Solve 6x^2-26x-20

$$6 { x }^{ 2 } -26x-20$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(3x^{2}-13x-10\right)$$

Consider 3x^{2}-13x-10. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-13 ab=3\left(-10\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -13.

$$a=-15 b=2$$

Rewrite 3x^{2}-13x-10 as \left(3x^{2}-15x\right)+\left(2x-10\right).

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

Factor out 3x in the first and 2 in the second group.

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

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

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

Rewrite the complete factored expression.

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

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.

$$6x^{2}-26x-20=0$$

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.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 6\left(-20\right)}}{2\times 6}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 6\left(-20\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-26\right)±\sqrt{676-24\left(-20\right)}}{2\times 6}$$

Multiply -24 times -20.

$$x=\frac{-\left(-26\right)±\sqrt{676+480}}{2\times 6}$$

Add 676 to 480.

$$x=\frac{-\left(-26\right)±\sqrt{1156}}{2\times 6}$$

Take the square root of 1156.

$$x=\frac{-\left(-26\right)±34}{2\times 6}$$

The opposite of -26 is 26.

$$x=\frac{26±34}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{26±34}{12}$$

Now solve the equation x=\frac{26±34}{12} when ± is plus. Add 26 to 34.

$$x=\frac{60}{12}$$

Divide 60 by 12.

$$x=5$$

Now solve the equation x=\frac{26±34}{12} when ± is minus. Subtract 34 from 26.

$$x=-\frac{8}{12}$$

Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.

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

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

$$6x^{2}-26x-20=6\left(x-5\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$6x^{2}-26x-20=6\left(x-5\right)\left(x+\frac{2}{3}\right)$$

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

$$6x^{2}-26x-20=6\left(x-5\right)\times \frac{3x+2}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6x^{2}-26x-20=2\left(x-5\right)\left(3x+2\right)$$

Evaluate

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