Can someone help me out here? x^6 - 7x^3 - 8 = 0 THere must be a simpler way to factoring this without having to use factor theorem! Someone please help?
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$\displaystyle x^6-7x^3-8=0\implies(x^3+1)(x^3-8)=0$ Continue.
Originally Posted by checkmarks Can someone help me out here? x^6 - 7x^3 - 8 = 0 THere must be a simpler way to factoring this without having to use factor theorem! Someone please help? It's a quadratic in $\displaystyle y=x^3$ RonL
ahh i thought of that. but i just assumed an exponent raised to an exponent, you would multiply...so (x^3)(x^3) = x^9 shows you how terrible i am at simple math thanks guys
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