Division of Polynomials: UNIZOR.COM - Math4Teens - Algebra - Fundamental...

Notes to a video lecture on http://www.unizor.com

Division of Polynomials

The formula for solutions of any quadratic equation is known.
For an equation
A·X²+B·X+C=0
the solutions are:

X1,2 =

−B±√B²−4AC 2A

There is a formula for a general cubic equation (Cardano formula), but it's very complex, and we will not list it here.

So, what can someone do to solve a cubic equation?
In some cases there is a possibility to guess one of the solutions. Then, using the polynomial division, we can reduce our cubic equation with one guessed solution to a quadratic one to find two other solutions.

Consider a cubic equation
X³ + 4X² −11X −30 = 0
As we know, any polynomial of the third order, like the one above, has three (generally speaking, complex) roots, that is values of an unknown that cause the value of the polynomial to be equal to zero. Assume these roots are
X₁ = a
X₂ = b
X₃ = c
Then, according to a corollary to a Fundamental Theorem of Algebra, our polynomial can be represented as
X³ + 4X² −11X −30 =
= (X−a)·(X−b)·(X−c)

Then, knowing one root (say, X₁ = a) we can construct the result of multiplication (X−b)·(X−c) by dividing our original polynomial by (X−a).
This process is similar to long division of numbers.

Notice that, if we multiply (X−a)·(X−b)·(X−c), the only element without an unknown will be −a·b·c.
Therefore,
−30 = −a·b·c.

If the free member of our equation is integer (and it is, it's equal to −30), it's reasonable to attempt, firstly, to find an integer root of our equation among its divisors.

Number 30 has only three divisors: 2, 3 and 5. Let's check if one of them (say, 5) is the root of a given polynomial.
5³ + 4·5² −11·5 −30 =
= 125 + 100 −55 − 11 = 40
So, X=5 is not a root.

How about X=−5?
(−5)³ + 4·(−5)² −11·(−5) −30 =
= −125 + 100 + 55 − 30 = 0
Lucky guess! We found a root a=−5 and now we can divide the original polynomial of the third order by
X−a = X−(−5) = X+5
obtaining the polynomial of the second order that should have two other roots b and c.

Here is this division, step by step.
Step 1:
Dividend: X³+4X²−11X−30
Divisor: X+5
Quotient: X²
Multiply: (X+5)·X²=X³+5X²
Remainder:
(X³+4X²−11X−30)−(X³+5X²)=
= −X²−11X−30

Step 2:
Dividend: −X²−11X−30
Divisor: X+5
Quotient: −X
Multiply: (X+5)·(−X)=−X²−5X
Remainder:
(−X²−11X−30) − (−X²−5X) =
= −6X−30

Step 3:
Dividend: −6X−30
Divisor: X+5
Quotient: −6
Multiply: (X+5)·(−6)=−6X−30
Remainder:
(−6X−30) − (−6X−30) = 0

Combine all quotients and get:
Dividend: X³+4X²−11X−30
Divisor: X+5
Quotient: X²−X−6
Multiply: (X+5)·(X²−X−6) =
= X³+4X²−11X−30
Remainder: zero

Now we can either attempt to find a root of the equation
X²−X−6 = 0
and perform a polynomial division or, considering this is an equation of the second order, just use the formula for its roots.

The guess and division approach is more educational, so we will choose it.
The free member of the polynomial we deal with now is −6, so we will try its divisors 2 or 3.
Start with a root b=2.
2²−2−6 = −4
So, 2 is not a root.
How about b=−2?
(−2)²−(−2)−6 = 4+2−6 = 0
Lucky guess! We found a second root b=−2 and now we can divide the original polynomial X²−X−6 of the second order by
X−b = X − (−2) = X + 2

Let's do a division to reduce our second order polynomial X²−X−6 as a product of (X+2) and a first order polynomial.
Step 1:
Dividend: X²−X−6
Divisor: X+2
Quotient: X
Multiply: (X+2)·X = X²+2X
Remainder:
(X²−X−6)−(X²+2X)=
= −3X−6

Step 2:
Dividend: −3X−6
Divisor: X+2
Quotient: −3
Multiply: (X+2)·(−3)=−3X−6
Remainder: (−3X−6)−(−3X−6) = 0

Combine all quotients and get:
Dividend: X²−X−6
Divisor: X+2
Quotient: X−3
Multiply: (X+2)·(X−3) =
= X²−X−6
Remainder: zero

By guessing and using the polynomial divisions we have two roots (X=−5 and X=−2) and representation of the original equation as
X³+4X²−11X−30 =
= (X+5)·(X+2)·(X−3)
The last multiplier reveals the third root of the original cubic equation X=3

CONCLUSION
Knowing one root X=a of a polynomial of the Nth order P^(N)(X), we can represent this polynomial as a product of (X−a) and a polynomial of the (N−1)th order P^(N−1)(X), thus simplifying a task of finding all the roots.