SOLUTION: Please solve(factor): x^3+9x^2+13x+5=0

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Question 177005: Please solve(factor): x^3+9x^2+13x+5=0

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Typically I do a quick graph to see what's what.
+graph%28+300%2C+200%2C+-8%2C+2%2C+-2%2C+2%2C+x%5E3%2B9x%5E2%2B13x%2B5%29+
Looks like x=-1 is a root.
Let's check,
x%5E3%2B9x%5E2%2B13x%2B5=0
-1%2B9-13%2B5=0
Yes it is, (x+1) is a factor.
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We can use polynomial long division to find the remaining quadratic function.
Showing polynomial long division is a little difficult.
The first part will show the factor (left hand column), then the factor times the divisor,
then the next line will show the subtraction.
Then repeat.
Hopefully it makes sense.
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...........___________________
(x+1)|x^3+9x^2+13x+5
x^2:...x^3+x^2
-----------------------------------
..................8x^2+13x+5
8x:............8x^2+8x
-----------------------------------
...............................5x+5
5:...........................5x+5
-----------------------------------
.....................................0
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%28x%5E3%2B9x%5E2%2B13x%2B5%29%2F%28x%2B1%29=x%5E2%2B8x%2B5
Let's use the quadratic formula to find the remaining roots.
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-8+%2B-+sqrt%28+%28-8%29%5E2-4%2A1%2A5+%29%29%2F%282%2A1%29+
x+=+%28-8+%2B-+sqrt%28+64-20+%29%29%2F%282%29+
x+=+%28-8+%2B-+sqrt%28+44%29%29%2F%282%29+
x+=+%28-8+%2B-+2%2Asqrt%28+11%29%29%2F%282%29+
x+=+-4+%2B-+sqrt%28+11%29+
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Now put it all together,
x%5E3%2B9x%5E2%2B13x%2B5=%28x%2B1%29%28x%2B4%2Bsqrt%2811%29%29%28x%2B4-sqrt%2811%29%29