We are asked to simplify x^(16)-1x-1 using two methods: and . Let's do it!
Long Division
We will perform polynomial long division to simplify the given .
x^(16)-1/x-1
Usually, before we start dividing, we complete the following two steps.
- Rewrite the dividend and the divisor in .
- If a is not present in the dividend, we add the term with a of 0.
In our case both the dividend and the divisor are already written in standard form. The of the dividend is 16 and it has only two terms, so there are a lot of missing terms.
x^(16)-1
⇕
x^(16)+0x^(15)+0x^(14)+...+0x^2+0x-1
Since there are so many missing terms, in this particular case we will not write them when performing long division. However, we have to keep in mind that these terms are in fact there! Other than that, we will follow the usual steps of polynomial long division.
- Divide the first term of the dividend by the first term of the divisor. The result is the first term of the .
- Multiply the first term of the quotient by the divisor and write the under the dividend.
- Subtract the product from the dividend. We will call the result of this operation the remainder.
- Repeat Step 1 through Step 3 treating the remainder as the dividend. Stop when the degree of the remainder is less than the degree of the divisor.
Let's perform the division!
l r x - 1 & |l x^(16)-1
r x^(15) r x-1 & |l x^(16)-1
r x^(15) rl x-1 & |l x^(16)-1 & x^(16)-x^(15)
r x^(15) r x-1 & |l x^(15)-1
r x^(15)+x^(14) r x-1 & |l x^(15)-1
rx^(15)+ x^(14) rl x-1 & |l x^(15)-1 & x^(15)-x^(14)
r x^(15)+x^(14) r x-1 & |l x^(14)-1
r x^(15)+x^(14)+x^(13) r x-1 & |l x^(14)-1
rx^(15)+x^(14)+ x^(13) rl x-1 & |l x^(14)-1 & x^(14)-x^(13)
r x^(15)+x^(14)+x^(13) r x-1 & |l x^(13)-1
After calculating three terms of the quotient, we can notice the repeating pattern.
- All terms of the quotient that we calculated so far have a coefficient equal to 1.
- With each step, the degree of the remainder decreases by 1 but the structure of the remainder, x^a-1, stays the same.
Based on these observations, we can assume that the pattern will continue. Therefore, we will eventually obtain x-1 as the remainder. Additionally, all of the terms of the quotient that are calculated in these steps will have a coefficient of 1.
r x^(15)+x^(14)+x^(13)+...+x^2+x r x-1 & |l x-1
r x^(15)+x^(14)+x^(13)+...+x^2+x+1 r x-1 & |l x-1
rx^(15)+x^(14)+x^(13)+...+x^2+x+ 1 rl x-1 & |l x-1 & x-1
r x^(15)+x^(14)+x^(13)+...+x^2+x+1 r x-1 & |l 0
The simplified form of the given expression that we found using long division is x^(15)+x^(14)+x^(13)+...+x^2+x+1.
Factoring
To simplify the given rational expression by factoring, we have to factor its and . However, in this case the denominator cannot be factored any further, so let's focus on the numerator. We will use the to factor the numerator.
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Difference of Squares Formula
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For any real numbers a and b,
a^2-b^2=(a+b)(a-b).
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Note that we will have to use this formula a few times to fully factor x^(16)-1. Let's do it!
x^(16)-1
x^(8* 2)-1
(x^8)^2-1
(x^8)^2-1^2
(x^8+1)(x^8-1)
(x^8+1)(x^4+1)(x^4-1)
(x^8+1)(x^4+1)((x^2)^2-1^2)
(x^8+1)(x^4+1)(x^2+1)(x^2-1)
(x^8+1)(x^4+1)(x^2+1)(x^2-1^2)
(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)
We have fully factored the numerator. Now we can substitute it into the original expression.
x^(16)-1/x-1 [0.8em]
⇕ [0.8em]
(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)/x-1
Finally, let's cancel out common factors and write our answer in standard form.
(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)/x-1
(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)/x-1
(x^8+1)(x^4+1)(x^2+1)(x+1)
(x^8+1)(x^4+1)(x^2(x+1)+1(x+1))
(x^8+1)(x^4+1)(x^3+x^2+x+1)
(x^8+1)(x^4(x^3+x^2+x+1)+1(x^3+x^2+x+1))
(x^8+1)(x^7+x^6+x^5+x^4+x^3+x^2+x+1)
x^8(x^7+x^6+x^5+x^4+x^3+x^2+x+1)+1(x^7+x^6+x^5+x^4+x^3+x^2+x+1)
x^(15)+x^(14)+x^(13)+x^(12)+x^(11)+x^(10)+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1
In most cases, it is expected to write polynomials in standard form. However, since our result has 15 terms and we are not asked to write it in standard form, we could as well use its factored form as the answer.
Conclusion
If the format of the answer does not matter, it is quicker to use factoring. The factoring method also allows us to calculate the factored result and its standard form. When using long division we receive the result only in standard form. Since the result is a polynomial with degree 15, it will be difficult to factor it. In this case, we would choose the factoring method as our preferred method.
