Solve 24-11x+x^2

$$24 - 11 x + x ^ { 2 }$$

Factor

$$\left(x-8\right)\left(x-3\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-11x+24$$

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 24=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-8 b=-3$$

Rewrite x^{2}-11x+24 as \left(x^{2}-8x\right)+\left(-3x+24\right).

$$\left(x^{2}-8x\right)+\left(-3x+24\right)$$

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

$$x\left(x-8\right)-3\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x-3\right)$$

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.

$$x^{2}-11x+24=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 24}}{2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 24}}{2}$$

Multiply -4 times 24.

$$x=\frac{-\left(-11\right)±\sqrt{121-96}}{2}$$

Add 121 to -96.

$$x=\frac{-\left(-11\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-\left(-11\right)±5}{2}$$

The opposite of -11 is 11.

$$x=\frac{11±5}{2}$$

Now solve the equation x=\frac{11±5}{2} when ± is plus. Add 11 to 5.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{11±5}{2} when ± is minus. Subtract 5 from 11.

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

Divide 6 by 2.

$$x=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and 3 for x_{2}.

$$x^{2}-11x+24=\left(x-8\right)\left(x-3\right)$$

Evaluate

$$\left(x-8\right)\left(x-3\right)$$