Solve 6y^2-90y+324=

$$6 y ^ { 2 } - 90 y + 324 =$$

Factor

$$6\left(y-9\right)\left(y-6\right)$$

Solution Steps

Factor out 6.

$$6\left(y^{2}-15y+54\right)$$

Consider y^{2}-15y+54. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+54. To find a and b, set up a system to be solved.

$$a+b=-15 ab=1\times 54=54$$

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 54.

$$-1,-54 -2,-27 -3,-18 -6,-9$$

Calculate the sum for each pair.

$$-1-54=-55 -2-27=-29 -3-18=-21 -6-9=-15$$

The solution is the pair that gives sum -15.

$$a=-9 b=-6$$

Rewrite y^{2}-15y+54 as \left(y^{2}-9y\right)+\left(-6y+54\right).

$$\left(y^{2}-9y\right)+\left(-6y+54\right)$$

Factor out y in the first and -6 in the second group.

$$y\left(y-9\right)-6\left(y-9\right)$$

Factor out common term y-9 by using distributive property.

$$\left(y-9\right)\left(y-6\right)$$

Rewrite the complete factored expression.

$$6\left(y-9\right)\left(y-6\right)$$

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.

$$6y^{2}-90y+324=0$$

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.

$$y=\frac{-\left(-90\right)±\sqrt{\left(-90\right)^{2}-4\times 6\times 324}}{2\times 6}$$

Square -90.

$$y=\frac{-\left(-90\right)±\sqrt{8100-4\times 6\times 324}}{2\times 6}$$

Multiply -4 times 6.

$$y=\frac{-\left(-90\right)±\sqrt{8100-24\times 324}}{2\times 6}$$

Multiply -24 times 324.

$$y=\frac{-\left(-90\right)±\sqrt{8100-7776}}{2\times 6}$$

Add 8100 to -7776.

$$y=\frac{-\left(-90\right)±\sqrt{324}}{2\times 6}$$

Take the square root of 324.

$$y=\frac{-\left(-90\right)±18}{2\times 6}$$

The opposite of -90 is 90.

$$y=\frac{90±18}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{90±18}{12}$$

Now solve the equation y=\frac{90±18}{12} when ± is plus. Add 90 to 18.

$$y=\frac{108}{12}$$

Divide 108 by 12.

$$y=9$$

Now solve the equation y=\frac{90±18}{12} when ± is minus. Subtract 18 from 90.

$$y=\frac{72}{12}$$

Divide 72 by 12.

$$y=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 9 for x_{1} and 6 for x_{2}.

$$6y^{2}-90y+324=6\left(y-9\right)\left(y-6\right)$$

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 6

$$x ^ 2 -15x +54 = 0$$

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

$$r + s = 15 rs = 54$$

Two numbers r and s sum up to 15 exactly when the average of the two numbers is \frac{1}{2}*15 = \frac{15}{2}. 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>

$$r = \frac{15}{2} - u s = \frac{15}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 54

$$(\frac{15}{2} - u) (\frac{15}{2} + u) = 54$$

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

$$\frac{225}{4} - u^2 = 54$$

Simplify the expression by subtracting \frac{225}{4} on both sides

$$-u^2 = 54-\frac{225}{4} = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =\frac{15}{2} - \frac{3}{2} = 6 s = \frac{15}{2} + \frac{3}{2} = 9$$

Evaluate

$$6\left(y-9\right)\left(y-6\right)$$