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July 30, 2014

July 30, 2014

Posted by **Iris** on Friday, December 30, 2011 at 5:40am.

x^4-5x^3+7x^2+3x-10

- Algebra 2 -
**Damon**, Friday, December 30, 2011 at 6:46amAlways try x = 1 and x = -1

If x = 1 the function is -4

but if x = -1 the function is zero.

Therefore (x+1) is a factor, so use long division.

(x+1)(x^3-6x^2+13x-10)

Now we need something that works with 10 like 10 and 1 or 5 and 2

Try 2 and -2

2 works so (x-2) is a factor. Divide

(x^3-6x^2+13x-10) by (x-2)

(x+1)(x-2)(x^2-4x+5)

the last one you must do by quadratic equation and the roots are complex

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

so

(x+1)(x-2)(x-2+i)(x-2-i)

and

x = -1, 2 , 2+i, 2-i

- Algebra 2 -
**drwls**, Friday, December 30, 2011 at 6:52amTry +/- 1,2,5 and 10, the integer factors of 10. That will apply the "rational roots" theorem.

You will see right away that x = -1 makes the polynomial zero. Therefore x+1 is a factor.

Divide x^4-5x^3+7x^2+3x-10 by x+1 for the other factor, a cubic polynomial.

That cubic factor is

x^3 -6x^2 +13x -10

x=2 makes that zero, so x-2 is a factor.

Divide x^3 -6x^2 +13x -10 by x-2 to get a quadratic factor. It will be

x^2 -4x + 5

Then see if you can factor that. The remaining roots are complex.

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