Use roots of a polynomial equation to find other roots.

Assignment

  • All vocabulary copied into notes
  • p177 13–19, 22–38 22–32 excluding 27

Rational Root Theorem

By now, you’ve likely run into a polynomial that you couldn’t factor, for example $2x^{3}-6x^{2}-x+5$. Short of turning to technology to either graph it or find the zeros directly, there is one more path: the rational root theorem.

If the polynomial has a rational root (so this doesn’t cover irrational or complex roots), that root can be written as $x= p/q$, where

  • $p$ is an integer factor of the constant term, and
  • $q$ is an integer factor of the leading coefficient.

If you look back at any of the polynomials you’ve factored, you’ll find they all fit. Any rational zero can be written as the quotient of $p$ and $q$, factors of the constant and leading coefficient.

We can use this knowledge to come up with a list of potential roots for a polynomial, like $2x^{3}-6x^{2}-x+5$. If this has a rational root, something we can factor out, it will be in the form $p/q$. To find the candidates, we need the factors of our constant and leading coefficient. And don’t forget about the negatives.

For $p$, we have $\pm5$ and $\pm1$. For $q$, $\pm2$ and $\pm1$. That means our list of potential rational zeros is

\[\pm\frac{5}{2}, \pm5, \pm\frac{1}{2}, \pm1\]

That is eight candidates to test. Eight values to evaluate with synthetic division. If we didn’t have access to graphing calculators, this is still huge news because we narrowed the list of potential zeros from infinity down to eight. That’s huge.

But we do have access to technology, so we’ll use it. Graphing this we see three zeros, one looks like it’s exactly at $x=1$, but the other two are decimal estimations. We’ll use synthetic division to factor out $(x-1)$ and hopefully arrive at a more exact answer for the other two.

After division, we end up with $(x-1)(2x^2-4x-5)$. The quadratic is still unfactorable, but we can use the quadratic formula to get the last two zeros. Our last two zeros, in exact form, are $\frac{2\pm\sqrt{14}}{2}$.

The Fundamental Theorem of Algebra

Straight from Wikipedia:

Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.

Despite the grand name, it’s a basic concept that is helpful when determining the zeros of a polynomial. If it’s degree 5, you can expect 5 zeros, complex ones included. If you are only concerned about real zeros, then instead you can expect up to 5.

As a simple example, graph $f(x)=x^2+1$. It does not cross the $x$-axis, so it has no real zeros. But you can still run it through the quadratic formula, or addition of squares, and get $\pm i$ as your two zeros.

Conjugate Root Theorem

You might have noticed that our roots involving radicals or $i$ come in pairs. Our first example above had $\frac{2+\sqrt{14}}{2}$ and $\frac{2-\sqrt{14}}{2}$, and the roots of $x^2+1$ were $i$ and $-i$.

Well, the polynomials we’ve been working with have all had rational coefficients, so no radicals or $i$ to be found. So when we do find roots involving them, their conjugates will naturally come along, too. Remember that when conjugates are multiplied, they “cancel” out the radical or imaginary part of the expression.

\((a+\sqrt{b})(a-\sqrt{b}) = a^2 +a\sqrt{b} - a\sqrt{b} - b = a^2 - b\) \((a+bi)(a-bi) = a^2 +abi - abi - (bi)^2 = a^2+b^2\)

So, if we had a root of $2+5i$, what’s the polynomial? Automatically, we know the other root is $2-5i$, so we can write our factors.

\[(x-(2+5i))(x-(2-5i))\]

You could distribute, but that’s a three by three and we want to save time when we can. We’ll use difference of squares to our advantage.

\[\begin{align*} (x-(2+5i))(x-(2-5i)) &= (x-2-5i)(x-2+5i) \\ &= ((x-2)-5i)((x-2)+5i) \\ &= (x-2)^2 - (5i)^2 \\ &= (x^2-4x+4) + 25 \\ &= x^2 - 4x + 29 \end{align*}\]