Question 1134045: Solve: 2x^4+3x^3+8x^2+15x-10=0
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! 2x^4 +3x^3 +8x^2 +15x -10 = 0
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there will be at most 4 solutions for x
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use the rational roots theorem since all the coefficients are integers
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the general form is
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a(n)x^n +a(n-1)x^(n-1) +.... +a(2)x^2 +a(1)x^1 +a(0) = 0
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let q = a(n) and p = a(0), then using rational roots theorem we consider factors of p and q
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q could equal to + or - 1, + or - 2
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p could be equal to + or - 1, + or - 2, + or - 5, + or - 10
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p/q = + or - 1, + or - 2, + or - 5, + or - 10, + or - 1/2, + or - 5/2
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now test these values with the given equation
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we find that x = 1/2 and x = -2 are rational solutions
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(x -1/2) * (x +2) = x^2 +1.5x -1 is a polynomial factor of the given equation
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(2x^4 +3x^3 +8x^2 +15x -10)/(x^2 +1.5x -1) = 2x^2+10
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2x^2+10 = 0
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2x^2 = -10
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x^2 = -5
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x = i * square root(5) or -i * square root(5)
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we have two rational roots and 2 complex roots
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x = 1/2, -2, i * square root(5), -i * square root(5)
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