Solve 80y^2-36y+4

$$80 y ^ { 2 } - 36 y + 4$$

Factor

$$4\left(4y-1\right)\left(5y-1\right)$$

Solution Steps

Factor out 4.

$$4\left(20y^{2}-9y+1\right)$$

Consider 20y^{2}-9y+1. Factor the expression by grouping. First, the expression needs to be rewritten as 20y^{2}+ay+by+1. To find a and b, set up a system to be solved.

$$a+b=-9 ab=20\times 1=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

Rewrite 20y^{2}-9y+1 as \left(20y^{2}-5y\right)+\left(-4y+1\right).

$$\left(20y^{2}-5y\right)+\left(-4y+1\right)$$

Factor out 5y in the first and -1 in the second group.

$$5y\left(4y-1\right)-\left(4y-1\right)$$

Factor out common term 4y-1 by using distributive property.

$$\left(4y-1\right)\left(5y-1\right)$$

Rewrite the complete factored expression.

$$4\left(4y-1\right)\left(5y-1\right)$$

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.

$$80y^{2}-36y+4=0$$

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.

$$y=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 80\times 4}}{2\times 80}$$

Square -36.

$$y=\frac{-\left(-36\right)±\sqrt{1296-4\times 80\times 4}}{2\times 80}$$

Multiply -4 times 80.

$$y=\frac{-\left(-36\right)±\sqrt{1296-320\times 4}}{2\times 80}$$

Multiply -320 times 4.

$$y=\frac{-\left(-36\right)±\sqrt{1296-1280}}{2\times 80}$$

Add 1296 to -1280.

$$y=\frac{-\left(-36\right)±\sqrt{16}}{2\times 80}$$

Take the square root of 16.

$$y=\frac{-\left(-36\right)±4}{2\times 80}$$

The opposite of -36 is 36.

$$y=\frac{36±4}{2\times 80}$$

Multiply 2 times 80.

$$y=\frac{36±4}{160}$$

Now solve the equation y=\frac{36±4}{160} when ± is plus. Add 36 to 4.

$$y=\frac{40}{160}$$

Reduce the fraction \frac{40}{160} to lowest terms by extracting and canceling out 40.

$$y=\frac{1}{4}$$

Now solve the equation y=\frac{36±4}{160} when ± is minus. Subtract 4 from 36.

$$y=\frac{32}{160}$$

Reduce the fraction \frac{32}{160} to lowest terms by extracting and canceling out 32.

$$y=\frac{1}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{4} for x_{1} and \frac{1}{5} for x_{2}.

$$80y^{2}-36y+4=80\left(y-\frac{1}{4}\right)\left(y-\frac{1}{5}\right)$$

Subtract \frac{1}{4} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$80y^{2}-36y+4=80\times \frac{4y-1}{4}\left(y-\frac{1}{5}\right)$$

Subtract \frac{1}{5} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$80y^{2}-36y+4=80\times \frac{4y-1}{4}\times \frac{5y-1}{5}$$

Multiply \frac{4y-1}{4} times \frac{5y-1}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$80y^{2}-36y+4=80\times \frac{\left(4y-1\right)\left(5y-1\right)}{4\times 5}$$

Multiply 4 times 5.

$$80y^{2}-36y+4=80\times \frac{\left(4y-1\right)\left(5y-1\right)}{20}$$

Cancel out 20, the greatest common factor in 80 and 20.

$$80y^{2}-36y+4=4\left(4y-1\right)\left(5y-1\right)$$

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 80

$$x ^ 2 -\frac{9}{20}x +\frac{1}{20} = 0$$

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

$$r + s = \frac{9}{20} rs = \frac{1}{20}$$

Two numbers r and s sum up to \frac{9}{20} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{20} = \frac{9}{40}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{9}{40} - u s = \frac{9}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{20}

$$(\frac{9}{40} - u) (\frac{9}{40} + u) = \frac{1}{20}$$

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

$$\frac{81}{1600} - u^2 = \frac{1}{20}$$

Simplify the expression by subtracting \frac{81}{1600} on both sides

$$-u^2 = \frac{1}{20}-\frac{81}{1600} = -\frac{1}{1600}$$

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

$$u^2 = \frac{1}{1600} u = \pm\sqrt{\frac{1}{1600}} = \pm \frac{1}{40}$$

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

$$r =\frac{9}{40} - \frac{1}{40} = 0.200 s = \frac{9}{40} + \frac{1}{40} = 0.250$$

Evaluate

$$80y^{2}-36y+4$$