Solve -12r^2+5r+3

$$- 12 r ^ { 2 } + 5 r + 3$$

Factor

$$\left(3-4r\right)\left(3r+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -12r^{2}+ar+br+3. To find a and b, set up a system to be solved.

$$a+b=5 ab=-12\times 3=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=9 b=-4$$

Rewrite -12r^{2}+5r+3 as \left(-12r^{2}+9r\right)+\left(-4r+3\right).

$$\left(-12r^{2}+9r\right)+\left(-4r+3\right)$$

Factor out 3r in -12r^{2}+9r.

$$3r\left(-4r+3\right)-4r+3$$

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

$$\left(-4r+3\right)\left(3r+1\right)$$

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.

$$-12r^{2}+5r+3=0$$

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.

$$r=\frac{-5±\sqrt{5^{2}-4\left(-12\right)\times 3}}{2\left(-12\right)}$$

Square 5.

$$r=\frac{-5±\sqrt{25-4\left(-12\right)\times 3}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$r=\frac{-5±\sqrt{25+48\times 3}}{2\left(-12\right)}$$

Multiply 48 times 3.

$$r=\frac{-5±\sqrt{25+144}}{2\left(-12\right)}$$

Add 25 to 144.

$$r=\frac{-5±\sqrt{169}}{2\left(-12\right)}$$

Take the square root of 169.

$$r=\frac{-5±13}{2\left(-12\right)}$$

Multiply 2 times -12.

$$r=\frac{-5±13}{-24}$$

Now solve the equation r=\frac{-5±13}{-24} when ± is plus. Add -5 to 13.

$$r=\frac{8}{-24}$$

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

$$r=-\frac{1}{3}$$

Now solve the equation r=\frac{-5±13}{-24} when ± is minus. Subtract 13 from -5.

$$r=-\frac{18}{-24}$$

Reduce the fraction \frac{-18}{-24} to lowest terms by extracting and canceling out 6.

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

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

$$-12r^{2}+5r+3=-12\left(r-\left(-\frac{1}{3}\right)\right)\left(r-\frac{3}{4}\right)$$

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

$$-12r^{2}+5r+3=-12\left(r+\frac{1}{3}\right)\left(r-\frac{3}{4}\right)$$

Add \frac{1}{3} to r by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-12r^{2}+5r+3=-12\times \frac{-3r-1}{-3}\left(r-\frac{3}{4}\right)$$

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

$$-12r^{2}+5r+3=-12\times \frac{-3r-1}{-3}\times \frac{-4r+3}{-4}$$

Multiply \frac{-3r-1}{-3} times \frac{-4r+3}{-4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-12r^{2}+5r+3=-12\times \frac{\left(-3r-1\right)\left(-4r+3\right)}{-3\left(-4\right)}$$

Multiply -3 times -4.

$$-12r^{2}+5r+3=-12\times \frac{\left(-3r-1\right)\left(-4r+3\right)}{12}$$

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

$$-12r^{2}+5r+3=-\left(-3r-1\right)\left(-4r+3\right)$$

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.

$$x ^ 2 -\frac{5}{12}x -\frac{1}{4} = 0$$

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

$$r + s = \frac{5}{12} rs = -\frac{1}{4}$$

Two numbers r and s sum up to \frac{5}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{12} = \frac{5}{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>

$$r = \frac{5}{24} - u s = \frac{5}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{4}

$$(\frac{5}{24} - u) (\frac{5}{24} + u) = -\frac{1}{4}$$

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

$$\frac{25}{576} - u^2 = -\frac{1}{4}$$

Simplify the expression by subtracting \frac{25}{576} on both sides

$$-u^2 = -\frac{1}{4}-\frac{25}{576} = -\frac{169}{576}$$

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

$$u^2 = \frac{169}{576} u = \pm\sqrt{\frac{169}{576}} = \pm \frac{13}{24}$$

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

$$r =\frac{5}{24} - \frac{13}{24} = -0.333 s = \frac{5}{24} + \frac{13}{24} = 0.750$$

Evaluate

$$\left(3-4r\right)\left(3r+1\right)$$