All right, we've trekked a little further up Polynomial Mountain and have come to another impasse. Now we've gotta find factors and roots of polynomials.
Hey, our polynomial buddies have caught up to us, and they seem to have calmed down a bit. They have a polynomial for us.
2x3 + x2 – 2x – 1
We start with our new discovery, the Remainder Theorem. It'll tell us if something is a factor of this polynomial. It can't tell us if we picked the right lottery numbers, though.
For instance, is x – 1 a factor? Is x – 1 one of the lotto numbers?
To answer the first question, we can plug x = 1 into the polynomial. If we get a 0, then x – 1 is a factor, because it divides the polynomial with no remainder. If we don't get 0, then it isn't.
P(1) = 2(1)3 + (1)2 – 2(1) – 1
= 2 + 1 – 2 – 1
= 0
Yep, this means that x – 1 is a factor of this polynomial. It tells us nothing about our chances for the Powerball. Good thing we didn't pin all our hopes on it.
Now what? Factors are multiplied together, so there must be at least one more factor left.
Our space-alien friends say we can use synthetic division to find the other factors.
We've gone ahead and laid it all out in one step. We can see that 2x3 + x2 – 2x – 1 divided by x – 1 is:
2x2 + 3x + 1
That's our other factor, but we can factor that a bit more to get:
2x2 + 3x + 1 =
(2x + 1)(x + 1)
This means our polynomial is actually equal to:
P(x) = (2x + 1)(x + 1)(x – 1)
Did you forget our first factor, x – 1? It's still there. We can now set each factor equal to zero to find the roots. That gives us 
, -1, and 1.
Excellent. That's everything we wanted to know, and all before teatime. How about we keep working our way up Polynomial Mountain with a few more problems?
Sample Problem
See if x + 1 is a factor of x4 + x3 + 3x2 – x + 1.
We plug and chug away, by shoving x = -1 into the polynomial.
P(-1) = (-1)4 + (-1)3 + 3(-1)2 – (-1) + 1
= 1 – 1 + 3 + 1 + 1
= 5
End of the line, hoss. A 5 is definitely not a 0, so x + 1 is not a factor for this polynomial. It doesn't divide through evenly. Mosey on to the next problem.
Sample Problem
Write a polynomial with the roots -3, 3, and 2.
Huh. Looks like we need to go backwards this time. Listen closely, and you might hear that Paul is dead.
If the roots are -3, 3, and 2, that means the factors are:
(x + 3), (x – 3), (x – 2)
Plug each root in for x, and you'll get zero, zero, zero every time. So our polynomial must be:
P(x) = (x + 3)(x – 3)(x – 2)
Which multiplies out to:
P(x) =(x2 – 9)(x – 2)
P(x) = x3 – 2x2 – 9x + 18
Wait, we just heard that Paul is, in fact, not dead. False alarm.
Sample Problem
Find all roots of x3 + 3x2 – x – 3.
Now this is a little tougher situation. What do we do if no factors are given to us?
Well, magic has a long, if unreliable, history of use in cases like this. However, most people aren't 7th level clerics capable of casting Divination, so maybe we should try something else.
All of our factors take the form of x – a. The a needs to multiply into the constant at the end of the polynomial, in this case -3. That means our factor (for the polynomial) needs a factor (from that constant).
The factors of the constant -3 are ±3 and ±1. We'll pick one and run it through the Remainder Theorem. A 3 sounds good. If P(3) = 0, then 3 is a root, which means x – 3 is a factor.
P(3) = (3)3 + 3(3)2 – (3) – 3
= 27 + 27 – 6
= 48
3 is not a root. Let's guess something else, like -3.
P(-3) = (-3)3 + 3(-3)2 – (-3) – 3
= -27 + 27 + 3 – 3
= 0
Okay. Step one: done. That means x + 3 is a factor. Now we can use synthetic division to find the other factors, and then all of the roots.
That gives us x2 – 1 as our other factor. We can factor that factor to get (x + 1)(x – 1).
This means our polynomial is:
P(x) = (x + 3)(x + 1)(x – 1)
That, dear Shmooper, gives us roots of -3, -1, and 1.
We did it. We're so smart (not to brag, or anything like that).
Sample Problem
Find all roots of x3 + 2x2 – 25x – 50, given that one root is 5.
If one root is 5, that means x – 5 is a factor. We can use synthetic division to find the rest.
Our other factor is x2 + 7x + 10. Further factoring furnishes fitting fanswers—er, answers.
(x + 2)(x + 5)
That means our full polynomial is:
P(x) = (x – 5)(x + 2)(x + 5)
The roots for this polynomial are x = 5, -2, and -5.
Finally. Good job, mathronaut. Onward and upward.
Example 1
What are all the roots of P(x) = (x – 1)(x + 3)(x + 2)? |
Example 2
What are the roots of P(x) = (x + 3)(x – 4)(x – 2)? |
Example 3
Write a polynomial with roots of -1, 1, and 2. |
Example 4
Write a polynomial with roots of -2, 2, and 5. |
Example 5
Find all roots of x3 – 4x2 – x + 4 given that one root is 4. |
Exercise 1
Find all roots of P(x) = (x – 3)(x + 3)(x – 1).
Exercise 2
Write a polynomial with roots of -5, 5, and 1.
Exercise 3
Find all the roots of x3 + 6x2 – x – 6.
Exercise 4
Find all the roots of x3 – x2 – 49x + 49.
Exercise 5
Find all the roots of x3 – 8x2 – 4x + 32.
Exercise 6
Find all the roots of x3 + x2 – 100x – 100.
Exercise 7
Find all the roots of x4 – 13x2 + 36.