Answer :
Sure! Let's go through the synthetic division step by step.
We are using synthetic division to test if [tex]\(-5\)[/tex] is a root of the polynomial [tex]\(P(x) = x^3 + 6x^2 - 7x - 60\)[/tex].
The coefficients of the polynomial are [tex]\([1, 6, -7, -60]\)[/tex], and our potential root is [tex]\(-5\)[/tex].
### Synthetic Division Steps
1. Write down the coefficients:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ \end{array} \][/tex]
2. Bring down the leading coefficient unchanged:
[tex]\[ \begin{array}{rrrr} & & & \\ \hline 1 & & & \\ \end{array} \][/tex]
So, [tex]\(a = 1\)[/tex].
3. Multiply the leading coefficient by the potential root [tex]\(-5\)[/tex] and write the result under the next coefficient:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & & \\ \hline 1 & & & \end{array} \][/tex]
4. Add the second coefficient to this result:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & & \\ \hline 1 & 1 & & \\ \end{array} \][/tex]
So, [tex]\(b = 1\)[/tex].
5. Repeat the process: multiply the result by the potential root [tex]\(-5\)[/tex]:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & -5 \cdot 1 & \\ \hline 1 & 1 & & \\ \end{array} \][/tex]
6. Add the third coefficient to this result:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & -5 & \\ \hline 1 & 1 & -12 & \\ \end{array} \][/tex]
So, [tex]\(c = -12\)[/tex].
7. Repeat the process: multiply the result by the potential root [tex]\(-5\)[/tex]:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & -5 & -5 \cdot (-12) \\ \hline 1 & 1 & -12 & \\ \end{array} \][/tex]
8. Add the fourth coefficient to this result:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & -5 & 60 \\ \hline 1 & 1 & -12 & 0 \\ \end{array} \][/tex]
So, [tex]\(d = 0\)[/tex].
So, summarizing the results:
[tex]\[ \begin{array}{l} a = 1 \\ d = 0 \end{array} \][/tex]
And the intermediate results are:
[tex]\[ b = 1 \\ c = -12 \][/tex]
Thus, the completed synthetic division problem looks like this:
[tex]\[ \begin{array}{c} -5 \left\lvert\, \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & -5 & 60 \\ \hline 1 & 1 & -12 & 0 \end{array}\right. \end{array} \][/tex]
We are using synthetic division to test if [tex]\(-5\)[/tex] is a root of the polynomial [tex]\(P(x) = x^3 + 6x^2 - 7x - 60\)[/tex].
The coefficients of the polynomial are [tex]\([1, 6, -7, -60]\)[/tex], and our potential root is [tex]\(-5\)[/tex].
### Synthetic Division Steps
1. Write down the coefficients:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ \end{array} \][/tex]
2. Bring down the leading coefficient unchanged:
[tex]\[ \begin{array}{rrrr} & & & \\ \hline 1 & & & \\ \end{array} \][/tex]
So, [tex]\(a = 1\)[/tex].
3. Multiply the leading coefficient by the potential root [tex]\(-5\)[/tex] and write the result under the next coefficient:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & & \\ \hline 1 & & & \end{array} \][/tex]
4. Add the second coefficient to this result:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & & \\ \hline 1 & 1 & & \\ \end{array} \][/tex]
So, [tex]\(b = 1\)[/tex].
5. Repeat the process: multiply the result by the potential root [tex]\(-5\)[/tex]:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & -5 \cdot 1 & \\ \hline 1 & 1 & & \\ \end{array} \][/tex]
6. Add the third coefficient to this result:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & -5 & \\ \hline 1 & 1 & -12 & \\ \end{array} \][/tex]
So, [tex]\(c = -12\)[/tex].
7. Repeat the process: multiply the result by the potential root [tex]\(-5\)[/tex]:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & -5 & -5 \cdot (-12) \\ \hline 1 & 1 & -12 & \\ \end{array} \][/tex]
8. Add the fourth coefficient to this result:
[tex]\[ \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & -5 & 60 \\ \hline 1 & 1 & -12 & 0 \\ \end{array} \][/tex]
So, [tex]\(d = 0\)[/tex].
So, summarizing the results:
[tex]\[ \begin{array}{l} a = 1 \\ d = 0 \end{array} \][/tex]
And the intermediate results are:
[tex]\[ b = 1 \\ c = -12 \][/tex]
Thus, the completed synthetic division problem looks like this:
[tex]\[ \begin{array}{c} -5 \left\lvert\, \begin{array}{rrrr} 1 & 6 & -7 & -60 \\ & -5 & -5 & 60 \\ \hline 1 & 1 & -12 & 0 \end{array}\right. \end{array} \][/tex]
