Factor the expression by grouping. First, the expression needs to be rewritten as $27x^{2}+ax+bx-25$. To find $a$ and $b$, set up a system to be solved.
$$a+b=30$$ $$ab=27\left(-25\right)=-675$$
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 $-675$.
Rewrite $27x^{2}+30x-25$ as $\left(27x^{2}-15x\right)+\left(45x-25\right)$.
$$\left(27x^{2}-15x\right)+\left(45x-25\right)$$
Factor out $3x$ in the first and $5$ in the second group.
$$3x\left(9x-5\right)+5\left(9x-5\right)$$
Factor out common term $9x-5$ by using distributive property.
$$\left(9x-5\right)\left(3x+5\right)$$
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 $ax^{2}+bx+c=0$.
$$27x^{2}+30x-25=0$$
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 $x=\frac{-30±60}{54}$ when $±$ is plus. Add $-30$ to $60$.
$$x=\frac{30}{54}$$
Reduce the fraction $\frac{30}{54}$ to lowest terms by extracting and canceling out $6$.
$$x=\frac{5}{9}$$
Now solve the equation $x=\frac{-30±60}{54}$ when $±$ is minus. Subtract $60$ from $-30$.
$$x=-\frac{90}{54}$$
Reduce the fraction $\frac{-90}{54}$ to lowest terms by extracting and canceling out $18$.
$$x=-\frac{5}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{5}{9}$ for $x_{1}$ and $-\frac{5}{3}$ for $x_{2}$.
Multiply $\frac{9x-5}{9}$ times $\frac{3x+5}{3}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
