Solve 6x^2-20x-16

$$6 x ^ { 2 } - 20 x - 16$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(3x^{2}-10x-8\right)$$

Consider 3x^{2}-10x-8. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-10 ab=3\left(-8\right)=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -10.

$$a=-12 b=2$$

Rewrite 3x^{2}-10x-8 as \left(3x^{2}-12x\right)+\left(2x-8\right).

$$\left(3x^{2}-12x\right)+\left(2x-8\right)$$

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

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

Factor out common term x-4 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$6x^{2}-20x-16=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 6\left(-16\right)}}{2\times 6}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 6\left(-16\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-20\right)±\sqrt{400-24\left(-16\right)}}{2\times 6}$$

Multiply -24 times -16.

$$x=\frac{-\left(-20\right)±\sqrt{400+384}}{2\times 6}$$

Add 400 to 384.

$$x=\frac{-\left(-20\right)±\sqrt{784}}{2\times 6}$$

Take the square root of 784.

$$x=\frac{-\left(-20\right)±28}{2\times 6}$$

The opposite of -20 is 20.

$$x=\frac{20±28}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{20±28}{12}$$

Now solve the equation x=\frac{20±28}{12} when ± is plus. Add 20 to 28.

$$x=\frac{48}{12}$$

Divide 48 by 12.

$$x=4$$

Now solve the equation x=\frac{20±28}{12} when ± is minus. Subtract 28 from 20.

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

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 4 for x_{1} and -\frac{2}{3} for x_{2}.

$$6x^{2}-20x-16=6\left(x-4\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$6x^{2}-20x-16=6\left(x-4\right)\left(x+\frac{2}{3}\right)$$

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

$$6x^{2}-20x-16=6\left(x-4\right)\times \frac{3x+2}{3}$$

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

$$6x^{2}-20x-16=2\left(x-4\right)\left(3x+2\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 6

$$x ^ 2 -\frac{10}{3}x -\frac{8}{3} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{10}{3} rs = -\frac{8}{3}$$

Two numbers r and s sum up to \frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{3} = \frac{5}{3}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{5}{3} - u s = \frac{5}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{8}{3}

$$(\frac{5}{3} - u) (\frac{5}{3} + u) = -\frac{8}{3}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{25}{9} - u^2 = -\frac{8}{3}$$

Simplify the expression by subtracting \frac{25}{9} on both sides

$$-u^2 = -\frac{8}{3}-\frac{25}{9} = -\frac{49}{9}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{49}{9} u = \pm\sqrt{\frac{49}{9}} = \pm \frac{7}{3}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{5}{3} - \frac{7}{3} = -0.667 s = \frac{5}{3} + \frac{7}{3} = 4$$

Evaluate

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