Solve 25x^2+5x-12

$$25 { x }^{ 2 } +5x-12$$

Factor

$$\left(5x-3\right)\left(5x+4\right)$$

Solution Steps

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.

$$a+b=5 ab=25\left(-12\right)=-300$$

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 -300.

$$-1,300 -2,150 -3,100 -4,75 -5,60 -6,50 -10,30 -12,25 -15,20$$

Calculate the sum for each pair.

$$-1+300=299 -2+150=148 -3+100=97 -4+75=71 -5+60=55 -6+50=44 -10+30=20 -12+25=13 -15+20=5$$

The solution is the pair that gives sum 5.

$$a=-15 b=20$$

Rewrite 25x^{2}+5x-12 as \left(25x^{2}-15x\right)+\left(20x-12\right).

$$\left(25x^{2}-15x\right)+\left(20x-12\right)$$

Factor out 5x in the first and 4 in the second group.

$$5x\left(5x-3\right)+4\left(5x-3\right)$$

Factor out common term 5x-3 by using distributive property.

$$\left(5x-3\right)\left(5x+4\right)$$

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.

$$25x^{2}+5x-12=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 25\left(-12\right)}}{2\times 25}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 25\left(-12\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-5±\sqrt{25-100\left(-12\right)}}{2\times 25}$$

Multiply -100 times -12.

$$x=\frac{-5±\sqrt{25+1200}}{2\times 25}$$

Add 25 to 1200.

$$x=\frac{-5±\sqrt{1225}}{2\times 25}$$

Take the square root of 1225.

$$x=\frac{-5±35}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-5±35}{50}$$

Now solve the equation x=\frac{-5±35}{50} when ± is plus. Add -5 to 35.

$$x=\frac{30}{50}$$

Reduce the fraction \frac{30}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{3}{5}$$

Now solve the equation x=\frac{-5±35}{50} when ± is minus. Subtract 35 from -5.

$$x=-\frac{40}{50}$$

Reduce the fraction \frac{-40}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{4}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{5} for x_{1} and -\frac{4}{5} for x_{2}.

$$25x^{2}+5x-12=25\left(x-\frac{3}{5}\right)\left(x-\left(-\frac{4}{5}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$25x^{2}+5x-12=25\left(x-\frac{3}{5}\right)\left(x+\frac{4}{5}\right)$$

Subtract \frac{3}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+5x-12=25\times \frac{5x-3}{5}\left(x+\frac{4}{5}\right)$$

Add \frac{4}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+5x-12=25\times \frac{5x-3}{5}\times \frac{5x+4}{5}$$

Multiply \frac{5x-3}{5} times \frac{5x+4}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+5x-12=25\times \frac{\left(5x-3\right)\left(5x+4\right)}{5\times 5}$$

Multiply 5 times 5.

$$25x^{2}+5x-12=25\times \frac{\left(5x-3\right)\left(5x+4\right)}{25}$$

Cancel out 25, the greatest common factor in 25 and 25.

$$25x^{2}+5x-12=\left(5x-3\right)\left(5x+4\right)$$

Evaluate

$$\left(5x-3\right)\left(5x+4\right)$$