Question

the function $f(x) = 2x^3 - 2x^2 + 9x - 9$ 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 possible rational roots of a polynomial \(a_nx^n + \dots + a_1x + a_0\) are \(\pm\frac{\text{factors of }a_0}{\text{factors of }a_n}\). For \(f(x)=2x^3 - 2x^2 + 9x - 9\), \(a_0=-9\) (factors: \(\pm1,\pm3,\pm9\)) and \(a_n = 2\) (factors: \(\pm1,\pm2\)). Possible roots: \(\pm1,\pm3,\pm9,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{9}{2}\).

Test \(x = 1\): \(f(1)=2 - 2 + 9 - 9 = 0\). So \(x = 1\) is a root.

Step2: Factor the Polynomial

Since \(x = 1\) is a root, \((x - 1)\) is a factor. Use polynomial division or synthetic division. Let's use synthetic division:

Coefficients: \(2\) (for \(x^3\)), \(-2\) (for \(x^2\)), \(9\) (for \(x\)), \(-9\) (constant).

Divide by \(x - 1\) (root \(1\)):

Bring down \(2\). Multiply by \(1\): \(2\). Add to \(-2\): \(0\). Multiply by \(1\): \(0\). Add to \(9\): \(9\). Multiply by \(1\): \(9\). Add to \(-9\): \(0\).

So the quotient is \(2x^2 + 0x + 9 = 2x^2 + 9\). Thus, \(f(x)=(x - 1)(2x^2 + 9)\).

Step3: Find Remaining Roots

Solve \(2x^2 + 9 = 0\). Subtract \(9\): \(2x^2=-9\). Divide by \(2\): \(x^2=-\frac{9}{2}\). Take square roots: \(x=\pm\frac{3i}{\sqrt{2}}=\pm\frac{3i\sqrt{2}}{2}\) (complex roots).

Answer:

The rational root is \(1\), and the complex roots are \(\frac{3i\sqrt{2}}{2}\) and \(-\frac{3i\sqrt{2}}{2}\). But the first rational root found is \(1\).