2.2 The Factor Theorem, Synthetic Division, and the Difference and Sum of Cubes

Synthetic Division

Synthetic Division is a shortcut for long division of polynomials. It saves time, energy, effort, and paper, and because math homework consumes a lot of all of these limited resources, it is in our best interest to use this efficient method of dividing polynomials.

We begin with an equation, the dividend. Ensure the dividend is in order of descending exponents.



Take the divisor, set it equal to zero, and isolate x.
0 = x + 2
-2 = x

Synthetic division will take four lines. On the left side of the second line, write -2 inside a _|. This will be the multiplier. On the first line, indented so that it is not over the -2, write the dividend with plenty of space between the terms.

In the end, we will have this:



How? Using a different diagram for easier understanding, we can see the process:



(A) Bring down the coefficients of each term, and the constant, in the same order as they are already in.

(B) The leftmost number is brought down and written. [It is the same as 5 + 0, as 0 is the previous term's coefficient (0x^4)].

(C) The sum is multiplied by -2 (or whatever number is in the corner) and written below the next number.

(D) The number you just wrote is added to the number above it, and written under the line.


The numbers, in order, are 5, -7, 8, and -14. The rightmost number is the remainder. The one to the left of it is the constant, the one to the left of that is the coefficient of the x-value, and the one to the left of that is the coefficient of the x^2 value. The quotient is 5x^2 - 7x + 6 with a remainder of -14.

Therefore the equation is
5x^3 + 3x^2 - 6x + 2 = (x - 2)(5x^2 - 7x + 6) - 14


This is the Burchat Method of Synthetic Division. Alternatively, we can use the Textbook Method of Synthetic Division (found on pg 98 of your textbook).

When using the TMoSD, we don't have to find the x-value that ensures x + n = 0 is true. Instead, n is taken in its current form and used. Then, in step (D), the numbers in the third line are subtracted from the numbers in the second line.



The Factor Theorem

The factor theorem is a shortcut to finding the factors of a polynomial.

It goes like this:
"If x = a is substituted into a polynomial for x and the remainder is 0, then (x - a) is a factor of the polynomial."

We can use the remainder theorem to determine whether the remainders of the factors are zero.

The equation is f(x) = x^3 + 2x^2 - x - 2.


f(3) = 3^3 + 2(3)^2 - 3 - 2
f(3) = 27 + 18 - 3 - 2
f(3) = 40
.: (x-3) is not a factor

f(2) = 2^3 + 2(2)^2 - 2 - 2
f(2) = 8 + 8 - 2 - 2
f(2) = 12
.: (x-2) is not a factor

f(1) = 1^3 + 2(1)^2 - 1 - 2
f(1) = 1 + 2 - 1 - 2
f(1) = 0
.: (x-1) is a factor

f(0) = 0^3 + 2(0)^2 - 0 - 2
f(0) = 0 + 0 - 0 - 2
f(0) = -2
.: x is not a factor

f(-1) = (-1)^3 + 2(-1)^2 - (-1) - 2
f(-1) = -1 + 2 + 1 - 2
f(-1) = 0
.: (x+1) is a factor

f(-2) = (-2)^3 + 2(-2)^2 - (-2) - 2
f(-2) = -8 + 8 + 2 - 2
.: (x+2) is a factor

f(-3) = (-3)^3 + 2(-3)^2 - (-3) - 2
f(-3) = -27 + 18 + 3 - 2
f(-3) = -8
.: (x+3) is not a factor.

The factors are (x + 1), (x - 1), and (x + 2).

We can check by multiplication:

(x + 1)(x - 1)(x + 2)
= (x^2 - x + x - 1)(x + 2)
= (x^2 - 1)(x + 2)
= x^3 + 2x^2 - x - 2


While it may seem that it could take forever to find the factors, because you could plug in any number as the a-value, you can cut down the number of possibilities. The -a-value is always a factor of the constant in the polynomial. The -a-values all, when multiplied by each other, equal the constant.


Why does this work? Remember that the zeroes are also the roots, or the x-intercepts, of the polynomial graph. Where y=0, the x-value (x=a) is the x-intercept.



Difference and Sum of Cubes

Like a difference of squares, there is an easy way to find the factors of a difference of cubes. In addition, we can do the same with a sum of cubes.

First, the polynomial must be identified as a sum or difference of cubes. The first step is to look between the two terms and determine which sign separates them. The following rules are only applicable if the sign is + or -, and there are separate but similar rules for addition and subtraction. Then the terms must be identified as cubes, which can be done by finding the cube root of each term and seeing if they are integers. This is also useful because the cube roots will be used in the process of finding the factors.

The following equation is used to find the difference of cubes:



The following equation is used to find the sum of cubes:



As you can see, the equations are very similar. It is important not to mix them up.


For the example, we will use the equation 64x^3 - 125.

Following the equation above,
a^3 = 64x^3
b^3 = 125

By finding the cube root of 64x^3 and 125, we can find the a and b values. In this case,
a = 4x
b = 5.

First, to find the difference of these cubes,

Sub a^3 = 64x^3, b^3 = 125, a = 4x, and b = 5 into a^3 - b^3 = (a - b)(a^2 + ab + b^2)

64x^3 - 125 = (4x - 5)[(4x)^2 + (4x)(5) + 5^2]
64x^3 - 125 = (4x - 5)(16x^2 + 20x + 25)

Next, to find the sum of these cubes,

Sub a^3 = 64x^3, b^3 = 125, a = 4x, and b = 5 into a^3 + b^3 = (a + b)(a^2 - ab + b^2)

64x^3 + 125 = (4x + b)[(4x)^2 - (4x)(5) + 5^2]
64x^3 - 125 = (4x + 5)(16x^2 - 20x + 25)

2.1 Division of Polynomials & the Remainder Theorem - Period 3

Hi everyone!

Chapter 1 on polynomial functions is finally over and starting from today we will look at polynomial equations and inequalities. In this lesson, we have learned how to reuse grade 4 mathematics – long division to divide a polynomial by a binomial and ‘remainder theorem’ to determine the remainder of the polynomial without dividing it.


So now let’s recall the long division we learned in grade 4:

Divide 6487 by 42.

Here, ‘42’ at the left will be the ‘divisor’;
‘6487’ at the centre will be the ‘dividend’;
‘154’ at the top will be the ‘quotient’ which the answer is needed to be in perfect columns; and
‘19’ at the bottom and has a letter ‘R’ representing it will the remainder of the expression.

In that case, 6487=154(42)+19

→That is, dividend=(divisor)(quotient)+remainder

Let’s try some real ones:
1. Divide x³-6x²+4x+1 by x-3
[Important point to note: the terms of the equation must always be in descending order during calculation] (The subtract signs along the side of the long division are just to indicate that subtraction is going on between the expressions.)

Therefore, we can say:


Method 1(provided by our teacher): x³-6x²+4x+1 = (x-3)(x²-3x-5)-14
Or
Method 2 (provided our textbook): x³-6x²+4x+1 = (x²-3x-5)(-14/(x-3))



How about those with some terms missing?

2. Divide x⁴-2x²+1 by x+2 (we can notice that the ’x³’ term and the ‘x’ term are missing!)
[In fact, we can insert terms with the value of ‘0’ into the equation. That is:
x⁴-2x²+1 → x⁴+0x³-2x²+0x+1]
This time, since there is no remainder both methods 1 and 2 can get the same form of answer.
Therefore, x⁴-2x²+1 divided by x+2 = (x+2)(x³-2x²+2x-4)
→ This is how long divisions of polynomials works.


Now, let’s try out the remainder theorem.


Before we start, let’s do an experiment.
a) Divide x³-4x²+3x-13 by x+4
b) Find f(-4)

b. f(x) = x³-4x²+3x-13
f(-4) = (-4)³-4(-4)²+3(-4)-13
=-153

→ Because both (a) and (b) get the same answer, we can deduce that:
When a polynomial function P(x) is divided by (x-b), the remainder is P(b), and when it is divided by ax-b, the remainder is P(b/a), where a and b are integers and a≠0.

Let's try one more example: Find the remainder of x³+2x²-13x+1 when it is divided by x-1
→ x-1=0
x=1
(1)³+2(1)²-13(1)+1
=1+2-13+1
=-9

Therefore, when x³+2x²-13x+1 is divided by x-1, the remainder will be -9.

This is the end of the lesson, hope you will find it helpful! :)
If there is anything with my entry, please feel free to tell me.