Question

find the polynmial f(x) that has the roots of -3 and 1 multiplicity of 2

Answer 1

Do you mean the root 1 has a multiplicity of two?

Answer 2

it just says the roots of -3 and 1 multiplicity of 2

Answer 3

Do you mean the root 1 has a multiplicity of two? <--- sounds like it

Answer 4

The root of a polynomial is when it crosses/touches the x-axis.

So whenever x = -3, f(x) = 0.
Whenever x = 1, f(x) = 0 also.

Now I will add 3 to both sides of x = -3 and get x + 3 = 0.
I can do that same for the other one: x - 1 = 0.

This is now familiar because it is in the form you should usually recognize roots. For example, take the parabola x^2-4. It factors out to be (x-2)(x+2). So, it's roots are x=2 and x=-2.

Therefore, f(x) must factor out to be (x+3)(x-1)^2.

Now, where did I get the ^2? This is the multiplicity. You can view it as a "double" root. It doesn't cross the x-axis, but only touches it.

A common "double" root is the graph y=x^2. It's a parabola, so it must have 2 roots. Well it does! 0 and 0. So 0 has a multiplicity of 2.

Anyway, therefore, your polynomial is f(x) = (x+3)(x-1)^2.

You can multiply it out to get a polynomial term by term, if you wish.

Answer 5

alright thank you so much