Solve x^2-15x+54

$${ x }^{ 2 } -15x+54$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+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 x^{2}-15x+54 as \left(x^{2}-9x\right)+\left(-6x+54\right).

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

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

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

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

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

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.

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

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 54}}{2}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 54}}{2}$$

Multiply -4 times 54.

$$x=\frac{-\left(-15\right)±\sqrt{225-216}}{2}$$

Add 225 to -216.

$$x=\frac{-\left(-15\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$x=\frac{-\left(-15\right)±3}{2}$$

The opposite of -15 is 15.

$$x=\frac{15±3}{2}$$

Now solve the equation x=\frac{15±3}{2} when ± is plus. Add 15 to 3.

$$x=\frac{18}{2}$$

Divide 18 by 2.

$$x=9$$

Now solve the equation x=\frac{15±3}{2} when ± is minus. Subtract 3 from 15.

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

Divide 12 by 2.

$$x=6$$

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}.

$$x^{2}-15x+54=\left(x-9\right)\left(x-6\right)$$

Evaluate

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