Fact 10.2.1. Strategy for Writing a Polynomial Equation.
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Use the roots to write the factors
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Use the multiplicities to put exponents on the factors
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Use other given information (usually a point) to determine the number multiplied out front
Section 10.2 Equation of Polynomials
Example 10.2.2.
Suppose we want to write the equation for a polynomial with roots of multiplicity \(2\) at \(-1\) and \(3\text{,}\) and a root of multiplicity \(5\) at \(1\text{,}\) and goes through the point \((2,-18)\text{.}\)
Even though this isn’t a word problem, it is a lot of information written as a sentence, so you might find it helpful to organize the information in a table, like the one below. One thing students often struggle with in the phrasing of the question is which number is the root and which number is the multiplicity. When we say "at", that tells you the location, which is the \(x\)-value. So, the phrase "root of multiplicity \(5\) at \(1\)" means that the \(x\)-value (the root) is \(1\text{,}\) and its multiplicity is \(5\text{.}\)
| Root | Factor | Multiplicity | Factor with Exponent |
| \(-1\) | \(2\) | ||
| \(3\) | \(2\) | ||
| \(1\) | \(5\) |
Now, we can use what we know about factors and multiplicities to finish filling out the table:
| Root | Factor | Multiplicity | Factor with Exponent |
| \(-1\) | \((x+1)\) | \(2\) | \((x+1)^2\) |
| \(3\) | \((x-3)\) | \(2\) | \((x-3)^2\) |
| \(1\) | \((x-1)\) | \(5\) | \((x-1)^5\) |
With that information, we can write most of our function:
\begin{equation*}
f(x)=A(x+1)^2(x-3)^2(x-1)^5
\end{equation*}
Now, we have to use the point we are given, \((2,-18)\text{,}\) to figure out what \(A\) is. We use the same strategy we’ve used for linear functions and for quadratic functions: plug in \(x=2\) and \(y=-18\) to solve for \(A\text{.}\)
\begin{align*}
f(x)\amp=A(x+1)^2(x-3)^2(x-1)^5\\
-18\amp=A(2+1)^2(2-3)^2(2-1)^5\\
-18\amp=A(3)^2(-1)^2(1)^5\\
-18\amp=A(9)(1)(1)\\
-18\amp=A(9)\\
-2\amp=A
\end{align*}
Now we have all of the information we need to write our final answer:
\begin{equation*}
f(x)=-2(x+1)^2(x-3)^2(x-1)^5
\end{equation*}
Checkpoint 10.2.5.
Write the equation for the polynomial function with a root of multiplicity \(4\) at \(-3\) and roots of multiplicity \(1\) at \(1\) and \(5\text{,}\) and goes through the point \((-2,42)\text{.}\)
Answer.
\(f(x)=\frac{1}{2}(x+3)^4(x-1)(x-5)\)
Solution.
Even though this isn’t a word problem, it is a lot of information written as a sentence, so you might find it helpful to organize the information in a table, like the one below.
| Root | Factor | Multiplicity | Factor with Exponent |
| \(-3\) | \(4\) | ||
| \(1\) | \(1\) | ||
| \(5\) | \(1\) |
Now, we can use what we know about factors and multiplicities to finish filling out the table:
| Root | Factor | Multiplicity | Factor with Exponent |
| \(-3\) | \((x+3)\) | \(4\) | \((x+3)^4\) |
| \(1\) | \((x-1)\) | \(1\) | \((x-1)^1=(x-1)\) |
| \(5\) | \((x-5)\) | \(1\) | \((x-5)^1=(x-5)\) |
With that information, we can write most of our function:
\begin{equation*}
f(x)=A(x+3)^4(x-1)(x-5)
\end{equation*}
Now, we have to use the point we are given, \((-2,42)\text{,}\) to figure out what \(A\) is. We use the same strategy we’ve used for linear functions and for quadratic functions: plug in \(x=-2\) and \(y=42\) to solve for \(A\text{.}\)
\begin{align*}
f(x)\amp=A(x+3)^4(x-1)(x-5)\\
42\amp=A(-2+3)^4(-2-1)(-2-5)\\
42\amp=A(1)^4(-3)(-7)\\
42\amp=A(21)\\
\frac{1}{2}\amp=A
\end{align*}
Now we have all of the information we need to write our final answer:
\begin{equation*}
f(x)=\frac{1}{2}(x+3)^4(x-1)(x-5)
\end{equation*}
