Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-30$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-17$$ $$ab=2\left(-30\right)=-60$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-60$.
Rewrite $2x^{2}-17x-30$ as $\left(2x^{2}-20x\right)+\left(3x-30\right)$.
$$\left(2x^{2}-20x\right)+\left(3x-30\right)$$
Factor out $2x$ in the first and $3$ in the second group.
$$2x\left(x-10\right)+3\left(x-10\right)$$
Factor out common term $x-10$ by using distributive property.
$$\left(x-10\right)\left(2x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}-17x-30=0$$
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 $x=\frac{17±23}{4}$ when $±$ is plus. Add $17$ to $23$.
$$x=\frac{40}{4}$$
Divide $40$ by $4$.
$$x=10$$
Now solve the equation $x=\frac{17±23}{4}$ when $±$ is minus. Subtract $23$ from $17$.
$$x=-\frac{6}{4}$$
Reduce the fraction $\frac{-6}{4}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{3}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $10$ for $x_{1}$ and $-\frac{3}{2}$ 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 $2$
$$x ^ 2 -\frac{17}{2}x -15 = 0$$
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 $rs = C$
$$r + s = \frac{17}{2} $$ $$ rs = -15$$
Two numbers $r$ and $s$ sum up to $\frac{17}{2}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{17}{2} = \frac{17}{4}$. 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 $u$.
$$r = \frac{17}{4} - u$$ $$s = \frac{17}{4} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -15$
$$(\frac{17}{4} - u) (\frac{17}{4} + u) = -15$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{289}{16} - u^2 = -15$$
Simplify the expression by subtracting $\frac{289}{16}$ on both sides
$$-u^2 = -15-\frac{289}{16} = -\frac{529}{16}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
