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4x^{4}-29x^{2}+25=0
To factor the expression, solve the equation where it equals to 0.
±\frac{25}{4},±\frac{25}{2},±25,±\frac{5}{4},±\frac{5}{2},±5,±\frac{1}{4},±\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 25 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.
x=1
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.
4x^{3}+4x^{2}-25x-25=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{4}-29x^{2}+25 by x-1 to get 4x^{3}+4x^{2}-25x-25. To factor the result, solve the equation where it equals to 0.
±\frac{25}{4},±\frac{25}{2},±25,±\frac{5}{4},±\frac{5}{2},±5,±\frac{1}{4},±\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 -25 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.
x=-1
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.
4x^{2}-25=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+4x^{2}-25x-25 by x+1 to get 4x^{2}-25. To factor the result, solve the equation where it equals to 0.
x=\frac{0±\sqrt{0^{2}-4\times 4\left(-25\right)}}{2\times 4}
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 4 for a, 0 for b, and -25 for c in the quadratic formula.
x=\frac{0±20}{8}
Do the calculations.
x=-\frac{5}{2} x=\frac{5}{2}
Solve the equation 4x^{2}-25=0 when ± is plus and when ± is minus.
\left(x-1\right)\left(2x-5\right)\left(x+1\right)\left(2x+5\right)
Rewrite the factored expression using the obtained roots.