Solve -9x^2+31x+20

$$- 9 x ^ { 2 } + 31 x + 20$$

Factor

$$\left(4-x\right)\left(9x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -9x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=31 ab=-9\times 20=-180$$

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 31.

$$a=36 b=-5$$

Rewrite -9x^{2}+31x+20 as \left(-9x^{2}+36x\right)+\left(-5x+20\right).

$$\left(-9x^{2}+36x\right)+\left(-5x+20\right)$$

Factor out 9x in the first and 5 in the second group.

$$9x\left(-x+4\right)+5\left(-x+4\right)$$

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

$$\left(-x+4\right)\left(9x+5\right)$$

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.

$$-9x^{2}+31x+20=0$$

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.

$$x=\frac{-31±\sqrt{31^{2}-4\left(-9\right)\times 20}}{2\left(-9\right)}$$

Square 31.

$$x=\frac{-31±\sqrt{961-4\left(-9\right)\times 20}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-31±\sqrt{961+36\times 20}}{2\left(-9\right)}$$

Multiply 36 times 20.

$$x=\frac{-31±\sqrt{961+720}}{2\left(-9\right)}$$

Add 961 to 720.

$$x=\frac{-31±\sqrt{1681}}{2\left(-9\right)}$$

Take the square root of 1681.

$$x=\frac{-31±41}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-31±41}{-18}$$

Now solve the equation x=\frac{-31±41}{-18} when ± is plus. Add -31 to 41.

$$x=\frac{10}{-18}$$

Reduce the fraction \frac{10}{-18} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{9}$$

Now solve the equation x=\frac{-31±41}{-18} when ± is minus. Subtract 41 from -31.

$$x=-\frac{72}{-18}$$

Divide -72 by -18.

$$x=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{9} for x_{1} and 4 for x_{2}.

$$-9x^{2}+31x+20=-9\left(x-\left(-\frac{5}{9}\right)\right)\left(x-4\right)$$

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

$$-9x^{2}+31x+20=-9\left(x+\frac{5}{9}\right)\left(x-4\right)$$

Add \frac{5}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-9x^{2}+31x+20=-9\times \frac{-9x-5}{-9}\left(x-4\right)$$

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

$$-9x^{2}+31x+20=\left(-9x-5\right)\left(x-4\right)$$

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.

$$x ^ 2 -\frac{31}{9}x -\frac{20}{9} = 0$$

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

$$r + s = \frac{31}{9} rs = -\frac{20}{9}$$

Two numbers r and s sum up to \frac{31}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{31}{9} = \frac{31}{18}. 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>

$$r = \frac{31}{18} - u s = \frac{31}{18} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{20}{9}

$$(\frac{31}{18} - u) (\frac{31}{18} + u) = -\frac{20}{9}$$

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

$$\frac{961}{324} - u^2 = -\frac{20}{9}$$

Simplify the expression by subtracting \frac{961}{324} on both sides

$$-u^2 = -\frac{20}{9}-\frac{961}{324} = -\frac{1681}{324}$$

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

$$u^2 = \frac{1681}{324} u = \pm\sqrt{\frac{1681}{324}} = \pm \frac{41}{18}$$

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

$$r =\frac{31}{18} - \frac{41}{18} = -0.556 s = \frac{31}{18} + \frac{41}{18} = 4$$

Evaluate

$$\left(4-x\right)\left(9x+5\right)$$