Factor the expression by grouping. First, the expression needs to be rewritten as $8x^{2}+ax+bx+9$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-18$$ $$ab=8\times 9=72$$
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 $72$.
Rewrite $8x^{2}-18x+9$ as $\left(8x^{2}-12x\right)+\left(-6x+9\right)$.
$$\left(8x^{2}-12x\right)+\left(-6x+9\right)$$
Factor out $4x$ in the first and $-3$ in the second group.
$$4x\left(2x-3\right)-3\left(2x-3\right)$$
Factor out common term $2x-3$ by using distributive property.
$$\left(2x-3\right)\left(4x-3\right)$$
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 $ax^{2}+bx+c=0$.
$$8x^{2}-18x+9=0$$
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 $x=\frac{18±6}{16}$ when $±$ is plus. Add $18$ to $6$.
$$x=\frac{24}{16}$$
Reduce the fraction $\frac{24}{16}$ to lowest terms by extracting and canceling out $8$.
$$x=\frac{3}{2}$$
Now solve the equation $x=\frac{18±6}{16}$ when $±$ is minus. Subtract $6$ from $18$.
$$x=\frac{12}{16}$$
Reduce the fraction $\frac{12}{16}$ to lowest terms by extracting and canceling out $4$.
$$x=\frac{3}{4}$$
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 $\frac{3}{4}$ for $x_{2}$.
Multiply $\frac{2x-3}{2}$ times $\frac{4x-3}{4}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
