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November 26, 2014

November 26, 2014

Posted by **ladybug** on Saturday, December 22, 2012 at 9:07pm.

f(x)=x^3-3x^2+16x-48

- calculus -
**drwls**, Sunday, December 23, 2012 at 3:13amTry + and - the prime factors of 48:

1, 2, 3, 4, 6, 8, 12 and 16.

When you find one that works, call it 'a' and divide the f(x) polymonial by (x-a) to get the quadratic factor. The other two roots can easily be found with that.

It looks to me like one root is +3. Therefore x-3 is a factor.

Now evaluate [x^3-3x^2+16x-48]/(x-3)

It should give you a quadratic equation with no remainder. Factor that.

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