Solve 10x^2-7x-12

$$10 { x }^{ 2 } -7x-12$$

Factor

$$\left(2x-3\right)\left(5x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-7 ab=10\left(-12\right)=-120$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -120.

$$1,-120 2,-60 3,-40 4,-30 5,-24 6,-20 8,-15 10,-12$$

Calculate the sum for each pair.

$$1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2$$

The solution is the pair that gives sum -7.

$$a=-15 b=8$$

Rewrite 10x^{2}-7x-12 as \left(10x^{2}-15x\right)+\left(8x-12\right).

$$\left(10x^{2}-15x\right)+\left(8x-12\right)$$

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

$$5x\left(2x-3\right)+4\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(5x+4\right)$$

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.

$$10x^{2}-7x-12=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 10\left(-12\right)}}{2\times 10}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 10\left(-12\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-7\right)±\sqrt{49-40\left(-12\right)}}{2\times 10}$$

Multiply -40 times -12.

$$x=\frac{-\left(-7\right)±\sqrt{49+480}}{2\times 10}$$

Add 49 to 480.

$$x=\frac{-\left(-7\right)±\sqrt{529}}{2\times 10}$$

Take the square root of 529.

$$x=\frac{-\left(-7\right)±23}{2\times 10}$$

The opposite of -7 is 7.

$$x=\frac{7±23}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{7±23}{20}$$

Now solve the equation x=\frac{7±23}{20} when ± is plus. Add 7 to 23.

$$x=\frac{30}{20}$$

Reduce the fraction \frac{30}{20} to lowest terms by extracting and canceling out 10.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{7±23}{20} when ± is minus. Subtract 23 from 7.

$$x=-\frac{16}{20}$$

Reduce the fraction \frac{-16}{20} to lowest terms by extracting and canceling out 4.

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

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

$$10x^{2}-7x-12=10\left(x-\frac{3}{2}\right)\left(x-\left(-\frac{4}{5}\right)\right)$$

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

$$10x^{2}-7x-12=10\left(x-\frac{3}{2}\right)\left(x+\frac{4}{5}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}-7x-12=10\times \frac{2x-3}{2}\left(x+\frac{4}{5}\right)$$

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

$$10x^{2}-7x-12=10\times \frac{2x-3}{2}\times \frac{5x+4}{5}$$

Multiply \frac{2x-3}{2} times \frac{5x+4}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10x^{2}-7x-12=10\times \frac{\left(2x-3\right)\left(5x+4\right)}{2\times 5}$$

Multiply 2 times 5.

$$10x^{2}-7x-12=10\times \frac{\left(2x-3\right)\left(5x+4\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10x^{2}-7x-12=\left(2x-3\right)\left(5x+4\right)$$

Evaluate

$$\left(2x-3\right)\left(5x+4\right)$$