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x^{3}-12x+16=0
Add 16 to both sides.
±16,±8,±4,±2,±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 16 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=2
Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.
x^{2}+2x-8=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-12x+16 by x-2 to get x^{2}+2x-8. Solve the equation where the result equals to 0.
x=\frac{-2±\sqrt{2^{2}-4\times 1\left(-8\right)}}{2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 2 for b, and -8 for c in the quadratic formula.
x=\frac{-2±6}{2}
Do the calculations.
x=-4 x=2
Solve the equation x^{2}+2x-8=0 when ± is plus and when ± is minus.
x=2 x=-4
List all found solutions.