Solve 2y^2-16y+24

$$2 y ^ { 2 } - 16 y + 24$$

Factor

$$2\left(y-6\right)\left(y-2\right)$$

Solution Steps

Factor out 2.

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

Consider y^{2}-8y+12. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+12. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

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

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

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

$$y\left(y-6\right)-2\left(y-6\right)$$

Factor out common term y-6 by using distributive property.

$$\left(y-6\right)\left(y-2\right)$$

Rewrite the complete factored expression.

$$2\left(y-6\right)\left(y-2\right)$$

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.

$$2y^{2}-16y+24=0$$

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.

$$y=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square -16.

$$y=\frac{-\left(-16\right)±\sqrt{256-4\times 2\times 24}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-16\right)±\sqrt{256-8\times 24}}{2\times 2}$$

Multiply -8 times 24.

$$y=\frac{-\left(-16\right)±\sqrt{256-192}}{2\times 2}$$

Add 256 to -192.

$$y=\frac{-\left(-16\right)±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

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

The opposite of -16 is 16.

$$y=\frac{16±8}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{16±8}{4}$$

Now solve the equation y=\frac{16±8}{4} when ± is plus. Add 16 to 8.

$$y=\frac{24}{4}$$

Divide 24 by 4.

$$y=6$$

Now solve the equation y=\frac{16±8}{4} when ± is minus. Subtract 8 from 16.

$$y=\frac{8}{4}$$

Divide 8 by 4.

$$y=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and 2 for x_{2}.

$$2y^{2}-16y+24=2\left(y-6\right)\left(y-2\right)$$

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 2

$$x ^ 2 -8x +12 = 0$$

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

$$r + s = 8 rs = 12$$

Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 4. 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>

$$r = 4 - u s = 4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 12

$$(4 - u) (4 + u) = 12$$

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

$$16 - u^2 = 12$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = 12-16 = -4$$

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.

$$r =4 - 2 = 2 s = 4 + 2 = 6$$

Evaluate

$$2\left(y-6\right)\left(y-2\right)$$