Solve 6x^2-33x-120=

$$6 x ^ { 2 } - 33 x - 120 =$$

Factor

$$3\left(x-8\right)\left(2x+5\right)$$

Solution Steps

Factor out 3.

$$3\left(2x^{2}-11x-40\right)$$

Consider 2x^{2}-11x-40. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\left(-40\right)=-80$$

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 -80.

$$1,-80 2,-40 4,-20 5,-16 8,-10$$

Calculate the sum for each pair.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

The solution is the pair that gives sum -11.

$$a=-16 b=5$$

Rewrite 2x^{2}-11x-40 as \left(2x^{2}-16x\right)+\left(5x-40\right).

$$\left(2x^{2}-16x\right)+\left(5x-40\right)$$

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

$$2x\left(x-8\right)+5\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(2x+5\right)$$

Rewrite the complete factored expression.

$$3\left(x-8\right)\left(2x+5\right)$$

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.

$$6x^{2}-33x-120=0$$

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.

$$x=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 6\left(-120\right)}}{2\times 6}$$

Square -33.

$$x=\frac{-\left(-33\right)±\sqrt{1089-4\times 6\left(-120\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-33\right)±\sqrt{1089-24\left(-120\right)}}{2\times 6}$$

Multiply -24 times -120.

$$x=\frac{-\left(-33\right)±\sqrt{1089+2880}}{2\times 6}$$

Add 1089 to 2880.

$$x=\frac{-\left(-33\right)±\sqrt{3969}}{2\times 6}$$

Take the square root of 3969.

$$x=\frac{-\left(-33\right)±63}{2\times 6}$$

The opposite of -33 is 33.

$$x=\frac{33±63}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{33±63}{12}$$

Now solve the equation x=\frac{33±63}{12} when ± is plus. Add 33 to 63.

$$x=\frac{96}{12}$$

Divide 96 by 12.

$$x=8$$

Now solve the equation x=\frac{33±63}{12} when ± is minus. Subtract 63 from 33.

$$x=-\frac{30}{12}$$

Reduce the fraction \frac{-30}{12} to lowest terms by extracting and canceling out 6.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and -\frac{5}{2} for x_{2}.

$$6x^{2}-33x-120=6\left(x-8\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$6x^{2}-33x-120=6\left(x-8\right)\left(x+\frac{5}{2}\right)$$

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

$$6x^{2}-33x-120=6\left(x-8\right)\times \frac{2x+5}{2}$$

Cancel out 2, the greatest common factor in 6 and 2.

$$6x^{2}-33x-120=3\left(x-8\right)\left(2x+5\right)$$

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

$$x ^ 2 -\frac{11}{2}x -20 = 0$$

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

$$r + s = \frac{11}{2} rs = -20$$

Two numbers r and s sum up to \frac{11}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{2} = \frac{11}{4}. 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>

$$r = \frac{11}{4} - u s = \frac{11}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -20

$$(\frac{11}{4} - u) (\frac{11}{4} + u) = -20$$

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

$$\frac{121}{16} - u^2 = -20$$

Simplify the expression by subtracting \frac{121}{16} on both sides

$$-u^2 = -20-\frac{121}{16} = -\frac{441}{16}$$

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

$$u^2 = \frac{441}{16} u = \pm\sqrt{\frac{441}{16}} = \pm \frac{21}{4}$$

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

$$r =\frac{11}{4} - \frac{21}{4} = -2.500 s = \frac{11}{4} + \frac{21}{4} = 8$$

Evaluate

$$3\left(x-8\right)\left(2x+5\right)$$