# High School: Number and Quantity

### The Complex Number System HSN-CN.C.8

8. Extend polynomial identities to the complex numbers. For example, rewrite x2 + 4 as (x + 2i)(x – 2i).

Students should know that complex numbers are everywhere. They're in parabolas and quadratics, but they can be found in polynomial identities as well, even in places where you'd never expect them to be. When we say everywhere, we mean it.

For instance, students should learn that it's possible to factor x2 + 4. They'll freak when you tell them. Give them a glass of water and a fluffy pillow and they'll be fine.

All they have to do is rewrite x2 + 4 as x2 – (-4). Now, as long as they're allowed to use imaginary numbers (and why wouldn't they be?) factoring the difference of perfect squares shouldn't be an issue. We should end up with (x + 2i)(x – 2i).

Students should also know that you aren't doing this to torture them. (Well, maybe a little.) They should know that we use factoring to help solve quadratic equations. That means we can set these equations to zero and solve rather than use the lengthy quadratic formula, which they're probably sick of already.

So it stands to reason that the solutions to the equation x+ 4 = 0 are 2i and -2i.

A nice, pretty little quadratic equation with real coefficients and then all of a sudden, out of nowhere—imaginary roots. Gotta love math; it keeps you guessing. Let's just hope your students feel the same.

#### Drills

1. Factor x2 + 9.

(x – 3i)(x + 3i)

When we rewrite as the difference of squares, our problem becomes x2 – (-9). Since the square root of -9 is 3i, choice (B) is correct. It's not (C) because that equals (x – 3i)2, and not the multiplication of conjugates. That's really all it takes.

2. Factor x2 + 121.

(x + 11i)(x – 11i)

When we rewrite the problem as the difference of perfect squares, it becomes x2 – (-121). Since , and the only correct pair of multiplied conjugates is in (A). While (C) also has a pair of conjugates, it incorrectly leaves the 11 part under the radical.

3. Solve x2 + 49 = 0 for x by factoring.

±7i

When we rewrite the equation, it becomes x2 – (-49) = 0. That factors as (x + 7i)(x – 7i). If we set each of those to equal zero, we have x = 7i and x = -7i. That means our answer is (B).

4. Solve x2 + 100 = 0 for x by factoring.

±10i

When we rewrite the equation, it becomes x2 – (-100) = 0. The factors are (x + 10i)(x – 10i). Since both of those factors equal zero, our answer becomes x = 10i and x = -10i.

5. Solve x2 + 6 = 0 for x by factoring.

When we rewrite the equation, it becomes x2 – (­-6) = 0. Since 6 isn't a perfect square, the square roots of -6 are . That means our answer is (C).

6. Solve x2 + 8 = 0 for x by factoring.

Rearranging this equation into the difference of two squares, we end up with x2 – (-8) = 0. This gives us  as the two factors. If we solve for x, we end up with . That's (B).

7. Which of the following quadratics has the solutions of ±i?

x2 + 1 = 0

We can disregard (A) and (C) right off the bat since solving those will give us real answers and we want imaginary ones. If we look at (B), we'll get (xi)(x + i) as our factors. Looks promising, right? Well… it is.

8. Which of the following quadratics has the solutions of ?

x2 + 12 = 0

Since we're looking for imaginary numbers, (B) won't help us here. We want the x values to be , or . If we plug  in for x, the only equation that works for is (D).

9. Which of the following quadratics has the solutions of ±5i?

x2 + 25 = 0

If we say that x = ±5i, then x2 = 25i2 = -25. That means that one of those equations will be true when we plug in -25 for x2. In our case, that equation happens to be (D).

10. Which of the following quadratics has the solutions ?