Solve 10x^2-19x-15

$$10 { x }^{ 2 } -19x-15$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=-19 ab=10\left(-15\right)=-150$$

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 -150.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Calculate the sum for each pair.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

The solution is the pair that gives sum -19.

$$a=-25 b=6$$

Rewrite 10x^{2}-19x-15 as \left(10x^{2}-25x\right)+\left(6x-15\right).

$$\left(10x^{2}-25x\right)+\left(6x-15\right)$$

Factor out 5x in the first and 3 in the second group.

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

Factor out common term 2x-5 by using distributive property.

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

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.

$$10x^{2}-19x-15=0$$

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.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 10\left(-15\right)}}{2\times 10}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 10\left(-15\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-19\right)±\sqrt{361-40\left(-15\right)}}{2\times 10}$$

Multiply -40 times -15.

$$x=\frac{-\left(-19\right)±\sqrt{361+600}}{2\times 10}$$

Add 361 to 600.

$$x=\frac{-\left(-19\right)±\sqrt{961}}{2\times 10}$$

Take the square root of 961.

$$x=\frac{-\left(-19\right)±31}{2\times 10}$$

The opposite of -19 is 19.

$$x=\frac{19±31}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{19±31}{20}$$

Now solve the equation x=\frac{19±31}{20} when ± is plus. Add 19 to 31.

$$x=\frac{50}{20}$$

Reduce the fraction \frac{50}{20} to lowest terms by extracting and canceling out 10.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{19±31}{20} when ± is minus. Subtract 31 from 19.

$$x=-\frac{12}{20}$$

Reduce the fraction \frac{-12}{20} to lowest terms by extracting and canceling out 4.

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

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

$$10x^{2}-19x-15=10\left(x-\frac{5}{2}\right)\left(x-\left(-\frac{3}{5}\right)\right)$$

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

$$10x^{2}-19x-15=10\left(x-\frac{5}{2}\right)\left(x+\frac{3}{5}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}-19x-15=10\times \frac{2x-5}{2}\left(x+\frac{3}{5}\right)$$

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

$$10x^{2}-19x-15=10\times \frac{2x-5}{2}\times \frac{5x+3}{5}$$

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

$$10x^{2}-19x-15=10\times \frac{\left(2x-5\right)\left(5x+3\right)}{2\times 5}$$

Multiply 2 times 5.

$$10x^{2}-19x-15=10\times \frac{\left(2x-5\right)\left(5x+3\right)}{10}$$

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

$$10x^{2}-19x-15=\left(2x-5\right)\left(5x+3\right)$$

Evaluate

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