Factor the expression by grouping. First, the expression needs to be rewritten as $9x^{2}+ax+bx+20$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-49$$ $$ab=9\times 20=180$$
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 $180$.
Rewrite $9x^{2}-49x+20$ as $\left(9x^{2}-45x\right)+\left(-4x+20\right)$.
$$\left(9x^{2}-45x\right)+\left(-4x+20\right)$$
Factor out $9x$ in the first and $-4$ in the second group.
$$9x\left(x-5\right)-4\left(x-5\right)$$
Factor out common term $x-5$ by using distributive property.
$$\left(x-5\right)\left(9x-4\right)$$
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 $ax^{2}+bx+c=0$.
$$9x^{2}-49x+20=0$$
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 $x=\frac{49±41}{18}$ when $±$ is plus. Add $49$ to $41$.
$$x=\frac{90}{18}$$
Divide $90$ by $18$.
$$x=5$$
Now solve the equation $x=\frac{49±41}{18}$ when $±$ is minus. Subtract $41$ from $49$.
$$x=\frac{8}{18}$$
Reduce the fraction $\frac{8}{18}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{4}{9}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $5$ for $x_{1}$ and $\frac{4}{9}$ for $x_{2}$.
