Factor the expression by grouping. First, the expression needs to be rewritten as $4x^{2}+ax+bx+5$. To find $a$ and $b$, set up a system to be solved.
$$a+b=12$$ $$ab=4\times 5=20$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $20$.
$$1,20$$ $$2,10$$ $$4,5$$
Calculate the sum for each pair.
$$1+20=21$$ $$2+10=12$$ $$4+5=9$$
The solution is the pair that gives sum $12$.
$$a=2$$ $$b=10$$
Rewrite $4x^{2}+12x+5$ as $\left(4x^{2}+2x\right)+\left(10x+5\right)$.
$$\left(4x^{2}+2x\right)+\left(10x+5\right)$$
Factor out $2x$ in the first and $5$ in the second group.
$$2x\left(2x+1\right)+5\left(2x+1\right)$$
Factor out common term $2x+1$ by using distributive property.
$$\left(2x+1\right)\left(2x+5\right)$$
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 $ax^{2}+bx+c=0$.
$$4x^{2}+12x+5=0$$
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 $x=\frac{-12±8}{8}$ when $±$ is plus. Add $-12$ to $8$.
$$x=-\frac{4}{8}$$
Reduce the fraction $\frac{-4}{8}$ to lowest terms by extracting and canceling out $4$.
$$x=-\frac{1}{2}$$
Now solve the equation $x=\frac{-12±8}{8}$ when $±$ is minus. Subtract $8$ from $-12$.
$$x=-\frac{20}{8}$$
Reduce the fraction $\frac{-20}{8}$ to lowest terms by extracting and canceling out $4$.
$$x=-\frac{5}{2}$$
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 $-\frac{5}{2}$ for $x_{2}$.
Multiply $\frac{2x+1}{2}$ times $\frac{2x+5}{2}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
