Solve 10x^2-11x-6=

$$10 { x }^{ 2 } -11x-6=$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-11 ab=10\left(-6\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -11.

$$a=-15 b=4$$

Rewrite 10x^{2}-11x-6 as \left(10x^{2}-15x\right)+\left(4x-6\right).

$$\left(10x^{2}-15x\right)+\left(4x-6\right)$$

Factor out 5x in the first and 2 in the second group.

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

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

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

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.

$$10x^{2}-11x-6=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 10\left(-6\right)}}{2\times 10}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 10\left(-6\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-11\right)±\sqrt{121-40\left(-6\right)}}{2\times 10}$$

Multiply -40 times -6.

$$x=\frac{-\left(-11\right)±\sqrt{121+240}}{2\times 10}$$

Add 121 to 240.

$$x=\frac{-\left(-11\right)±\sqrt{361}}{2\times 10}$$

Take the square root of 361.

$$x=\frac{-\left(-11\right)±19}{2\times 10}$$

The opposite of -11 is 11.

$$x=\frac{11±19}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{11±19}{20}$$

Now solve the equation x=\frac{11±19}{20} when ± is plus. Add 11 to 19.

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

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{11±19}{20} when ± is minus. Subtract 19 from 11.

$$x=-\frac{8}{20}$$

Reduce the fraction \frac{-8}{20} to lowest terms by extracting and canceling out 4.

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

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{2}{5} for x_{2}.

$$10x^{2}-11x-6=10\left(x-\frac{3}{2}\right)\left(x-\left(-\frac{2}{5}\right)\right)$$

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

$$10x^{2}-11x-6=10\left(x-\frac{3}{2}\right)\left(x+\frac{2}{5}\right)$$

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

$$10x^{2}-11x-6=10\times \frac{2x-3}{2}\left(x+\frac{2}{5}\right)$$

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

$$10x^{2}-11x-6=10\times \frac{2x-3}{2}\times \frac{5x+2}{5}$$

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

$$10x^{2}-11x-6=10\times \frac{\left(2x-3\right)\left(5x+2\right)}{2\times 5}$$

Multiply 2 times 5.

$$10x^{2}-11x-6=10\times \frac{\left(2x-3\right)\left(5x+2\right)}{10}$$

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

$$10x^{2}-11x-6=\left(2x-3\right)\left(5x+2\right)$$

Evaluate

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