Factor the expression by grouping. First, the expression needs to be rewritten as $2a^{2}+pa+qa+2$. To find $p$ and $q$, set up a system to be solved.
$$p+q=5$$ $$pq=2\times 2=4$$
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 $4$.
$$1,4$$ $$2,2$$
Calculate the sum for each pair.
$$1+4=5$$ $$2+2=4$$
The solution is the pair that gives sum $5$.
$$p=1$$ $$q=4$$
Rewrite $2a^{2}+5a+2$ as $\left(2a^{2}+a\right)+\left(4a+2\right)$.
$$\left(2a^{2}+a\right)+\left(4a+2\right)$$
Factor out $a$ in the first and $2$ in the second group.
$$a\left(2a+1\right)+2\left(2a+1\right)$$
Factor out common term $2a+1$ by using distributive property.
$$\left(2a+1\right)\left(a+2\right)$$
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 $ax^{2}+bx+c=0$.
$$2a^{2}+5a+2=0$$
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 $a=\frac{-5±3}{4}$ when $±$ is plus. Add $-5$ to $3$.
$$a=-\frac{2}{4}$$
Reduce the fraction $\frac{-2}{4}$ to lowest terms by extracting and canceling out $2$.
$$a=-\frac{1}{2}$$
Now solve the equation $a=\frac{-5±3}{4}$ when $±$ is minus. Subtract $3$ from $-5$.
$$a=-\frac{8}{4}$$
Divide $-8$ by $4$.
$$a=-2$$
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 $-2$ for $x_{2}$.
Cancel out $2$, the greatest common factor in $2$ and $2$.
$$2a^{2}+5a+2=\left(2a+1\right)\left(a+2\right)$$
Steps Using Direct Factoring Method
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.This is achieved by dividing both sides of the equation by $2$
$$x ^ 2 +\frac{5}{2}x +1 = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = -\frac{5}{2} $$ $$ rs = 1$$
Two numbers $r$ and $s$ sum up to $-\frac{5}{2}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{5}{2} = -\frac{5}{4}$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
$$r = -\frac{5}{4} - u$$ $$s = -\frac{5}{4} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 1$
$$(-\frac{5}{4} - u) (-\frac{5}{4} + u) = 1$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{25}{16} - u^2 = 1$$
Simplify the expression by subtracting $\frac{25}{16}$ on both sides
$$-u^2 = 1-\frac{25}{16} = -\frac{9}{16}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
