Solve 6y^2+y-12

$$6 y ^ { 2 } + y - 12$$

Factor

$$\left(3y-4\right)\left(2y+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6y^{2}+ay+by-12. To find a and b, set up a system to be solved.

$$a+b=1 ab=6\left(-12\right)=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 1.

$$a=-8 b=9$$

Rewrite 6y^{2}+y-12 as \left(6y^{2}-8y\right)+\left(9y-12\right).

$$\left(6y^{2}-8y\right)+\left(9y-12\right)$$

Factor out 2y in the first and 3 in the second group.

$$2y\left(3y-4\right)+3\left(3y-4\right)$$

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

$$\left(3y-4\right)\left(2y+3\right)$$

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.

$$6y^{2}+y-12=0$$

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.

$$y=\frac{-1±\sqrt{1^{2}-4\times 6\left(-12\right)}}{2\times 6}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\times 6\left(-12\right)}}{2\times 6}$$

Multiply -4 times 6.

$$y=\frac{-1±\sqrt{1-24\left(-12\right)}}{2\times 6}$$

Multiply -24 times -12.

$$y=\frac{-1±\sqrt{1+288}}{2\times 6}$$

Add 1 to 288.

$$y=\frac{-1±\sqrt{289}}{2\times 6}$$

Take the square root of 289.

$$y=\frac{-1±17}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{-1±17}{12}$$

Now solve the equation y=\frac{-1±17}{12} when ± is plus. Add -1 to 17.

$$y=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation y=\frac{-1±17}{12} when ± is minus. Subtract 17 from -1.

$$y=-\frac{18}{12}$$

Reduce the fraction \frac{-18}{12} to lowest terms by extracting and canceling out 6.

$$y=-\frac{3}{2}$$

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

$$6y^{2}+y-12=6\left(y-\frac{4}{3}\right)\left(y-\left(-\frac{3}{2}\right)\right)$$

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

$$6y^{2}+y-12=6\left(y-\frac{4}{3}\right)\left(y+\frac{3}{2}\right)$$

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

$$6y^{2}+y-12=6\times \frac{3y-4}{3}\left(y+\frac{3}{2}\right)$$

Add \frac{3}{2} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6y^{2}+y-12=6\times \frac{3y-4}{3}\times \frac{2y+3}{2}$$

Multiply \frac{3y-4}{3} times \frac{2y+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6y^{2}+y-12=6\times \frac{\left(3y-4\right)\left(2y+3\right)}{3\times 2}$$

Multiply 3 times 2.

$$6y^{2}+y-12=6\times \frac{\left(3y-4\right)\left(2y+3\right)}{6}$$

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

$$6y^{2}+y-12=\left(3y-4\right)\left(2y+3\right)$$

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

$$x ^ 2 +\frac{1}{6}x -2 = 0$$

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

$$r + s = -\frac{1}{6} rs = -2$$

Two numbers r and s sum up to -\frac{1}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{6} = -\frac{1}{12}. 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>

$$r = -\frac{1}{12} - u s = -\frac{1}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -2

$$(-\frac{1}{12} - u) (-\frac{1}{12} + u) = -2$$

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

$$\frac{1}{144} - u^2 = -2$$

Simplify the expression by subtracting \frac{1}{144} on both sides

$$-u^2 = -2-\frac{1}{144} = -\frac{289}{144}$$

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

$$u^2 = \frac{289}{144} u = \pm\sqrt{\frac{289}{144}} = \pm \frac{17}{12}$$

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

$$r =-\frac{1}{12} - \frac{17}{12} = -1.500 s = -\frac{1}{12} + \frac{17}{12} = 1.333$$

Evaluate

$$6y^{2}+y-12$$