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4x^{4}-33x^{2}+8=0
To factor the expression, solve the equation where it equals to 0.
±2,±4,±8,±1,±\frac{1}{2},±\frac{1}{4}
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 8 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.
x=\frac{1}{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.
2x^{3}+x^{2}-16x-8=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{4}-33x^{2}+8 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 2x^{3}+x^{2}-16x-8. To factor the result, solve the equation where it equals to 0.
±4,±8,±2,±1,±\frac{1}{2}
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.
x=-\frac{1}{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}-8=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+x^{2}-16x-8 by 2\left(x+\frac{1}{2}\right)=2x+1 to get x^{2}-8. To factor the result, solve the equation where it equals to 0.
x=\frac{0±\sqrt{0^{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, 0 for b, and -8 for c in the quadratic formula.
x=\frac{0±4\sqrt{2}}{2}
Do the calculations.
x=-2\sqrt{2} x=2\sqrt{2}
Solve the equation x^{2}-8=0 when ± is plus and when ± is minus.
\left(2x-1\right)\left(2x+1\right)\left(x^{2}-8\right)
Rewrite the factored expression using the obtained roots. Polynomial x^{2}-8 is not factored since it does not have any rational roots.