Solve 5y^2-44y+96

$$5 y ^ { 2 } - 44 y + 96$$

Factor

$$\left(y-4\right)\left(5y-24\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5y^{2}+ay+by+96. To find a and b, set up a system to be solved.

$$a+b=-44 ab=5\times 96=480$$

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 480.

$$-1,-480 -2,-240 -3,-160 -4,-120 -5,-96 -6,-80 -8,-60 -10,-48 -12,-40 -15,-32 -16,-30 -20,-24$$

Calculate the sum for each pair.

$$-1-480=-481 -2-240=-242 -3-160=-163 -4-120=-124 -5-96=-101 -6-80=-86 -8-60=-68 -10-48=-58 -12-40=-52 -15-32=-47 -16-30=-46 -20-24=-44$$

The solution is the pair that gives sum -44.

$$a=-24 b=-20$$

Rewrite 5y^{2}-44y+96 as \left(5y^{2}-24y\right)+\left(-20y+96\right).

$$\left(5y^{2}-24y\right)+\left(-20y+96\right)$$

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

$$y\left(5y-24\right)-4\left(5y-24\right)$$

Factor out common term 5y-24 by using distributive property.

$$\left(5y-24\right)\left(y-4\right)$$

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.

$$5y^{2}-44y+96=0$$

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.

$$y=\frac{-\left(-44\right)±\sqrt{\left(-44\right)^{2}-4\times 5\times 96}}{2\times 5}$$

Square -44.

$$y=\frac{-\left(-44\right)±\sqrt{1936-4\times 5\times 96}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-44\right)±\sqrt{1936-20\times 96}}{2\times 5}$$

Multiply -20 times 96.

$$y=\frac{-\left(-44\right)±\sqrt{1936-1920}}{2\times 5}$$

Add 1936 to -1920.

$$y=\frac{-\left(-44\right)±\sqrt{16}}{2\times 5}$$

Take the square root of 16.

$$y=\frac{-\left(-44\right)±4}{2\times 5}$$

The opposite of -44 is 44.

$$y=\frac{44±4}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{44±4}{10}$$

Now solve the equation y=\frac{44±4}{10} when ± is plus. Add 44 to 4.

$$y=\frac{48}{10}$$

Reduce the fraction \frac{48}{10} to lowest terms by extracting and canceling out 2.

$$y=\frac{24}{5}$$

Now solve the equation y=\frac{44±4}{10} when ± is minus. Subtract 4 from 44.

$$y=\frac{40}{10}$$

Divide 40 by 10.

$$y=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{24}{5} for x_{1} and 4 for x_{2}.

$$5y^{2}-44y+96=5\left(y-\frac{24}{5}\right)\left(y-4\right)$$

Subtract \frac{24}{5} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5y^{2}-44y+96=5\times \frac{5y-24}{5}\left(y-4\right)$$

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

$$5y^{2}-44y+96=\left(5y-24\right)\left(y-4\right)$$

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 5

$$x ^ 2 -\frac{44}{5}x +\frac{96}{5} = 0$$

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

$$r + s = \frac{44}{5} rs = \frac{96}{5}$$

Two numbers r and s sum up to \frac{44}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{44}{5} = \frac{22}{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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{22}{5} - u s = \frac{22}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{96}{5}

$$(\frac{22}{5} - u) (\frac{22}{5} + u) = \frac{96}{5}$$

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

$$\frac{484}{25} - u^2 = \frac{96}{5}$$

Simplify the expression by subtracting \frac{484}{25} on both sides

$$-u^2 = \frac{96}{5}-\frac{484}{25} = -\frac{4}{25}$$

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

$$u^2 = \frac{4}{25} u = \pm\sqrt{\frac{4}{25}} = \pm \frac{2}{5}$$

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

$$r =\frac{22}{5} - \frac{2}{5} = 4.000 s = \frac{22}{5} + \frac{2}{5} = 4.800$$

Evaluate

$$\left(y-4\right)\left(5y-24\right)$$