Solve -12x^2+x+20

$$- 12 x ^ { 2 } + x + 20$$

Factor

$$\left(-4x-5\right)\left(3x-4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -12x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=1 ab=-12\times 20=-240$$

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 -240.

$$-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16$$

Calculate the sum for each pair.

$$-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1$$

The solution is the pair that gives sum 1.

$$a=16 b=-15$$

Rewrite -12x^{2}+x+20 as \left(-12x^{2}+16x\right)+\left(-15x+20\right).

$$\left(-12x^{2}+16x\right)+\left(-15x+20\right)$$

Factor out -4x in the first and -5 in the second group.

$$-4x\left(3x-4\right)-5\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(-4x-5\right)$$

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.

$$-12x^{2}+x+20=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-12\right)\times 20}}{2\left(-12\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-12\right)\times 20}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-1±\sqrt{1+48\times 20}}{2\left(-12\right)}$$

Multiply 48 times 20.

$$x=\frac{-1±\sqrt{1+960}}{2\left(-12\right)}$$

Add 1 to 960.

$$x=\frac{-1±\sqrt{961}}{2\left(-12\right)}$$

Take the square root of 961.

$$x=\frac{-1±31}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-1±31}{-24}$$

Now solve the equation x=\frac{-1±31}{-24} when ± is plus. Add -1 to 31.

$$x=\frac{30}{-24}$$

Reduce the fraction \frac{30}{-24} to lowest terms by extracting and canceling out 6.

$$x=-\frac{5}{4}$$

Now solve the equation x=\frac{-1±31}{-24} when ± is minus. Subtract 31 from -1.

$$x=-\frac{32}{-24}$$

Reduce the fraction \frac{-32}{-24} to lowest terms by extracting and canceling out 8.

$$x=\frac{4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{4} for x_{1} and \frac{4}{3} for x_{2}.

$$-12x^{2}+x+20=-12\left(x-\left(-\frac{5}{4}\right)\right)\left(x-\frac{4}{3}\right)$$

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

$$-12x^{2}+x+20=-12\left(x+\frac{5}{4}\right)\left(x-\frac{4}{3}\right)$$

Add \frac{5}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-12x^{2}+x+20=-12\times \frac{-4x-5}{-4}\left(x-\frac{4}{3}\right)$$

Subtract \frac{4}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-12x^{2}+x+20=-12\times \frac{-4x-5}{-4}\times \frac{-3x+4}{-3}$$

Multiply \frac{-4x-5}{-4} times \frac{-3x+4}{-3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-12x^{2}+x+20=-12\times \frac{\left(-4x-5\right)\left(-3x+4\right)}{-4\left(-3\right)}$$

Multiply -4 times -3.

$$-12x^{2}+x+20=-12\times \frac{\left(-4x-5\right)\left(-3x+4\right)}{12}$$

Cancel out 12, the greatest common factor in -12 and 12.

$$-12x^{2}+x+20=-\left(-4x-5\right)\left(-3x+4\right)$$

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.

$$x ^ 2 -\frac{1}{12}x -\frac{5}{3} = 0$$

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

$$r + s = \frac{1}{12} rs = -\frac{5}{3}$$

Two numbers r and s sum up to \frac{1}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{12} = \frac{1}{24}. 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>

$$r = \frac{1}{24} - u s = \frac{1}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{3}

$$(\frac{1}{24} - u) (\frac{1}{24} + u) = -\frac{5}{3}$$

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

$$\frac{1}{576} - u^2 = -\frac{5}{3}$$

Simplify the expression by subtracting \frac{1}{576} on both sides

$$-u^2 = -\frac{5}{3}-\frac{1}{576} = -\frac{961}{576}$$

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

$$u^2 = \frac{961}{576} u = \pm\sqrt{\frac{961}{576}} = \pm \frac{31}{24}$$

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

$$r =\frac{1}{24} - \frac{31}{24} = -1.250 s = \frac{1}{24} + \frac{31}{24} = 1.333$$

Evaluate

$$20+x-12x^{2}$$