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