Solve -72q^2+12q+40=

$$- 72 q ^ { 2 } + 12 q + 40 =$$

Factor

$$4\left(5-6q\right)\left(3q+2\right)$$

Solution Steps

Factor out 4.

$$4\left(-18q^{2}+3q+10\right)$$

Consider -18q^{2}+3q+10. Factor the expression by grouping. First, the expression needs to be rewritten as -18q^{2}+aq+bq+10. To find a and b, set up a system to be solved.

$$a+b=3 ab=-18\times 10=-180$$

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 -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 3.

$$a=15 b=-12$$

Rewrite -18q^{2}+3q+10 as \left(-18q^{2}+15q\right)+\left(-12q+10\right).

$$\left(-18q^{2}+15q\right)+\left(-12q+10\right)$$

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

$$3q\left(-6q+5\right)+2\left(-6q+5\right)$$

Factor out common term -6q+5 by using distributive property.

$$\left(-6q+5\right)\left(3q+2\right)$$

Rewrite the complete factored expression.

$$4\left(-6q+5\right)\left(3q+2\right)$$

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.

$$-72q^{2}+12q+40=0$$

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.

$$q=\frac{-12±\sqrt{12^{2}-4\left(-72\right)\times 40}}{2\left(-72\right)}$$

Square 12.

$$q=\frac{-12±\sqrt{144-4\left(-72\right)\times 40}}{2\left(-72\right)}$$

Multiply -4 times -72.

$$q=\frac{-12±\sqrt{144+288\times 40}}{2\left(-72\right)}$$

Multiply 288 times 40.

$$q=\frac{-12±\sqrt{144+11520}}{2\left(-72\right)}$$

Add 144 to 11520.

$$q=\frac{-12±\sqrt{11664}}{2\left(-72\right)}$$

Take the square root of 11664.

$$q=\frac{-12±108}{2\left(-72\right)}$$

Multiply 2 times -72.

$$q=\frac{-12±108}{-144}$$

Now solve the equation q=\frac{-12±108}{-144} when ± is plus. Add -12 to 108.

$$q=\frac{96}{-144}$$

Reduce the fraction \frac{96}{-144} to lowest terms by extracting and canceling out 48.

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

Now solve the equation q=\frac{-12±108}{-144} when ± is minus. Subtract 108 from -12.

$$q=-\frac{120}{-144}$$

Reduce the fraction \frac{-120}{-144} to lowest terms by extracting and canceling out 24.

$$q=\frac{5}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{3} for x_{1} and \frac{5}{6} for x_{2}.

$$-72q^{2}+12q+40=-72\left(q-\left(-\frac{2}{3}\right)\right)\left(q-\frac{5}{6}\right)$$

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

$$-72q^{2}+12q+40=-72\left(q+\frac{2}{3}\right)\left(q-\frac{5}{6}\right)$$

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

$$-72q^{2}+12q+40=-72\times \frac{-3q-2}{-3}\left(q-\frac{5}{6}\right)$$

Subtract \frac{5}{6} from q by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-72q^{2}+12q+40=-72\times \frac{-3q-2}{-3}\times \frac{-6q+5}{-6}$$

Multiply \frac{-3q-2}{-3} times \frac{-6q+5}{-6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-72q^{2}+12q+40=-72\times \frac{\left(-3q-2\right)\left(-6q+5\right)}{-3\left(-6\right)}$$

Multiply -3 times -6.

$$-72q^{2}+12q+40=-72\times \frac{\left(-3q-2\right)\left(-6q+5\right)}{18}$$

Cancel out 18, the greatest common factor in -72 and 18.

$$-72q^{2}+12q+40=-4\left(-3q-2\right)\left(-6q+5\right)$$

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.

$$x ^ 2 -\frac{1}{6}x -\frac{5}{9} = 0$$

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

$$r + s = \frac{1}{6} rs = -\frac{5}{9}$$

Two numbers r and s sum up to \frac{1}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{6} = \frac{1}{12}. 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>

$$r = \frac{1}{12} - u s = \frac{1}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{9}

$$(\frac{1}{12} - u) (\frac{1}{12} + u) = -\frac{5}{9}$$

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

$$\frac{1}{144} - u^2 = -\frac{5}{9}$$

Simplify the expression by subtracting \frac{1}{144} on both sides

$$-u^2 = -\frac{5}{9}-\frac{1}{144} = -\frac{9}{16}$$

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

$$u^2 = \frac{9}{16} u = \pm\sqrt{\frac{9}{16}} = \pm \frac{3}{4}$$

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

$$r =\frac{1}{12} - \frac{3}{4} = -0.667 s = \frac{1}{12} + \frac{3}{4} = 0.833$$

Evaluate

$$40+12q-72q^{2}$$