##### I need help with an algebra problem

 Algebra Tutor: None Selected Time limit: 1 Day

Use the Factor Theorem to determine a polynomial equation, of lowest degree, that has only the indicated roots:

1/3 is a double root,  -2 is a double root.

The answer should be 9x^4 +30x^3 +13x^2 -20x +4 = 0

I just need to know how to do the work to get to that answer. Thanks

Nov 6th, 2015

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The factor theorem states that a polynomial $f(x)$ has a factor $(x - k)$ if and only if$f(k)=0$ (i.e. $k$ is a root).

1/3 is a double root, -2 is a double root

So the factors are

(x - 1/3),

(x - 1/3), (because it is a double root we have to include it twice)

(x - (-2)) = (x +2) and

(x - (-2)) = (x +2) (because it is a double root we have to include it twice)

So to get the polynomial multiply the factors.

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I will submit the remaining  answer within 10 minutes. Sorry for the delay.

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Nov 6th, 2015

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$\\ f(x)=\bigg(x-\frac{1}{3}\bigg)\bigg(x-\frac{1}{3}\bigg)(x+2)(x+2)\\ \\ f(x)=\bigg(x-\frac{1}{3}\bigg)^2(x+2)^2\\ \\ f(x)=\bigg(x^2-\frac{2}{3}x+\frac{1}{9}\bigg)(x^2+4x+4)\\ \\ f(x)=x^4+4x^3+4x^2-\frac{2}{3}x^3-\frac{8}{3}x^2-\frac{8}{3}x+\frac{1}{9}x^2+\frac{4}{9}x+\frac{4}{9}\\ \\ f(x)=x^4+\frac{12-2}{3}x^3+\frac{36x^2-24x^2+x^2}{9}+\frac{-24x+4x}{9}+\frac{4}{9}\\ \\ f(x)=x^4-\frac{10}{3}x^3+\frac{13}{9}x^2-\frac{20}{9}x+\frac{4}{9}\\$

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Nov 6th, 2015

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I made a mistake in the last step. It should be +10/3 x^3 and not -10/3 x^3. Please correct it.

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To get the answer that you have mentioned, put f(x) = 0 and multiply throughout by 9. Multiplying by a constant will not change the zeroes or roots.

$\\ f(x)=0\\ \\ x^4+\frac{10}{3}x^3+\frac{13}{9}x^2-\frac{20}{9}x+\frac{4}{9}=0\\ \\ Multiply\:\:\:both\:\:\:sides\:\:\:by\:\:\:9\\ \\ 9\bigg(x^4+\frac{10}{3}x^3+\frac{13}{9}x^2-\frac{20}{9}x+\frac{4}{9}\bigg)=0\\ \\ 9x^4+30x^3+13x^2-20x+4=0$

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Nov 6th, 2015

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Nov 6th, 2015
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