Solve 1-26a^2+25a^4

$$1 - 26 a ^ { 2 } + 25 a ^ { 4 }$$

Factor

$$\left(a-1\right)\left(5a-1\right)\left(a+1\right)\left(5a+1\right)$$

Solution Steps

To factor the expression, solve the equation where it equals to 0.

$$25a^{4}-26a^{2}+1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 25. List all candidates \frac{p}{q}.

$$±\frac{1}{25},±\frac{1}{5},±1$$

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.

$$a=1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide 25a^{4}-26a^{2}+1 by a-1 to get 25a^{3}+25a^{2}-a-1. To factor the result, solve the equation where it equals to 0.

$$25a^{3}+25a^{2}-a-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 25. List all candidates \frac{p}{q}.

$$±\frac{1}{25},±\frac{1}{5},±1$$

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.

$$a=-1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide 25a^{3}+25a^{2}-a-1 by a+1 to get 25a^{2}-1. To factor the result, solve the equation where it equals to 0.

$$25a^{2}-1=0$$

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 25 for a, 0 for b, and -1 for c in the quadratic formula.

$$a=\frac{0±\sqrt{0^{2}-4\times 25\left(-1\right)}}{2\times 25}$$

Do the calculations.

$$a=\frac{0±10}{50}$$

Solve the equation 25a^{2}-1=0 when ± is plus and when ± is minus.

$$a=-\frac{1}{5} a=\frac{1}{5}$$

Rewrite the factored expression using the obtained roots.

$$\left(a-1\right)\left(5a-1\right)\left(a+1\right)\left(5a+1\right)$$

Evaluate

$$25a^{4}-26a^{2}+1$$