Posted by Iris on Friday, December 30, 2011 at 5:40am.
Hey how would you factor this polynomial to find the zeroes of this function?
x^45x^3+7x^2+3x10

Algebra 2  Damon, Friday, December 30, 2011 at 6:46am
Always 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^36x^2+13x10)
Now we need something that works with 10 like 10 and 1 or 5 and 2
Try 2 and 2
2 works so (x2) is a factor. Divide
(x^36x^2+13x10) by (x2)
(x+1)(x2)(x^24x+5)
the last one you must do by quadratic equation and the roots are complex
x = (2+i) and x = (2i)
so
(x+1)(x2)(x2+i)(x2i)
and
x = 1, 2 , 2+i, 2i

Algebra 2  drwls, Friday, December 30, 2011 at 6:52am
Try +/ 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^45x^3+7x^2+3x10 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 x2 is a factor.
Divide x^3 6x^2 +13x 10 by x2 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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