Solve -18x^2+320x+72

$$- 18 x ^ { 2 } + 320 x + 72$$

Factor

$$2\left(18-x\right)\left(9x+2\right)$$

Solution Steps

Factor out 2.

$$2\left(-9x^{2}+160x+36\right)$$

Consider -9x^{2}+160x+36. Factor the expression by grouping. First, the expression needs to be rewritten as -9x^{2}+ax+bx+36. To find a and b, set up a system to be solved.

$$a+b=160 ab=-9\times 36=-324$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -324.

$$-1,324 -2,162 -3,108 -4,81 -6,54 -9,36 -12,27 -18,18$$

Calculate the sum for each pair.

$$-1+324=323 -2+162=160 -3+108=105 -4+81=77 -6+54=48 -9+36=27 -12+27=15 -18+18=0$$

The solution is the pair that gives sum 160.

$$a=162 b=-2$$

Rewrite -9x^{2}+160x+36 as \left(-9x^{2}+162x\right)+\left(-2x+36\right).

$$\left(-9x^{2}+162x\right)+\left(-2x+36\right)$$

Factor out 9x in the first and 2 in the second group.

$$9x\left(-x+18\right)+2\left(-x+18\right)$$

Factor out common term -x+18 by using distributive property.

$$\left(-x+18\right)\left(9x+2\right)$$

Rewrite the complete factored expression.

$$2\left(-x+18\right)\left(9x+2\right)$$

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.

$$-18x^{2}+320x+72=0$$

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.

$$x=\frac{-320±\sqrt{320^{2}-4\left(-18\right)\times 72}}{2\left(-18\right)}$$

Square 320.

$$x=\frac{-320±\sqrt{102400-4\left(-18\right)\times 72}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$x=\frac{-320±\sqrt{102400+72\times 72}}{2\left(-18\right)}$$

Multiply 72 times 72.

$$x=\frac{-320±\sqrt{102400+5184}}{2\left(-18\right)}$$

Add 102400 to 5184.

$$x=\frac{-320±\sqrt{107584}}{2\left(-18\right)}$$

Take the square root of 107584.

$$x=\frac{-320±328}{2\left(-18\right)}$$

Multiply 2 times -18.

$$x=\frac{-320±328}{-36}$$

Now solve the equation x=\frac{-320±328}{-36} when ± is plus. Add -320 to 328.

$$x=\frac{8}{-36}$$

Reduce the fraction \frac{8}{-36} to lowest terms by extracting and canceling out 4.

$$x=-\frac{2}{9}$$

Now solve the equation x=\frac{-320±328}{-36} when ± is minus. Subtract 328 from -320.

$$x=-\frac{648}{-36}$$

Divide -648 by -36.

$$x=18$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{9} for x_{1} and 18 for x_{2}.

$$-18x^{2}+320x+72=-18\left(x-\left(-\frac{2}{9}\right)\right)\left(x-18\right)$$

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

$$-18x^{2}+320x+72=-18\left(x+\frac{2}{9}\right)\left(x-18\right)$$

Add \frac{2}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-18x^{2}+320x+72=-18\times \frac{-9x-2}{-9}\left(x-18\right)$$

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

$$-18x^{2}+320x+72=2\left(-9x-2\right)\left(x-18\right)$$

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.

$$x ^ 2 -\frac{160}{9}x -4 = 0$$

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

$$r + s = \frac{160}{9} rs = -4$$

Two numbers r and s sum up to \frac{160}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{160}{9} = \frac{80}{9}. 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>

$$r = \frac{80}{9} - u s = \frac{80}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(\frac{80}{9} - u) (\frac{80}{9} + u) = -4$$

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

$$\frac{6400}{81} - u^2 = -4$$

Simplify the expression by subtracting \frac{6400}{81} on both sides

$$-u^2 = -4-\frac{6400}{81} = -\frac{6724}{81}$$

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

$$u^2 = \frac{6724}{81} u = \pm\sqrt{\frac{6724}{81}} = \pm \frac{82}{9}$$

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

$$r =\frac{80}{9} - \frac{82}{9} = -0.222 s = \frac{80}{9} + \frac{82}{9} = 18$$

Evaluate

$$72+320x-18x^{2}$$