cubic function factored with zeroes

[Guest]

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) ?

 

[stapel]

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

Show that the expression for f(x) becomes
f(x) = a(x-x₁)(x-x₂)(x-x₃)

Is this the exact text of the exercise and instructions?

Thank you.

Eliz.

 

[Guest]

Is this the exact text of the exercise and instructions?

Thank you.

Eliz.

Well, no, because I need to show this within another exercise... the reason I know I need to do this is that the solutions manual shows this step, but doesn't explain how it accomplishes this step. The exercise needs to be proven, and this is one of the steps. However, the solutions manual doesn't explain how it works. It only says:

If the graph has three x-intercepts x₁, x₂, and x₃, then the expression for f(x) must factor as f(x) = a(x - x₁)(x - x₂)(x - x₃)

That's all it says for that step.

 

[stapel]

It only says:

If the graph has three x-intercepts x₁, x₂, and x₃, then the expression for f(x) must factor as f(x) = a(x - x₁)(x - x₂)(x - x₃)

That's all it says for that step.

This is just a basic property of polynomials. If you're needing a name, I think it's called the "Factor Theorem".

Eliz.

 

[Guest]

Yes, I know it's a basic property... that since x₁, x₂, and x₃ are zeroes, then you can factor them out. But how do I know that a is left there?

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

where p is another polynomial?

oh wait, 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?