Solve 2y^2-4y-6

$$2 y ^ { 2 } - 4 y - 6$$

Factor

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

Solution Steps

Factor out 2.

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

Consider y^{2}-2y-3. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-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 y^{2}-2y-3 as \left(y^{2}-3y\right)+\left(y-3\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$2y^{2}-4y-6=0$$

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.

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

Square -4.

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-4\right)±\sqrt{16-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$y=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 2}$$

Add 16 to 48.

$$y=\frac{-\left(-4\right)±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$y=\frac{-\left(-4\right)±8}{2\times 2}$$

The opposite of -4 is 4.

$$y=\frac{4±8}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{4±8}{4}$$

Now solve the equation y=\frac{4±8}{4} when ± is plus. Add 4 to 8.

$$y=\frac{12}{4}$$

Divide 12 by 4.

$$y=3$$

Now solve the equation y=\frac{4±8}{4} when ± is minus. Subtract 8 from 4.

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

Divide -4 by 4.

$$y=-1$$

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}.

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

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

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

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 2

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

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

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