3.3: Polynomial Equations
 Page ID
 238
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So far we have learned how to find the roots of a polynomial equation. If we have an equation that involves only polynomials we follow the steps:
 Bring all the terms over to the left hand side of the equation so that the right hand side of the equation is a 0.
 Get rid of denominators by multiplying by the least common denominator.
 If there is a common factor for all the terms, factor immediately. Otherwise, multiply the terms out.
 Use a calculator to locate roots.
 Use the Rational Root Theorem and synthetic division to exactly determine the roots.
Example 1
Find all the rational solutions of
\[\dfrac{(2x^3  5)}{4} = x  x^2.\]
Solution:

\(\dfrac{(2x^3  5)}{4}  x + x^2 = 0\)

\((2x^3  5) 4x + 4x^2 = 0\)

\(2x^3 + 4x^2  4x  5 = 0\)

From the graph, we see that there is a root between 3 and 2 and a root between 0 and 1 and a root between 1 and 2.

Since the only possible rational roots are 1, 1, 5, 5, .5, .5, 2.5, 2.5, the possible rational roots are \(\dfrac{5}{2}\) and .5. Neither of these two are roots, hence there are no rational roots.
Example 2
Solve
\[x[x^2(2x + 3) + 10x + 17] + 5 = 2.\]

\(x[x^2(2x + 3) + 10x + 17] + 3 = 0\)

\(2x^4 + 3x^3 + 10x^2 + 17x + 3 = 0\)

We see that there is a root between 2 and 1 and between 1 and 0.

Our only possible roots are \(\frac{1}{2}\) and \(\frac{3}{2}\).

Using synthetic division, we see that \(\frac{3}{2}\) is a root, and the remainder is
\[2x^3 + 10x + 2 = 2(x^3 + 5x + 1)\]
which has no rational roots. Hence the rational root is \(\frac{3}{2}\) and using the calculator we see that the irrational root is 0.198.
Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.