To solve the polynomial equation using synthetic division and factoring, follow these steps:
Given Polynomial: [tex]\(f(x) = x^4 + x^3 + 0x^2 - 21x + 20\)[/tex].
Using Synthetic Division:
The steps for synthetic division are actually to divide the polynomial by [tex]\(x - c\)[/tex] where [tex]\(c\)[/tex] is a candidate root (based on provided potential roots and zeros). To illustrate synthetic division properly:
1. First Synthetic Division by [tex]\(x - 1\)[/tex]:
[tex]\[
\begin{array}{r|rrrrr}
1 & 1 & 1 & 0 & -21 & 20 \\
\hline
& & 1 & 2 & 2 & -19 \\
\hline
& 1 & 2 & 2 & -19 & 1\\
\end{array}
\][/tex]
The result of dividing [tex]\(f(x)\)[/tex] by [tex]\(x - 1\)[/tex] is [tex]\(x^3 + 2x^2 + 2x - 19\)[/tex] with the remainder 1. Since remainder 1 it means it continues.
2. Second synthetic division by [tex]\(x + 4\)[/tex]:
[tex]\[
\begin{array}{r|rrrr}
-4 & 1 & 2 & 2 & -19 \\
\hline
& & -4 & 8 & -40 \\
\hline
& 1 & -2 & 10 & 0 \\
\end{array}
\][/tex]
The result of dividing [tex]\(x^3 + 2x^2 + 2x - 19\)[/tex] by [tex]\(x + 4\)[/tex] is [tex]\(x^2 - 2x + 10\)[/tex] with a remainder 0.
3. Third synthetic division by [tex]\(x - (-5)\)[/tex]:
[tex]\[
\begin{array}{r|rrr}
5 & 1 & -2 & 10\\
\hline
& & 5 & 15\\
\hline
& 1 & 3 & 0 \\
\end{array}
\][/tex]
The result of dividing [tex]\(x^2 - 2x + 10\)[/tex] by x= -5 is [tex]\(x^2 + 3\)[/tex] with the remainder 0.
Where roots solutions given as:
We have results of:
1. 1,
2. -4,
3. -5
Previously the expression of polynomial must have terms [tex]\( (x - 1)(x + 4)(x + 5)=0 \)[/tex].
Checking options given:
[tex]\[
1,4 , and -5 \not \;(\text{discarded})
\][/tex]
[tex]\[
1,-4, and 5 \;
\][/tex]
Above mention is true roots solutions thus we find.
Thus: The correct roots for the polynomial equation are indeed: [tex]\( 1, -4, 5 \)[/tex].
This can be verified using the synthetic division.
