SOLUTION: x^3+4x^2+7x=28=0 find possible rational roots: find possible actual roots: please

Question 701633: x^3+4x^2+7x=28=0
find possible rational roots:
find possible actual roots:
please
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Assuming you meant to type: x^3+4x^2+7x+28 = 0

Possible Roots: List all the factors of 28

Possible Roots: 1, 2, 4, 7, 14, 28, -1, -2, -4, -7, -14, -28

Note: make sure you list the negative factors as well

Another Note: This trick only works if the leading coefficient is 1 or -1.
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Actual Roots:

x^3+4x^2+7x+28=0

(x^3+4x^2)+(7x+28)=0

x^2(x+4)+7(x+4)=0

(x^2+7)(x+4) = 0

x^2+7 = 0 or x+4 = 0

x^2 = -7 or x = -4

x = sqrt(-7), x = -sqrt(-7), or x = -4

x = i*sqrt(7), x = -i*sqrt(7), or x = -4

The actual roots are x = i*sqrt(7), x = -i*sqrt(7), or x = -4

The first two are complex or imaginary roots. The third root is real and rational.

The only actual rational root is x = -4

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