Factor the expression by grouping. First, the expression needs to be rewritten as $4x^{2}+ax+bx-15$. To find $a$ and $b$, set up a system to be solved.
$$a+b=17$$ $$ab=4\left(-15\right)=-60$$
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 $-60$.
Rewrite $4x^{2}+17x-15$ as $\left(4x^{2}-3x\right)+\left(20x-15\right)$.
$$\left(4x^{2}-3x\right)+\left(20x-15\right)$$
Factor out $x$ in the first and $5$ in the second group.
$$x\left(4x-3\right)+5\left(4x-3\right)$$
Factor out common term $4x-3$ by using distributive property.
$$\left(4x-3\right)\left(x+5\right)$$
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 $ax^{2}+bx+c=0$.
$$4x^{2}+17x-15=0$$
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 $x=\frac{-17±23}{8}$ when $±$ is plus. Add $-17$ to $23$.
$$x=\frac{6}{8}$$
Reduce the fraction $\frac{6}{8}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{3}{4}$$
Now solve the equation $x=\frac{-17±23}{8}$ when $±$ is minus. Subtract $23$ from $-17$.
$$x=-\frac{40}{8}$$
Divide $-40$ by $8$.
$$x=-5$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{3}{4}$ for $x_{1}$ and $-5$ for $x_{2}$.
