Question

the function $f(x) = 5x^3 + 10x^2 + 9x + 18$ has at least one rational root. use the rational root theorem to find that root, then proceed to find all complex roots. (note: roots may be integer, rational, irrational, and/or complex.)

answer attempt 1 out of 2

there is one root :

Explanation:

Step1: Apply Rational Root Theorem

The Rational Root Theorem states that any possible rational root, \( \frac{p}{q} \), of a polynomial \( a_nx^n + a_{n - 1}x^{n - 1}+\dots+a_1x + a_0 \) is a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \). For \( f(x)=5x^3 + 10x^2+9x + 18 \), the constant term \( a_0 = 18 \) and the leading coefficient \( a_n=5 \). The factors of \( 18 \) are \( \pm1,\pm2,\pm3,\pm6,\pm9,\pm18 \) and the factors of \( 5 \) are \( \pm1,\pm5 \). So the possible rational roots are \( \pm1,\pm2,\pm3,\pm6,\pm9,\pm18,\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{3}{5},\pm\frac{6}{5},\pm\frac{9}{5},\pm\frac{18}{5} \).

We test these possible roots by substituting into \( f(x) \). Let's start with \( x=-2 \):

\( f(-2)=5(-2)^3 + 10(-2)^2+9(-2)+18=5(-8)+10(4)-18 + 18=-40 + 40-18 + 18 = 0 \). So \( x = - 2 \) is a rational root.

Step2: Factor the Polynomial

Since \( x=-2 \) is a root, \( (x + 2) \) is a factor of \( f(x) \). We perform polynomial long division or use synthetic division to factor \( f(x) \). Using synthetic division with root \( -2 \):

The coefficients of \( f(x) \) are \( 5,10,9,18 \).

Bring down the \( 5 \). Multiply \( 5\times(-2)=-10 \), add to \( 10 \): \( 10-10 = 0 \). Multiply \( 0\times(-2)=0 \), add to \( 9 \): \( 9+0 = 9 \). Multiply \( 9\times(-2)=-18 \), add to \( 18 \): \( 18-18 = 0 \). So the quotient polynomial is \( 5x^2+0x + 9=5x^2+9 \). Thus, \( f(x)=(x + 2)(5x^2+9) \).

Step3: Find the Remaining Roots

Set \( 5x^2+9 = 0 \). Then \( 5x^2=-9 \), \( x^2=-\frac{9}{5} \), so \( x=\pm\frac{3i}{\sqrt{5}}=\pm\frac{3i\sqrt{5}}{5} \).

Answer:

The rational root is \( - 2 \), and the complex roots are \( -2,\frac{3i\sqrt{5}}{5},-\frac{3i\sqrt{5}}{5} \)