Micheal L.

asked • 02/12/21

Solve (x^4 - x^3 + 4x + 2) / (x^2 + 3) Using synthetic division

Please help, I have been stuck on this question for almost five hours. ASAP PLEASE

2 Answers By Expert Tutors

By:

Robert C. answered • 02/12/21

Tutor
New to Wyzant

Part-time university math professor

See tutors like this
See tutors like this

With the limited formatting available with this editor it’s better to just copy and paste from wikipedia(below). Ask me if you need any points explained, for adapting it to your problem.


Robert Clark


==============================================================

Expanded synthetic division[edit]


This method generalizes to division by any monic polynomial with only a slight modification with changes in bold. Using the same steps as before, perform the following division:


\frac{x^3 - 12x^2 - 42}{x^2 + x - 3}

We concern ourselves only with the coefficients. Write the coefficients of the polynomial to be divided at the top.


{\displaystyle {\begin{array}{|rrrr}\ 1&-12&0&-42\end{array}}}


Negate the coefficients of the divisor. 

\begin{array}{rrr} -1x^2 &-1x &+3 \end{array}

Write in every coefficient but the first one on the left in an upward right diagonal (see next diagram).

{\displaystyle {\begin{array}{cc}{\begin{array}{rr}\\&3\\-1&\\\end{array}}&{\begin{array}{|rrrr}\ 1&-12&0&-42\\&&&\\&&&\\\hline \end{array}}\end{array}}}


Note the change of sign from  1 to −1 and from −3 to 3 . "Drop" the first coefficient after the bar to the last row.

\begin{array}{cc} \begin{array}{rr} \\ &3 \\ -1& \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & & & \\ & & & \\ \hline 1 & & & \\ \end{array} \end{array}


Multiply the dropped number by the diagonal before the bar, and place the resulting entries diagonally to the right from the dropped entry.

\begin{array}{cc} \begin{array}{rr} \\ &3 \\ -1& \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & & 3 & \\ & -1 & & \\ \hline 1 & & & \\ \end{array} \end{array}


Perform an addition in the next column.


\begin{array}{cc} \begin{array}{rr} \\ &3 \\ -1& \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & & 3 & \\ & -1 & & \\ \hline 1 & -13 & & \\ \end{array} \end{array}


Repeat the previous two steps until you would go past the entries at the top with the next diagonal.

\begin{array}{cc} \begin{array}{rr} \\ &3 \\ -1& \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & & 3 & -39 \\ & -1 & 13 & \\ \hline 1 & -13 & 16 & \\ \end{array} \end{array}


Then simply add up any remaining columns.

\begin{array}{cc} \begin{array}{rr} \\ &3 \\ -1& \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & & 3 & -39 \\ & -1 & 13 & \\ \hline 1 & -13 & 16 & -81 \\ \end{array} \end{array}


Count the terms to the left of the bar. Since there are two, the remainder has degree one and this is the two right-most terms under the bar. Mark the separation with a vertical bar.


\begin{array}{rr|rr} 1 & -13 & 16 & -81 \end{array}

The terms are written with increasing degree from right to left beginning with degree zero for both the remainder and the result.


\begin{array}{rr|rr} 1x & -13 & 16x & -81 \end{array}

The result of our division is:


\frac{x^3 - 12x^2 - 42}{x^2 + x - 3} = x - 13 + \frac{16x - 81}{x^2 + x - 3}


===================================================

Upvote • 1 Downvote
More
Report

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.