Rational root theorem example | Math, Algebra 2, Algebra ... - Free Printable

The problem involves factoring the polynomial \( f(x) = 2x^4 + 0x^3 - 19x^2 + 0x + 9 \). Let's break down the solution step by step.

Step 1: Identify Possible Rational Roots

The Rational Root Theorem states that any possible rational root of the polynomial \( f(x) \) is of the form \( \frac{p}{q} \), where:
- \( p \) is a factor of the constant term (here, \( 9 \)),
- \( q \) is a factor of the leading coefficient (here, \( 2 \)).

The factors of \( 9 \) are \( \pm 1, \pm 3, \pm 9 \).
The factors of \( 2 \) are \( \pm 1, \pm 2 \).

Thus, the possible rational roots are:
\[
\pm 1, \pm \frac{1}{2}, \pm 3, \pm \frac{3}{2}, \pm 9, \pm \frac{9}{2}
\]

Step 2: Test Possible Roots

We test these possible roots by substituting them into \( f(x) \).

#### Test \( x = 1 \):
\[
f(1) = 2(1)^4 + 0(1)^3 - 19(1)^2 + 0(1) + 9 = 2 - 19 + 9 = -8 \neq 0
\]
So, \( x = 1 \) is not a root.

#### Test \( x = -1 \):
\[
f(-1) = 2(-1)^4 + 0(-1)^3 - 19(-1)^2 + 0(-1) + 9 = 2 - 19 + 9 = -8 \neq 0
\]
So, \( x = -1 \) is not a root.

#### Test \( x = 3 \):
\[
f(3) = 2(3)^4 + 0(3)^3 - 19(3)^2 + 0(3) + 9 = 2(81) - 19(9) + 9 = 162 - 171 + 9 = 0
\]
So, \( x = 3 \) is a root.

Step 3: Perform Polynomial Division

Since \( x = 3 \) is a root, \( x - 3 \) is a factor of \( f(x) \). We perform synthetic division of \( f(x) \) by \( x - 3 \).

#### Synthetic Division for \( x = 3 \):
\[
\begin{array}{r|rrrrr}
3 & 2 & 0 & -19 & 0 & 9 \\
& & 6 & 18 & -3 & -9 \\
\hline
& 2 & 6 & -1 & -3 & 0 \\
\end{array}
\]
The quotient is \( 2x^3 + 6x^2 - x - 3 \), and the remainder is 0. Thus:
\[
f(x) = (x - 3)(2x^3 + 6x^2 - x - 3)
\]

Step 4: Factor the Cubic Polynomial

Let \( g(x) = 2x^3 + 6x^2 - x - 3 \). We test possible rational roots for \( g(x) \).

#### Test \( x = -3 \):
\[
g(-3) = 2(-3)^3 + 6(-3)^2 - (-3) - 3 = 2(-27) + 6(9) + 3 - 3 = -54 + 54 + 3 - 3 = 0
\]
So, \( x = -3 \) is a root.

#### Synthetic Division for \( x = -3 \):
\[
\begin{array}{r|rrrr}
-3 & 2 & 6 & -1 & -3 \\
& & -6 & 0 & 3 \\
\hline
& 2 & 0 & -1 & 0 \\
\end{array}
\]
The quotient is \( 2x^2 - 1 \), and the remainder is 0. Thus:
\[
g(x) = (x + 3)(2x^2 - 1)
\]

Step 5: Combine All Factors

Substituting back, we have:
\[
f(x) = (x - 3)(x + 3)(2x^2 - 1)
\]

Step 6: Factor the Quadratic Polynomial

The quadratic \( 2x^2 - 1 \) can be factored as:
\[
2x^2 - 1 = 2\left(x^2 - \frac{1}{2}\right) = 2\left(x - \frac{1}{\sqrt{2}}\right)\left(x + \frac{1}{\sqrt{2}}\right)
\]
However, for simplicity, we often leave it as \( 2x^2 - 1 \) or factor it further using radicals:
\[
2x^2 - 1 = (x - \frac{\sqrt{2}}{2})(x + \frac{\sqrt{2}}{2})
\]

Final Answer

The complete factorization of \( f(x) \) is:
\[
f(x) = (x - 3)(x + 3)(x - \frac{\sqrt{2}}{2})(x + \frac{\sqrt{2}}{2})
\]

However, if we want to keep it in a simpler form without radicals:
\[
f(x) = (x - 3)(x + 3)(2x^2 - 1)
\]

Thus, the final answer is:
\[
\boxed{(x - 3)(x + 3)(2x^2 - 1)}
\]