Solve 25y^2-80y-80

$$25 y ^ { 2 } - 80 y - 80$$

Factor

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

Solution Steps

Factor out 5.

$$5\left(5y^{2}-16y-16\right)$$

Consider 5y^{2}-16y-16. Factor the expression by grouping. First, the expression needs to be rewritten as 5y^{2}+ay+by-16. To find a and b, set up a system to be solved.

$$a+b=-16 ab=5\left(-16\right)=-80$$

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. List all such integer pairs that give product -80.

$$1,-80 2,-40 4,-20 5,-16 8,-10$$

Calculate the sum for each pair.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

The solution is the pair that gives sum -16.

$$a=-20 b=4$$

Rewrite 5y^{2}-16y-16 as \left(5y^{2}-20y\right)+\left(4y-16\right).

$$\left(5y^{2}-20y\right)+\left(4y-16\right)$$

Factor out 5y in the first and 4 in the second group.

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

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

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

Rewrite the complete factored expression.

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

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.

$$25y^{2}-80y-80=0$$

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.

$$y=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 25\left(-80\right)}}{2\times 25}$$

Square -80.

$$y=\frac{-\left(-80\right)±\sqrt{6400-4\times 25\left(-80\right)}}{2\times 25}$$

Multiply -4 times 25.

$$y=\frac{-\left(-80\right)±\sqrt{6400-100\left(-80\right)}}{2\times 25}$$

Multiply -100 times -80.

$$y=\frac{-\left(-80\right)±\sqrt{6400+8000}}{2\times 25}$$

Add 6400 to 8000.

$$y=\frac{-\left(-80\right)±\sqrt{14400}}{2\times 25}$$

Take the square root of 14400.

$$y=\frac{-\left(-80\right)±120}{2\times 25}$$

The opposite of -80 is 80.

$$y=\frac{80±120}{2\times 25}$$

Multiply 2 times 25.

$$y=\frac{80±120}{50}$$

Now solve the equation y=\frac{80±120}{50} when ± is plus. Add 80 to 120.

$$y=\frac{200}{50}$$

Divide 200 by 50.

$$y=4$$

Now solve the equation y=\frac{80±120}{50} when ± is minus. Subtract 120 from 80.

$$y=-\frac{40}{50}$$

Reduce the fraction \frac{-40}{50} to lowest terms by extracting and canceling out 10.

$$y=-\frac{4}{5}$$

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 -\frac{4}{5} for x_{2}.

$$25y^{2}-80y-80=25\left(y-4\right)\left(y-\left(-\frac{4}{5}\right)\right)$$

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

$$25y^{2}-80y-80=25\left(y-4\right)\left(y+\frac{4}{5}\right)$$

Add \frac{4}{5} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25y^{2}-80y-80=25\left(y-4\right)\times \frac{5y+4}{5}$$

Cancel out 5, the greatest common factor in 25 and 5.

$$25y^{2}-80y-80=5\left(y-4\right)\left(5y+4\right)$$

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.This is achieved by dividing both sides of the equation by 25

$$x ^ 2 -\frac{16}{5}x -\frac{16}{5} = 0$$

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

$$r + s = \frac{16}{5} rs = -\frac{16}{5}$$

Two numbers r and s sum up to \frac{16}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{16}{5} = \frac{8}{5}. 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>

$$r = \frac{8}{5} - u s = \frac{8}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{16}{5}

$$(\frac{8}{5} - u) (\frac{8}{5} + u) = -\frac{16}{5}$$

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

$$\frac{64}{25} - u^2 = -\frac{16}{5}$$

Simplify the expression by subtracting \frac{64}{25} on both sides

$$-u^2 = -\frac{16}{5}-\frac{64}{25} = -\frac{144}{25}$$

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

$$u^2 = \frac{144}{25} u = \pm\sqrt{\frac{144}{25}} = \pm \frac{12}{5}$$

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

$$r =\frac{8}{5} - \frac{12}{5} = -0.800 s = \frac{8}{5} + \frac{12}{5} = 4$$

Evaluate

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