Solve 18x^2-18x-80

$$18 x ^ { 2 } - 18 x - 80$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(9x^{2}-9x-40\right)$$

Consider 9x^{2}-9x-40. Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-9 ab=9\left(-40\right)=-360$$

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 -360.

$$1,-360 2,-180 3,-120 4,-90 5,-72 6,-60 8,-45 9,-40 10,-36 12,-30 15,-24 18,-20$$

Calculate the sum for each pair.

$$1-360=-359 2-180=-178 3-120=-117 4-90=-86 5-72=-67 6-60=-54 8-45=-37 9-40=-31 10-36=-26 12-30=-18 15-24=-9 18-20=-2$$

The solution is the pair that gives sum -9.

$$a=-24 b=15$$

Rewrite 9x^{2}-9x-40 as \left(9x^{2}-24x\right)+\left(15x-40\right).

$$\left(9x^{2}-24x\right)+\left(15x-40\right)$$

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

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

Factor out common term 3x-8 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$18x^{2}-18x-80=0$$

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.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 18\left(-80\right)}}{2\times 18}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 18\left(-80\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-18\right)±\sqrt{324-72\left(-80\right)}}{2\times 18}$$

Multiply -72 times -80.

$$x=\frac{-\left(-18\right)±\sqrt{324+5760}}{2\times 18}$$

Add 324 to 5760.

$$x=\frac{-\left(-18\right)±\sqrt{6084}}{2\times 18}$$

Take the square root of 6084.

$$x=\frac{-\left(-18\right)±78}{2\times 18}$$

The opposite of -18 is 18.

$$x=\frac{18±78}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{18±78}{36}$$

Now solve the equation x=\frac{18±78}{36} when ± is plus. Add 18 to 78.

$$x=\frac{96}{36}$$

Reduce the fraction \frac{96}{36} to lowest terms by extracting and canceling out 12.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{18±78}{36} when ± is minus. Subtract 78 from 18.

$$x=-\frac{60}{36}$$

Reduce the fraction \frac{-60}{36} to lowest terms by extracting and canceling out 12.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{8}{3} for x_{1} and -\frac{5}{3} for x_{2}.

$$18x^{2}-18x-80=18\left(x-\frac{8}{3}\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$18x^{2}-18x-80=18\left(x-\frac{8}{3}\right)\left(x+\frac{5}{3}\right)$$

Subtract \frac{8}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-18x-80=18\times \frac{3x-8}{3}\left(x+\frac{5}{3}\right)$$

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

$$18x^{2}-18x-80=18\times \frac{3x-8}{3}\times \frac{3x+5}{3}$$

Multiply \frac{3x-8}{3} times \frac{3x+5}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-18x-80=18\times \frac{\left(3x-8\right)\left(3x+5\right)}{3\times 3}$$

Multiply 3 times 3.

$$18x^{2}-18x-80=18\times \frac{\left(3x-8\right)\left(3x+5\right)}{9}$$

Cancel out 9, the greatest common factor in 18 and 9.

$$18x^{2}-18x-80=2\left(3x-8\right)\left(3x+5\right)$$

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 18

$$x ^ 2 -1x -\frac{40}{9} = 0$$

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

$$r + s = 1 rs = -\frac{40}{9}$$

Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{2}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{1}{2} - u s = \frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{40}{9}

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -\frac{40}{9}$$

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

$$\frac{1}{4} - u^2 = -\frac{40}{9}$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -\frac{40}{9}-\frac{1}{4} = -\frac{169}{36}$$

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

$$u^2 = \frac{169}{36} u = \pm\sqrt{\frac{169}{36}} = \pm \frac{13}{6}$$

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

$$r =\frac{1}{2} - \frac{13}{6} = -1.667 s = \frac{1}{2} + \frac{13}{6} = 2.667$$

Evaluate

$$18x^{2}-18x-80$$