Solve x^2-15x+44

$${ x }^{ 2 } -15x+44$$

Factor

$$\left(x-11\right)\left(x-4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+44. To find a and b, set up a system to be solved.

$$a+b=-15 ab=1\times 44=44$$

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 44.

$$-1,-44 -2,-22 -4,-11$$

Calculate the sum for each pair.

$$-1-44=-45 -2-22=-24 -4-11=-15$$

The solution is the pair that gives sum -15.

$$a=-11 b=-4$$

Rewrite x^{2}-15x+44 as \left(x^{2}-11x\right)+\left(-4x+44\right).

$$\left(x^{2}-11x\right)+\left(-4x+44\right)$$

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

$$x\left(x-11\right)-4\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(x-4\right)$$

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.

$$x^{2}-15x+44=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 44}}{2}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 44}}{2}$$

Multiply -4 times 44.

$$x=\frac{-\left(-15\right)±\sqrt{225-176}}{2}$$

Add 225 to -176.

$$x=\frac{-\left(-15\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-\left(-15\right)±7}{2}$$

The opposite of -15 is 15.

$$x=\frac{15±7}{2}$$

Now solve the equation x=\frac{15±7}{2} when ± is plus. Add 15 to 7.

$$x=\frac{22}{2}$$

Divide 22 by 2.

$$x=11$$

Now solve the equation x=\frac{15±7}{2} when ± is minus. Subtract 7 from 15.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 11 for x_{1} and 4 for x_{2}.

$$x^{2}-15x+44=\left(x-11\right)\left(x-4\right)$$

Evaluate

$$\left(x-11\right)\left(x-4\right)$$