Lesson Factoring the binomials x^n+a^n for odd degrees

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Factoring the binomials x%5En%2Ba%5En for odd degrees


If you are familiar with the shortcut multiplication formula for the sum of cubes, then you just know this factorization formula

a%5E3%2Bb%5E3 = %28a%2Bb%29%2A%28a%5E2+-+ab+%2B+b%5E2%29                                                                      (1)

(see the lesson The sum of cubes formula under the current topic in this site).
There is general factorization formula for the sum of any odd degrees

a%5En+%2Bb%5En  =  .     (2)

This formula is valid for all real numbers a and b and for any odd integer index n greater than or equal to 3.
The above formula (1) is the particular case of it. In this lesson we prove the general formula (2).
To prove it let us perform direct calculations:

=     (Now apply the distributive property of addition and multiplication operations.)

               (One of two pairs of brackets is opened and the result is presented in two lines.)
+ =          (Now open the resting brackets and apply the commutative property of multiplication operation)

                       (Note that the co-indexed terms in the upper and the lover lines have the opposite signs.)
    + =               (Therefore, you can cancel these like terms.)

              (The crossing lines show canceling the like terms.)

    + = a%5En+%2B+b%5En.         (You got the final result.)

The formula is proved.
The distributive and commutative properties of addition and multiplication operations were used in the proof.

Formula (2) is valid not only to numbers. It is valid for the polynomials with real coefficients too. For example,

x%5En+%2Ba%5En  =  .     (3)

for any real number a and for any odd integer index n greater than or equal to 3. You can check validity of the formula (5) directly by performing all relevant
calculations: opening the brackets, multiplying the terms and canceling the like terms. You will get the same result, because addition and multiplication operations
for the polynomials with real coefficients have the same distributive and commutative properties as for real numbers.

Formula (3) provides factorization the binomial x%5En+%2Ba%5En into the product of the linear binomial x+%2Ba and the polynomial
for odd indexes n.
This explicit factorization shows that the binomial x%5En+%2Ba%5En is divided by the linear binomial x+%2Ba for odd indexes n.

Note the important special case of the formula (5) when the value of a equal to 1:

x%5En+%2B1  =  %28x%2B1%29%2A%28x%5E%28n-1%29+-+x%5E%28n-2%29+%2B+x%5E%28n-3%29+%2B+ellipsis+-+x+%2B+1%29.                                     (4)

Again, formula (4) shows that the binomial x%5En+%2B1 is divided by the linear binomial x+-1 for odd indexes n.
The quotient is the polynomial %28x-1%29%2A%28x%5E%28n-1%29+-+x%5E%28n-2%29+%2B+x%5E%28n-3%29+%2B+ellipsis+-+x+%2B+1%29.
Let us write it explicitly:

%28x%5En+%2B1%29%2F%28x%2B1%29  =  x%5E%28n-1%29+-+x%5E%28n-2%29+%2B+x%5E%28n-3%29+-+ellipsis+-+x+%2B+1.                                               (5)


Now, let us ask the question, whether the binomial   x%5En+%2B+a%5En   is divisible by the binomial   x+%2B+a   at even integer index   n?
The answer is "No".   The binomial  x%5En+%2B+a%5En  is NOT divisible by the binomial  x+%2B+a  at even integer index  n,  if the real number   a   is different from zero.
To prove it, let us assume for a moment that the binomial   x%5En+%2B+a%5En   is divisible by the binomial   x+%2B+a   at even integer index   n. That is
x%5En+%2B+a%5En = %28x+%2B+a%29%2Ag%28x%29,

where   g%28x%29   is some other polynomial with real coefficients.
Then the polynomial   %28x+%2B+a%29%2Ag%28x%29   on the the right side of this equation is equal to zero at the value of   x+=+-a.   But, at the even integer index   n, the left side of this
equation is equal to   2%2Aa%5En   at this value of   x,   which is different from zero,   if   a   is not equal to   0.   This contradiction proves that the binomial   x%5En+%2B+a%5En   is not
divisible by the binomial   x+%2B+a   at even integer index   n.

Summary



1. The formula is valid

      a%5En+%2Bb%5En  =  

      for all real numbers a and b and for odd integer index n greater than or equal to 3.


2. For odd integer index n greater than or equal to 3 and for any real number a the binomial   x%5En+%2Ba%5En   is divided by the linear binomial   x+%2Ba.
      The formula is valid

      x%5En+%2Ba%5En  =  .

      This formula is factoring the binomial   x%5En+%2Ba%5En   into the product of the linear binomial   x+%2Ba and the polynomial   .
      The quotient of division the binomial   x%5En+%2Ba%5En   by the binomial   x+%2Ba   is the polynomial   .


3. For odd integer index n greater than or equal to 3 the binomial   x%5En%2B1   is divided by the linear binomial   x%2B1.
      The formula is valid

      x%5En%2B1  =  %28x%2B1%29%2A%28x%5E%28n-1%29+-+x%5E%28n-2%29+%2B+x%5E%28n-3%29+%2B+ellipsis+-+x+%2B+1%29.

      This formula is factoring the binomial   x%5En%2B1   into the product of the linear binomial   x%2B1 and the polynomial   x%5E%28n-1%29+-+x%5E%28n-2%29+%2B+x%5E%28n-3%29+-+ellipsis+-+x+%2B+1%29.
      The quotient of division the binomial   x%5En%2B1   by the binomial   x%2B1   is the polynomial   x%5E%28n-1%29+-+x%5E%28n-2%29+%2B+x%5E%28n-3%29+-+ellipsis+-+x+%2B+1%29:
      %28x%5En+%2B1%29%2F%28x%2B1%29  =  x%5E%28n-1%29+-+x%5E%28n-2%29+%2B+x%5E%28n-3%29+-+ellipsis+-+x+%2B+1.


4. The binomial   x%5En%2Ba%5En   is not divisible by the binomial   x%2Ba   for even integer index   n.


To complete this lesson, we present examples that show how everything works.

Example 1

Factor the binomial   x%5E5+%2B+32.

Solution
Note that   x%5E5+%2B+32 = x%5E5+%2B+2%5E5.
In accordance with the formula (5),   x%5E5+%2B+2%5E5%29 = %28x%2B2%29%2A%28x%5E4+-+2%2Ax%5E3+%2B+4%2Ax%5E2+-+8%2Ax+%2B16%29%29.

So, the answer is   x%5E5+%2B+32 = %28x%2B2%29%2A%28x%5E4+-+2%2Ax%5E3+%2B+4%2Ax%5E2+-+8%2Ax+%2B16%29%29.


Example 2

Simplify the rational expression 1%2F%281-x%2Bx%5E2-x%5E3%2Bx%5E4-x%5E5%2Bx%5E6%29.

Solution
Multiply the numerator and the denominator of this rational fraction by x%2B1 and apply the formula (4). You get

1%2F%281-x%2Bx%5E2-x%5E3%2Bx%5E4-x%5E5%2Bx%5E6%29 = %28x%2B1%29%2F%28%28x%2B1%29%2A%281-x%2Bx%5E2-x%5E3%2Bx%5E4-x%5E5%2Bx%5E6%29%29 = %28x%2B1%29%2F%28x%5E7%2B1%29.


Example 3

Rationalize the fraction by making its denominator free of roots.

Solution
Multiply the numerator and the denominator of the fraction by %28root%287%2C3%29%29%2B1 and apply the formula (4). You get

= = %28root%287%2C3%29%2B1%29%2F%28%28root%287%2C3%29%29%5E7%2B1%29 = %28root%287%2C3%29%2B1%29%2F%283%2B1%29 = %28root%287%2C3%29%2B1%29%2F4.


For similar lessons see Factoring the binomials   x%5En-a%5En   under the current topic in this site.

For divisibility and factoring of a general polynomial  f%28x%29  by a binomial  x%2Ba  see the lesson  Divisibility of polynomial  f%28x%29  by binomial  x-a  under the current topic in this site.

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