Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-5$$ $$ab=2\times 3=6$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $6$.
$$-1,-6$$ $$-2,-3$$
Calculate the sum for each pair.
$$-1-6=-7$$ $$-2-3=-5$$
The solution is the pair that gives sum $-5$.
$$a=-3$$ $$b=-2$$
Rewrite $2x^{2}-5x+3$ as $\left(2x^{2}-3x\right)+\left(-2x+3\right)$.
$$\left(2x^{2}-3x\right)+\left(-2x+3\right)$$
Factor out $x$ in the first and $-1$ in the second group.
$$x\left(2x-3\right)-\left(2x-3\right)$$
Factor out common term $2x-3$ by using distributive property.
$$\left(2x-3\right)\left(x-1\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}-5x+3=0$$
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 $x=\frac{5±1}{4}$ when $±$ is plus. Add $5$ to $1$.
$$x=\frac{6}{4}$$
Reduce the fraction $\frac{6}{4}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{3}{2}$$
Now solve the equation $x=\frac{5±1}{4}$ when $±$ is minus. Subtract $1$ from $5$.
$$x=\frac{4}{4}$$
Divide $4$ by $4$.
$$x=1$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{3}{2}$ for $x_{1}$ and $1$ for $x_{2}$.
Cancel out $2$, the greatest common factor in $2$ and $2$.
$$2x^{2}-5x+3=\left(2x-3\right)\left(x-1\right)$$
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 $2$
$$x ^ 2 -\frac{5}{2}x +\frac{3}{2} = 0$$
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 $rs = C$
$$r + s = \frac{5}{2} $$ $$ rs = \frac{3}{2}$$
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 $rs = \frac{3}{2}$
