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solve the polynomial inequality x^3+x^2+64x+64greater than 0
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You need to evaluate the following inequality, such that:
`x^3 + x^2 + 64x + 64 > 0`
You need to convert the inequality into an equation, such that:
`x^3 + x^2 + 64x + 64 = 0`
You may create two groups of terms to perform factorization, such that:
`(x^3 + x^2) + (64x+ 64) = 0`
You may factor out` x^2` and 64 such that:
`x^2(x + 1) + 64(x + 1) = 0`
You may factor out (x + 1), such that:
`(x + 1)(x^2 + 64) = 0`
Using the zero product rule yields:
`x + 1 = 0 => x = -1`
`x^2 + 64 > 0 for x in R`
You need to check what are x values that makes the inequality `(x + 1)(x^2 + 64) > 0` to hold. You need to remember that a product of two positive factors or two negative factors is positive.
Since `x^2 + 64 > 0` , hence `x + 1` needs to be also positive. You shoudl notice that `x + 1 > 0 ` for `x > -1` .
Hence, evaluating the interval solution to the given inequality, yields `x > -1 => x in (-1,oo).`
Posted by sciencesolve on October 21, 2013 at 6:15 PM (Answer #1)
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