Solve 4y^2+4y-8

$$4 y ^ { 2 } + 4 y - 8$$

Factor

$$4\left(y-1\right)\left(y+2\right)$$

Solution Steps

Factor out 4.

$$4\left(y^{2}+y-2\right)$$

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

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

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. The only such pair is the system solution.

$$a=-1 b=2$$

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

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

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

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

Factor out common term y-1 by using distributive property.

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

Rewrite the complete factored expression.

$$4\left(y-1\right)\left(y+2\right)$$

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.

$$4y^{2}+4y-8=0$$

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.

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

Square 4.

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

Multiply -4 times 4.

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

Multiply -16 times -8.

$$y=\frac{-4±\sqrt{16+128}}{2\times 4}$$

Add 16 to 128.

$$y=\frac{-4±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$y=\frac{-4±12}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{-4±12}{8}$$

Now solve the equation y=\frac{-4±12}{8} when ± is plus. Add -4 to 12.

$$y=\frac{8}{8}$$

Divide 8 by 8.

$$y=1$$

Now solve the equation y=\frac{-4±12}{8} when ± is minus. Subtract 12 from -4.

$$y=-\frac{16}{8}$$

Divide -16 by 8.

$$y=-2$$

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

$$4y^{2}+4y-8=4\left(y-1\right)\left(y-\left(-2\right)\right)$$

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

$$4y^{2}+4y-8=4\left(y-1\right)\left(y+2\right)$$

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 4

$$x ^ 2 +1x -2 = 0$$

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

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

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

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

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

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

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

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

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -2-\frac{1}{4} = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =-\frac{1}{2} - \frac{3}{2} = -2 s = -\frac{1}{2} + \frac{3}{2} = 1$$

Evaluate

$$4\left(y-1\right)\left(y+2\right)$$