Question

the table shows some quotients of the polynomial ( x^n - 1 ) divided by the linear factor ( x - 1 ). complete parts a through c below.

a. use long division or synthetic division to find the missing quotients to complete the table.
(simplify your answer.)

dividend	divisor	quotient
( x^3 - 1 )	( x - 1 )	( x^2 + x + 1 )
( x^4 - 1 )	( x - 1 )	\\( \square \\)

(another table below with more rows for ( x^4 -1, x^5 -1, x^6 -1 ) as dividends, ( x -1 ) as divisor, and missing quotients)

(video, textbook, get more help, clear all, check answer buttons on the side)

Explanation:

Step1: Recall polynomial division

We need to divide \(x^4 - 1\) by \(x - 1\). We know that \(x^4 - 1\) can be factored as \((x^2 + 1)(x^2 - 1)\) (difference of squares), and \(x^2 - 1\) can be further factored as \((x + 1)(x - 1)\). So \(x^4 - 1=(x^2 + 1)(x + 1)(x - 1)\). When we divide by \(x - 1\), we can cancel out the \(x - 1\) factor.

Step2: Perform the division

Dividing \(x^4 - 1\) by \(x - 1\), we can also use polynomial long division or synthetic division. Let's use the factored form. After canceling \(x - 1\) from the numerator and denominator, we have \((x^2 + 1)(x + 1)\). Expanding \((x^2 + 1)(x + 1)\) gives \(x^3 + x^2 + x + 1\)? Wait, no, wait. Wait, \(x^4 - 1=(x - 1)(x^3 + x^2 + x + 1)\)? Wait, let's do polynomial long division. Divide \(x^4+0x^3 + 0x^2+0x - 1\) by \(x - 1\).

- First term: \(x^4\div x = x^3\). Multiply \(x - 1\) by \(x^3\) to get \(x^4 - x^3\). Subtract from \(x^4+0x^3\) gives \(x^3\).
- Next term: \(x^3\div x = x^2\). Multiply \(x - 1\) by \(x^2\) to get \(x^3 - x^2\). Subtract from \(x^3+0x^2\) gives \(x^2\).
- Next term: \(x^2\div x = x\). Multiply \(x - 1\) by \(x\) to get \(x^2 - x\). Subtract from \(x^2+0x\) gives \(x\).
- Next term: \(x\div x = 1\). Multiply \(x - 1\) by \(1\) to get \(x - 1\). Subtract from \(x - 1\) gives \(0\). So the quotient is \(x^3 + x^2 + x + 1\)? Wait, no, wait, earlier factoring was wrong. Wait, \(x^4 - 1=(x - 1)(x^3 + x^2 + x + 1)\) is correct. Wait, but let's check with the previous pattern. The first two quotients: for \(x^2 - 1\div(x - 1)=x + 1\), for \(x^3 - 1\div(x - 1)=x^2 + x + 1\), so for \(x^n - 1\div(x - 1)\), the quotient is \(x^{n - 1}+x^{n - 2}+\dots+x + 1\). So for \(x^4 - 1\div(x - 1)\), the quotient should be \(x^3 + x^2 + x + 1\)? Wait, no, wait the table has for \(x^2 - 1\) quotient \(x + 1\), for \(x^3 - 1\) quotient \(x^2 + x + 1\), so for \(x^4 - 1\), the quotient should be \(x^3 + x^2 + x + 1\)? Wait, but let's do the division again. Wait, \(x^4 - 1=(x - 1)(x^3 + x^2 + x + 1)\), yes. So when we divide \(x^4 - 1\) by \(x - 1\), the quotient is \(x^3 + x^2 + x + 1\)? Wait, no, wait the user's table: the first row of dividend is \(x^2 - 1\), divisor \(x - 1\), quotient \(x + 1\); second row dividend \(x^3 - 1\), divisor \(x - 1\), quotient \(x^2 + x + 1\); third row dividend \(x^4 - 1\), divisor \(x - 1\), quotient? Let's see the pattern. For \(x^n - 1\div(x - 1)\), quotient is \(x^{n - 1}+x^{n - 2}+\dots+x + 1\). So for \(n = 4\), quotient is \(x^3 + x^2 + x + 1\)? Wait, but let's check with polynomial long division:

Divide \(x^4 - 1\) by \(x - 1\):

1. \(x^4\div x = x^3\). Multiply \(x - 1\) by \(x^3\): \(x^4 - x^3\). Subtract from \(x^4 - 1\): \((x^4 - 1)-(x^4 - x^3)=x^3 - 1\).
2. \(x^3\div x = x^2\). Multiply \(x - 1\) by \(x^2\): \(x^3 - x^2\). Subtract from \(x^3 - 1\): \((x^3 - 1)-(x^3 - x^2)=x^2 - 1\).
3. \(x^2\div x = x\). Multiply \(x - 1\) by \(x\): \(x^2 - x\). Subtract from \(x^2 - 1\): \((x^2 - 1)-(x^2 - x)=x - 1\).
4. \(x\div x = 1\). Multiply \(x - 1\) by \(1\): \(x - 1\). Subtract from \(x - 1\): \(0\). So the quotient is \(x^3 + x^2 + x + 1\). Wait, but let's check the degrees. The dividend \(x^4 - 1\) is degree 4, divisor \(x - 1\) is degree 1, so quotient should be degree 3. \(x^3 + x^2 + x + 1\) is degree 3, correct.

Wait, but maybe I made a mistake. Wait, the user's table: the first dividend is \(x^2 - 1\) (degree 2), divisor degree 1, quotient degree 1 (\(x + 1\)); second dividend \(x^3 - 1\) (degree 3), divisor degree 1, quotient degree 2 (\(x^2 + x + 1\)); so third dividend \(x^4 - 1\) (degree 4), divisor degree 1, quotient degree 3, which is \(x^3 + x^2 + x + 1\). Yes, that makes sense. So the quotient for \(x^4 - 1\) divided by \(x - 1\) is \(x^3 + x^2 + x + 1\).

Answer:

\(x^3 + x^2 + x + 1\)