Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-20$. To find $a$ and $b$, set up a system to be solved.
$$a+b=3$$ $$ab=2\left(-20\right)=-40$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-40$.
$$-1,40$$ $$-2,20$$ $$-4,10$$ $$-5,8$$
Calculate the sum for each pair.
$$-1+40=39$$ $$-2+20=18$$ $$-4+10=6$$ $$-5+8=3$$
The solution is the pair that gives sum $3$.
$$a=-5$$ $$b=8$$
Rewrite $2x^{2}+3x-20$ as $\left(2x^{2}-5x\right)+\left(8x-20\right)$.
$$\left(2x^{2}-5x\right)+\left(8x-20\right)$$
Factor out $x$ in the first and $4$ in the second group.
$$x\left(2x-5\right)+4\left(2x-5\right)$$
Factor out common term $2x-5$ by using distributive property.
$$\left(2x-5\right)\left(x+4\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}+3x-20=0$$
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 $x=\frac{-3±13}{4}$ when $±$ is plus. Add $-3$ to $13$.
$$x=\frac{10}{4}$$
Reduce the fraction $\frac{10}{4}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{5}{2}$$
Now solve the equation $x=\frac{-3±13}{4}$ when $±$ is minus. Subtract $13$ from $-3$.
$$x=-\frac{16}{4}$$
Divide $-16$ by $4$.
$$x=-4$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{5}{2}$ for $x_{1}$ and $-4$ for $x_{2}$.
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 $2$
$$x ^ 2 +\frac{3}{2}x -10 = 0$$
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 $rs = C$
$$r + s = -\frac{3}{2} $$ $$ rs = -10$$
Two numbers $r$ and $s$ sum up to $-\frac{3}{2}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{3}{2} = -\frac{3}{4}$. 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 $u$.
$$r = -\frac{3}{4} - u$$ $$s = -\frac{3}{4} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -10$
$$(-\frac{3}{4} - u) (-\frac{3}{4} + u) = -10$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{9}{16} - u^2 = -10$$
Simplify the expression by subtracting $\frac{9}{16}$ on both sides
$$-u^2 = -10-\frac{9}{16} = -\frac{169}{16}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
