## DeadShot Group Title How would I set this up to solve it? Find the polynomial function with roots 1, 7, and -3 of multiplicity 2. one year ago one year ago

1. ParthKohli Group Title

All the roots of multiplicity $$2$$?

2. ParthKohli Group Title

And here's your example: If you want a polynomial with root $$2$$ of multiplicity $$3$$,$(x - 2)^3$

so, root -3 of multiplicity 2 would be $(x - (-3))^2$ right?

4. ParthKohli Group Title

That is correct :-)

5. ParthKohli Group Title

So is it only $$-3$$ with multiplicity $$2$$?

I have no idea, the question on my homework says "Find the polynomial function with roots 1, 7, and -3 of multiplicity 2.", the teacher didn't say anything else about it.

so, $(x-1) (x-7) (x-(-3))^2$ is the correct way to set it up?

8. ParthKohli Group Title

Yes, that's correct! But if we have all those with multiplicity $$2$$,$(x - 1)^2(x - 7)^2(x + 3)^2$

ok, so I'll do all of them with multiplicity of two. $(x-1)^2 (x-7)^2 (x-(-3))^2$ and then solve it.

10. ParthKohli Group Title

I dunno, it's your choice. It can be either.

So, it becomes $(x^2 + 1) (x^2 + 49) (x^2 + 9)$ right?

12. ParthKohli Group Title

I think it's just $$-3$$ with multiplicity of $$2$$.

13. ParthKohli Group Title

They don't usually give such hard problems.

oh, so then it's $(x-1) (x-7) (x^2 + 9)$ right?

15. ParthKohli Group Title

Hey wait, don't we have the same pics?

16. ash2326 Group Title

It should be $(x-1) (x-7) (x + 3)^2$

17. ash2326 Group Title

$(x+3)^2\ne x^2+9$ $(x+3)^2=x^2+2\times 3\times x+3^2$

18. ParthKohli Group Title

Why is OpenStudy so slow right now?

so, then it's $(x-1) (x-7) (x^2 + 6x + 9)$ ?

20. hartnn Group Title

thats correct, but i think you need to multiply them out for final answer, and bring it in the form, ax^4+bx^3+cx^2+dx+e

21. ParthKohli Group Title

Yes.

so then the answer is $(x^4 + 6x^3 - x^3 - 7x^3- x^2 - 7x^2 -6x - 42x - 9 - 63)$ simplified, it is $(x^4 - 2x^3 - 8x^2 - 48x - 72)$ right?

23. ParthKohli Group Title

Yup