Consider $1-2a+a^{2}$. Use the perfect square formula, $p^{2}-2pq+q^{2}=\left(p-q\right)^{2}$, where $p=a$ and $q=1$.
$$\left(a-1\right)^{2}$$
Rewrite the complete factored expression.
$$8\left(a-1\right)^{2}$$
Steps Using Square Of Binomial
This trinomial has the form of a trinomial square, perhaps multiplied by a common factor. Trinomial squares can be factored by finding the square roots of the leading and trailing terms.
$$factor(8a^{2}-16a+8)$$
Find the greatest common factor of the coefficients.
$$gcf(8,-16,8)=8$$
Factor out $8$.
$$8\left(a^{2}-2a+1\right)$$
The trinomial square is the square of the binomial that is the sum or difference of the square roots of the leading and trailing terms, with the sign determined by the sign of the middle term of the trinomial square.
$$8\left(a-1\right)^{2}$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$8a^{2}-16a+8=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
