More on Fundamental Theorem of Algebra

Objective: Reinforce Fundamental Theorem & related problem solving.

Warmup: factor $f\left(x\right)=2x^4-2x^3-10x^2-2x-12$ given that $i$ is a zero.

Solution

If $i$ is a zero, since all the coefficients are real, we know that $\bar{i}$ (the complex conjugate of $i$) is also a zero. $\bar{i}=-i$. So $x-i$ and $x+i$ are both factors of $f$.

Factoring by division says we should divide $f$ by those factors. Their product is $(x-i)(x+i)=x^2+1$. Divide $f$ by that product (why?):

So far we have factored $f$ as

Continuing to factor:

Zeros (or roots) are:

What is the degree of $f$?

Which zeros (or roots) are $x$-intercepts?

Fundamental Theorem of Algebra

If $f$ is a polynomial of degree $n$ ($n>0$) then $f$ has at least one complex (maybe real) root.

Because every root (or zero) $r$ corresponds to a factor $x-r$ (why?), we can divide $\frac{f}{x-r}$ to get a polynomial of degree $n-1$. This also has a root, and so on. This shows that really a polynomial of degree $n$ has $n$ complex roots, though some may be repeated.

Repeated zero. What's that about? Consider ${(x-2)}^2$. It has a zero of 2. That corresponds to the factor $x-2$. If we divide our polynomial by that factor we get

That also has a zero of 2, representing the “other” factor, also $x-2$. Because the factor $x-2$ is repeated, we say that the zero 2 is repeated or that it has multiplicity 2. Think of each zero corresponding to a linear factor.

• Problem 17. Is 0 a zero? Plug it in:

  ${\displaystyle \begin{array}{rcl}{x}^4-x^2-3x+3& & \\ 0^4-0^2-3\cdot 0+3& & \\ 3& & \mathrm{no}!\end{array}}$

• Problem 27. Find the zeros of $x^3-x^2+49x-49$. First factor:

  ${\displaystyle \begin{array}{rcl}& & \text{x3-x2+49*x-49}\\ & =& (x^2+49)(x-1)\\ & =& (x+7i)(x-7i)(x-1)\end{array}}$

  So zeros are

  $-7i$, $7i$, 1

• Problem 41: Write a polynomial of least degree with real coefficients, and a coefficient of one, given these zeros: $3i$, $-3i$, 5.

  Solution: If they didn't say $-3i$ was a zero we'd know it anyway. (Why?)

  ${\displaystyle \begin{array}{rcl}& & (x-3i)(x+3i)(x-5)\\ & =& (x^2+9)(x-5)\\ & =& x^3-5x^2+9x-45\end{array}}$

• Problem 55. Need to solve

  ${\displaystyle \begin{array}{rcl}312.76& =& -.131t^3+5.033t^2-23.2t+233\\ 0& =& -.131t^3+5.033t^2-23.2t+233-312.76\\ 0& =& -.131t^3+5.033t^2-23.2t-79.76\end{array}}$

  Now plug the right side into the

  Y=

  of the calculator, graph, and use

  2nd

  calc

  zero to find the zeros.