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Section 12.2 Equation of Polynomials

We learned how to write the equation of quadratic functions in Section 11.2, and our strategy for polynomials will be very similar. The only difference is that there can be lots of roots, not just two.

Example 12.20.

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 \(5\text{.}\)

Table 12.21.
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:

Table 12.22.
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 12.23.

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.

Table 12.24.
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:

Table 12.25.
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*}