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2x^{4}-5x^{3}-8x^{2}+25x-10=0
To factor the expression, solve the equation where it equals to 0.
±5,±10,±\frac{5}{2},±1,±2,±\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 -10 and q divides the leading coefficient 2. 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.
2x^{3}-x^{2}-10x+5=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{4}-5x^{3}-8x^{2}+25x-10 by x-2 to get 2x^{3}-x^{2}-10x+5. To factor the result, solve the equation where it equals to 0.
±\frac{5}{2},±5,±\frac{1}{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 5 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}-5=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-x^{2}-10x+5 by 2\left(x-\frac{1}{2}\right)=2x-1 to get x^{2}-5. To factor the result, solve the equation where it equals to 0.
x=\frac{0±\sqrt{0^{2}-4\times 1\left(-5\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 -5 for c in the quadratic formula.
x=\frac{0±2\sqrt{5}}{2}
Do the calculations.
x=-\sqrt{5} x=\sqrt{5}
Solve the equation x^{2}-5=0 when ± is plus and when ± is minus.
\left(x-2\right)\left(2x-1\right)\left(x^{2}-5\right)
Rewrite the factored expression using the obtained roots. Polynomial x^{2}-5 is not factored since it does not have any rational roots.