Solve 110-21a+a^2

$$110 - 21 a + a ^ { 2 }$$

Factor

$$\left(a-11\right)\left(a-10\right)$$

Solution Steps

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

$$a^{2}-21a+110$$

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+110. To find p and q, set up a system to be solved.

$$p+q=-21 pq=1\times 110=110$$

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 110.

$$-1,-110 -2,-55 -5,-22 -10,-11$$

Calculate the sum for each pair.

$$-1-110=-111 -2-55=-57 -5-22=-27 -10-11=-21$$

The solution is the pair that gives sum -21.

$$p=-11 q=-10$$

Rewrite a^{2}-21a+110 as \left(a^{2}-11a\right)+\left(-10a+110\right).

$$\left(a^{2}-11a\right)+\left(-10a+110\right)$$

Factor out a in the first and -10 in the second group.

$$a\left(a-11\right)-10\left(a-11\right)$$

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

$$\left(a-11\right)\left(a-10\right)$$

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.

$$a^{2}-21a+110=0$$

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.

$$a=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 110}}{2}$$

Square -21.

$$a=\frac{-\left(-21\right)±\sqrt{441-4\times 110}}{2}$$

Multiply -4 times 110.

$$a=\frac{-\left(-21\right)±\sqrt{441-440}}{2}$$

Add 441 to -440.

$$a=\frac{-\left(-21\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$a=\frac{-\left(-21\right)±1}{2}$$

The opposite of -21 is 21.

$$a=\frac{21±1}{2}$$

Now solve the equation a=\frac{21±1}{2} when ± is plus. Add 21 to 1.

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

Divide 22 by 2.

$$a=11$$

Now solve the equation a=\frac{21±1}{2} when ± is minus. Subtract 1 from 21.

$$a=\frac{20}{2}$$

Divide 20 by 2.

$$a=10$$

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 10 for x_{2}.

$$a^{2}-21a+110=\left(a-11\right)\left(a-10\right)$$

Evaluate

$$\left(a-11\right)\left(a-10\right)$$