Solve 6r^2-12r-18

$$6 r ^ { 2 } - 12 r - 18$$

Factor

$$6\left(r-3\right)\left(r+1\right)$$

Solution Steps

Factor out 6.

$$6\left(r^{2}-2r-3\right)$$

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

$$a+b=-2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

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

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

Factor out r in r^{2}-3r.

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

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

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

Rewrite the complete factored expression.

$$6\left(r-3\right)\left(r+1\right)$$

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.

$$6r^{2}-12r-18=0$$

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.

$$r=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 6\left(-18\right)}}{2\times 6}$$

Square -12.

$$r=\frac{-\left(-12\right)±\sqrt{144-4\times 6\left(-18\right)}}{2\times 6}$$

Multiply -4 times 6.

$$r=\frac{-\left(-12\right)±\sqrt{144-24\left(-18\right)}}{2\times 6}$$

Multiply -24 times -18.

$$r=\frac{-\left(-12\right)±\sqrt{144+432}}{2\times 6}$$

Add 144 to 432.

$$r=\frac{-\left(-12\right)±\sqrt{576}}{2\times 6}$$

Take the square root of 576.

$$r=\frac{-\left(-12\right)±24}{2\times 6}$$

The opposite of -12 is 12.

$$r=\frac{12±24}{2\times 6}$$

Multiply 2 times 6.

$$r=\frac{12±24}{12}$$

Now solve the equation r=\frac{12±24}{12} when ± is plus. Add 12 to 24.

$$r=\frac{36}{12}$$

Divide 36 by 12.

$$r=3$$

Now solve the equation r=\frac{12±24}{12} when ± is minus. Subtract 24 from 12.

$$r=-\frac{12}{12}$$

Divide -12 by 12.

$$r=-1$$

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

$$6r^{2}-12r-18=6\left(r-3\right)\left(r-\left(-1\right)\right)$$

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

$$6r^{2}-12r-18=6\left(r-3\right)\left(r+1\right)$$

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.This is achieved by dividing both sides of the equation by 6

$$x ^ 2 -2x -3 = 0$$

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

$$r + s = 2 rs = -3$$

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

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -3

$$(1 - u) (1 + u) = -3$$

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

$$1 - u^2 = -3$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -3-1 = -4$$

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

$$r =1 - 2 = -1 s = 1 + 2 = 3$$

Evaluate

$$6\left(r-3\right)\left(r+1\right)$$