Solve -5x^2+40x-80

$$- 5 x ^ { 2 } + 40 x - 80$$

Factor

$$-5\left(x-4\right)^{2}$$

Solution Steps

Factor out 5.

$$5\left(-x^{2}+8x-16\right)$$

Consider -x^{2}+8x-16. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=8 ab=-\left(-16\right)=16$$

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 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 8.

$$a=4 b=4$$

Rewrite -x^{2}+8x-16 as \left(-x^{2}+4x\right)+\left(4x-16\right).

$$\left(-x^{2}+4x\right)+\left(4x-16\right)$$

Factor out -x in the first and 4 in the second group.

$$-x\left(x-4\right)+4\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x+4\right)$$

Rewrite the complete factored expression.

$$5\left(x-4\right)\left(-x+4\right)$$

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.

$$-5x^{2}+40x-80=0$$

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.

$$x=\frac{-40±\sqrt{40^{2}-4\left(-5\right)\left(-80\right)}}{2\left(-5\right)}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\left(-5\right)\left(-80\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-40±\sqrt{1600+20\left(-80\right)}}{2\left(-5\right)}$$

Multiply 20 times -80.

$$x=\frac{-40±\sqrt{1600-1600}}{2\left(-5\right)}$$

Add 1600 to -1600.

$$x=\frac{-40±\sqrt{0}}{2\left(-5\right)}$$

Take the square root of 0.

$$x=\frac{-40±0}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-40±0}{-10}$$

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

$$-5x^{2}+40x-80=-5\left(x-4\right)\left(x-4\right)$$

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.

$$x ^ 2 -8x +16 = 0$$

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

$$r + s = 8 rs = 16$$

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 = 16

$$(4 - u) (4 + u) = 16$$

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

$$16 - u^2 = 16$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = 16-16 = 0$$

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.

$$r = s = 4$$

Evaluate

$$-5\left(x-4\right)^{2}$$