Consider $14a-5-8a^{2}$. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-8a^{2}+14a-5$$
Factor the expression by grouping. First, the expression needs to be rewritten as $-8a^{2}+pa+qa-5$. To find $p$ and $q$, set up a system to be solved.
$$p+q=14$$ $$pq=-8\left(-5\right)=40$$
Since $pq$ is positive, $p$ and $q$ have the same sign. Since $p+q$ is positive, $p$ and $q$ are both positive. List all such integer pairs that give product $40$.
$$1,40$$ $$2,20$$ $$4,10$$ $$5,8$$
Calculate the sum for each pair.
$$1+40=41$$ $$2+20=22$$ $$4+10=14$$ $$5+8=13$$
The solution is the pair that gives sum $14$.
$$p=10$$ $$q=4$$
Rewrite $-8a^{2}+14a-5$ as $\left(-8a^{2}+10a\right)+\left(4a-5\right)$.
$$\left(-8a^{2}+10a\right)+\left(4a-5\right)$$
Factor out $2a$ in the first and $-1$ in the second group.
$$2a\left(-4a+5\right)-\left(-4a+5\right)$$
Factor out common term $-4a+5$ by using distributive property.
$$\left(-4a+5\right)\left(2a-1\right)$$
Rewrite the complete factored expression.
$$4\left(-4a+5\right)\left(2a-1\right)$$
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 $ax^{2}+bx+c=0$.
$$-32a^{2}+56a-20=0$$
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.
Now solve the equation $a=\frac{-56±24}{-64}$ when $±$ is plus. Add $-56$ to $24$.
$$a=-\frac{32}{-64}$$
Reduce the fraction $\frac{-32}{-64}$ to lowest terms by extracting and canceling out $32$.
$$a=\frac{1}{2}$$
Now solve the equation $a=\frac{-56±24}{-64}$ when $±$ is minus. Subtract $24$ from $-56$.
$$a=-\frac{80}{-64}$$
Reduce the fraction $\frac{-80}{-64}$ to lowest terms by extracting and canceling out $16$.
$$a=\frac{5}{4}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{1}{2}$ for $x_{1}$ and $\frac{5}{4}$ for $x_{2}$.
Multiply $\frac{-2a+1}{-2}$ times $\frac{-4a+5}{-4}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
