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Precalculus
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True or false: In the equation F(x) = (3x+5) (x2 - 6x + 9)^2 = 0, 3 has a multiplicity of 2.
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Well, if you factored the second portion of that, you would have (x^2-6x+9) become (x-3)(x-3). As you can see, if I set both of those factors equal to 0, I would get x = 3 twice. That means yes, the multiplicity is 2.
Thanks so much. :)
Yep, np :3
isn't there a mistake in the answer... as the quadratic is squared... so its \[F(x) = (3x + 5)(x^2 -6x + 9)^2 = (3x +5)((x -3)^2)^2 = (3x + 5)(x -3)^4\] so isn't the multiplicity of the root x = 3 actually 4
I did find a mistake in my answer, because the answer wasn't a multiplicity of 2 but 4. Thanks for clearing that up for me.
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