) = a(x-x1)(x-x2)(x-x3)

is that right?

7. Apr 5, 2006

### Gale

right, the fact that f is cubic means there is exactly 3 intercepts. Whatever the degree of the polynomial.. thats the number of zeros. and yes, you need a factor of a because when you expand the equation the x^3 factor needs a factor because the original had one, and in fact, this factor must be the same.

8. Apr 6, 2006

### HallsofIvy

Staff Emeritus
No, the fact that f is cubic means that there are no more that 3 intercepts- which is what is needed here. f(x)= x3 has only one x-intercept.

9. Apr 6, 2006

### Gale

well, the other two are imaginary...

10. Apr 7, 2006

### VietDao29

Err, what exactly do you mean by "imaginary"? Do you mean "multiplicity" instead?

11. Apr 7, 2006

### HallsofIvy

Staff Emeritus
Imaginary "roots", not "intercepts"!

12. Apr 7, 2006

### Gale

haha, uh, my bad? practically the same thing anyway... yeah, i'm shutting up now.

13. Apr 8, 2006

### topsquark

Fiddlesticks. I don't have my books with me right now so I can't give a name to the theorem, but there is a theorem that states that if a polynomial function has a zero (aka x-intercept in this case) at x = x1 that the polynomial has a linear factor x - x1. This is directly related to (or can be shown by, depending on how you want to look at it) the fact that if you divide your polynomial by x - x1 you MUST get a zero remainder if x1 is a zero of the polynomial.

In the absence of the proof, probably the simplest way to show this is to do the division and require the remainder to be zero. You should be able to work backward and construct conditions on the a, b, c, d that proves that x - x1 is a factor.

-Dan