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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