Factor the expression by grouping. First, the expression needs to be rewritten as $25x^{2}+ax+bx+12$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-35$$ $$ab=25\times 12=300$$
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 $300$.
Rewrite $25x^{2}-35x+12$ as $\left(25x^{2}-20x\right)+\left(-15x+12\right)$.
$$\left(25x^{2}-20x\right)+\left(-15x+12\right)$$
Factor out $5x$ in the first and $-3$ in the second group.
$$5x\left(5x-4\right)-3\left(5x-4\right)$$
Factor out common term $5x-4$ by using distributive property.
$$\left(5x-4\right)\left(5x-3\right)$$
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 $ax^{2}+bx+c=0$.
$$25x^{2}-35x+12=0$$
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 $x=\frac{35±5}{50}$ when $±$ is plus. Add $35$ to $5$.
$$x=\frac{40}{50}$$
Reduce the fraction $\frac{40}{50}$ to lowest terms by extracting and canceling out $10$.
$$x=\frac{4}{5}$$
Now solve the equation $x=\frac{35±5}{50}$ when $±$ is minus. Subtract $5$ from $35$.
$$x=\frac{30}{50}$$
Reduce the fraction $\frac{30}{50}$ to lowest terms by extracting and canceling out $10$.
$$x=\frac{3}{5}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{4}{5}$ for $x_{1}$ and $\frac{3}{5}$ for $x_{2}$.
Multiply $\frac{5x-4}{5}$ times $\frac{5x-3}{5}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
