Solve 6r^2-10r-4

$$6 r ^ { 2 } - 10 r - 4$$

Factor

$$2\left(r-2\right)\left(3r+1\right)$$

Solution Steps

Factor out 2.

$$2\left(3r^{2}-5r-2\right)$$

Consider 3r^{2}-5r-2. Factor the expression by grouping. First, the expression needs to be rewritten as 3r^{2}+ar+br-2. To find a and b, set up a system to be solved.

$$a+b=-5 ab=3\left(-2\right)=-6$$

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. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite 3r^{2}-5r-2 as \left(3r^{2}-6r\right)+\left(r-2\right).

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

Factor out 3r in 3r^{2}-6r.

$$3r\left(r-2\right)+r-2$$

Factor out common term r-2 by using distributive property.

$$\left(r-2\right)\left(3r+1\right)$$

Rewrite the complete factored expression.

$$2\left(r-2\right)\left(3r+1\right)$$

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.

$$6r^{2}-10r-4=0$$

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.

$$r=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

Square -10.

$$r=\frac{-\left(-10\right)±\sqrt{100-4\times 6\left(-4\right)}}{2\times 6}$$

Multiply -4 times 6.

$$r=\frac{-\left(-10\right)±\sqrt{100-24\left(-4\right)}}{2\times 6}$$

Multiply -24 times -4.

$$r=\frac{-\left(-10\right)±\sqrt{100+96}}{2\times 6}$$

Add 100 to 96.

$$r=\frac{-\left(-10\right)±\sqrt{196}}{2\times 6}$$

Take the square root of 196.

$$r=\frac{-\left(-10\right)±14}{2\times 6}$$

The opposite of -10 is 10.

$$r=\frac{10±14}{2\times 6}$$

Multiply 2 times 6.

$$r=\frac{10±14}{12}$$

Now solve the equation r=\frac{10±14}{12} when ± is plus. Add 10 to 14.

$$r=\frac{24}{12}$$

Divide 24 by 12.

$$r=2$$

Now solve the equation r=\frac{10±14}{12} when ± is minus. Subtract 14 from 10.

$$r=-\frac{4}{12}$$

Reduce the fraction \frac{-4}{12} to lowest terms by extracting and canceling out 4.

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

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

$$6r^{2}-10r-4=6\left(r-2\right)\left(r-\left(-\frac{1}{3}\right)\right)$$

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

$$6r^{2}-10r-4=6\left(r-2\right)\left(r+\frac{1}{3}\right)$$

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

$$6r^{2}-10r-4=6\left(r-2\right)\times \frac{3r+1}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6r^{2}-10r-4=2\left(r-2\right)\left(3r+1\right)$$

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

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

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

$$r + s = \frac{5}{3} rs = -\frac{2}{3}$$

Two numbers r and s sum up to \frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{3} = \frac{5}{6}. 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>

$$r = \frac{5}{6} - u s = \frac{5}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{3}

$$(\frac{5}{6} - u) (\frac{5}{6} + u) = -\frac{2}{3}$$

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

$$\frac{25}{36} - u^2 = -\frac{2}{3}$$

Simplify the expression by subtracting \frac{25}{36} on both sides

$$-u^2 = -\frac{2}{3}-\frac{25}{36} = -\frac{49}{36}$$

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

$$u^2 = \frac{49}{36} u = \pm\sqrt{\frac{49}{36}} = \pm \frac{7}{6}$$

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

$$r =\frac{5}{6} - \frac{7}{6} = -0.333 s = \frac{5}{6} + \frac{7}{6} = 2$$

Evaluate

$$2\left(r-2\right)\left(3r+1\right)$$