use synthetic division to solve $(x^{4}-1)\\div(x - 1)$. what is the quotient?\n$x^{3}-x^{2}+x…

Explanation:

Step1: Set up synthetic division

For ((x^{4}-1)\div(x - 1)), the divisor is (x-1), so (c = 1). The dividend (x^{4}-1) can be written as (x^{4}+0x^{3}+0x^{2}+0x - 1). The coefficients are (1,0,0,0,-1).

Step2: Bring down the first coefficient

Bring down the first coefficient (1). $$ \begin{array}{r|rrrrr} 1&1&0&0&0&-1\ &&&\ \hline &1&&&& \end{array} $$

Step3: Multiply and add

Multiply (1) (the number just brought down) by (1) (the value of (c)) to get (1). Add to the next coefficient (0 + 1=1). $$ \begin{array}{r|rrrrr} 1&1&0&0&0&-1\ &&1&&&\ \hline &1&1&&& \end{array} $$ Multiply (1) (the new number) by (1) to get (1). Add to the next coefficient (0+1 = 1). $$ \begin{array}{r|rrrrr} 1&1&0&0&0&-1\ &&1&1&&\ \hline &1&1&1&& \end{array} $$ Multiply (1) (the new number) by (1) to get (1). Add to the next coefficient (0 + 1=1). $$ \begin{array}{r|rrrrr} 1&1&0&0&0&-1\ &&1&1&1&\ \hline &1&1&1&1& \end{array} $$ Multiply (1) (the new number) by (1) to get (1). Add to the last coefficient (-1+1 = 0). $$ \begin{array}{r|rrrrr} 1&1&0&0&0&-1\ &&1&1&1&1\ \hline &1&1&1&1&0 \end{array} $$ The coefficients of the quotient polynomial are (1,1,1,1) and the degree of the quotient is (4 - 1=3). So the quotient is (x^{3}+x^{2}+x + 1).

Answer:

(x^{3}+x^{2}+x + 1) (corresponds to the third option)