Solve a^2-23a+112

$${ a }^{ 2 } -23a+112$$

Factor

$$\left(a-16\right)\left(a-7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+112. To find p and q, set up a system to be solved.

$$p+q=-23 pq=1\times 112=112$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 112.

$$-1,-112 -2,-56 -4,-28 -7,-16 -8,-14$$

Calculate the sum for each pair.

$$-1-112=-113 -2-56=-58 -4-28=-32 -7-16=-23 -8-14=-22$$

The solution is the pair that gives sum -23.

$$p=-16 q=-7$$

Rewrite a^{2}-23a+112 as \left(a^{2}-16a\right)+\left(-7a+112\right).

$$\left(a^{2}-16a\right)+\left(-7a+112\right)$$

Factor out a in the first and -7 in the second group.

$$a\left(a-16\right)-7\left(a-16\right)$$

Factor out common term a-16 by using distributive property.

$$\left(a-16\right)\left(a-7\right)$$

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.

$$a^{2}-23a+112=0$$

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.

$$a=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 112}}{2}$$

Square -23.

$$a=\frac{-\left(-23\right)±\sqrt{529-4\times 112}}{2}$$

Multiply -4 times 112.

$$a=\frac{-\left(-23\right)±\sqrt{529-448}}{2}$$

Add 529 to -448.

$$a=\frac{-\left(-23\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$a=\frac{-\left(-23\right)±9}{2}$$

The opposite of -23 is 23.

$$a=\frac{23±9}{2}$$

Now solve the equation a=\frac{23±9}{2} when ± is plus. Add 23 to 9.

$$a=\frac{32}{2}$$

Divide 32 by 2.

$$a=16$$

Now solve the equation a=\frac{23±9}{2} when ± is minus. Subtract 9 from 23.

$$a=\frac{14}{2}$$

Divide 14 by 2.

$$a=7$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 16 for x_{1} and 7 for x_{2}.

$$a^{2}-23a+112=\left(a-16\right)\left(a-7\right)$$

Evaluate

$$\left(a-16\right)\left(a-7\right)$$