Solve y^2-9y+18

$$y ^ { 2 } - 9 y + 18$$

Factor

$$\left(y-6\right)\left(y-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+18. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 18=18$$

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 18.

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

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -9.

$$a=-6 b=-3$$

Rewrite y^{2}-9y+18 as \left(y^{2}-6y\right)+\left(-3y+18\right).

$$\left(y^{2}-6y\right)+\left(-3y+18\right)$$

Factor out y in the first and -3 in the second group.

$$y\left(y-6\right)-3\left(y-6\right)$$

Factor out common term y-6 by using distributive property.

$$\left(y-6\right)\left(y-3\right)$$

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.

$$y^{2}-9y+18=0$$

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.

$$y=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 18}}{2}$$

Square -9.

$$y=\frac{-\left(-9\right)±\sqrt{81-4\times 18}}{2}$$

Multiply -4 times 18.

$$y=\frac{-\left(-9\right)±\sqrt{81-72}}{2}$$

Add 81 to -72.

$$y=\frac{-\left(-9\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$y=\frac{-\left(-9\right)±3}{2}$$

The opposite of -9 is 9.

$$y=\frac{9±3}{2}$$

Now solve the equation y=\frac{9±3}{2} when ± is plus. Add 9 to 3.

$$y=\frac{12}{2}$$

Divide 12 by 2.

$$y=6$$

Now solve the equation y=\frac{9±3}{2} when ± is minus. Subtract 3 from 9.

$$y=\frac{6}{2}$$

Divide 6 by 2.

$$y=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and 3 for x_{2}.

$$y^{2}-9y+18=\left(y-6\right)\left(y-3\right)$$

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.

$$x ^ 2 -9x +18 = 0$$

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

$$r + s = 9 rs = 18$$

Two numbers r and s sum up to 9 exactly when the average of the two numbers is \frac{1}{2}*9 = \frac{9}{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>

$$r = \frac{9}{2} - u s = \frac{9}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 18

$$(\frac{9}{2} - u) (\frac{9}{2} + u) = 18$$

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

$$\frac{81}{4} - u^2 = 18$$

Simplify the expression by subtracting \frac{81}{4} on both sides

$$-u^2 = 18-\frac{81}{4} = -\frac{9}{4}$$

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.

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

Evaluate

$$\left(y-6\right)\left(y-3\right)$$