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All the roots of multiplicity \(2\)?
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?
That is correct :-)
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?
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.
I dunno, it's your choice. It can be either.
So, it becomes \[(x^2 + 1) (x^2 + 49) (x^2 + 9)\] right?
I think it's just \(-3\) with multiplicity of \(2\).
They don't usually give such hard problems.
oh, so then it's \[(x-1) (x-7) (x^2 + 9)\] right?
Hey wait, don't we have the same pics?
It should be \[(x-1) (x-7) (x + 3)^2\]
\[(x+3)^2\ne x^2+9\] \[(x+3)^2=x^2+2\times 3\times x+3^2\]
Why is OpenStudy so slow right now?
so, then it's \[(x-1) (x-7) (x^2 + 6x + 9)\] ?
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
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?