Solve -2y^2+32y-128

$$- 2 y ^ { 2 } + 32 y - 128$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(-y^{2}+16y-64\right)$$

Consider -y^{2}+16y-64. Factor the expression by grouping. First, the expression needs to be rewritten as -y^{2}+ay+by-64. To find a and b, set up a system to be solved.

$$a+b=16 ab=-\left(-64\right)=64$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 64.

$$1,64 2,32 4,16 8,8$$

Calculate the sum for each pair.

$$1+64=65 2+32=34 4+16=20 8+8=16$$

The solution is the pair that gives sum 16.

$$a=8 b=8$$

Rewrite -y^{2}+16y-64 as \left(-y^{2}+8y\right)+\left(8y-64\right).

$$\left(-y^{2}+8y\right)+\left(8y-64\right)$$

Factor out -y in the first and 8 in the second group.

$$-y\left(y-8\right)+8\left(y-8\right)$$

Factor out common term y-8 by using distributive property.

$$\left(y-8\right)\left(-y+8\right)$$

Rewrite the complete factored expression.

$$2\left(y-8\right)\left(-y+8\right)$$

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.

$$-2y^{2}+32y-128=0$$

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.

$$y=\frac{-32±\sqrt{32^{2}-4\left(-2\right)\left(-128\right)}}{2\left(-2\right)}$$

Square 32.

$$y=\frac{-32±\sqrt{1024-4\left(-2\right)\left(-128\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$y=\frac{-32±\sqrt{1024+8\left(-128\right)}}{2\left(-2\right)}$$

Multiply 8 times -128.

$$y=\frac{-32±\sqrt{1024-1024}}{2\left(-2\right)}$$

Add 1024 to -1024.

$$y=\frac{-32±\sqrt{0}}{2\left(-2\right)}$$

Take the square root of 0.

$$y=\frac{-32±0}{2\left(-2\right)}$$

Multiply 2 times -2.

$$y=\frac{-32±0}{-4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and 8 for x_{2}.

$$-2y^{2}+32y-128=-2\left(y-8\right)\left(y-8\right)$$

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.

$$x ^ 2 -16x +64 = 0$$

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

$$r + s = 16 rs = 64$$

Two numbers r and s sum up to 16 exactly when the average of the two numbers is \frac{1}{2}*16 = 8. 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>

$$r = 8 - u s = 8 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 64

$$(8 - u) (8 + u) = 64$$

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

$$64 - u^2 = 64$$

Simplify the expression by subtracting 64 on both sides

$$-u^2 = 64-64 = 0$$

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

$$u^2 = 0 u = 0$$

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

$$r = s = 8$$

Evaluate

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