Question

the function f(x) = 2x³ + 2x² - 7x - 7 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 \(f(x)=2x^3 + 2x^2 - 7x - 7\), \(a_0=-7\) (factors: \(\pm1, \pm7\)) and \(a_n = 2\) (factors: \(\pm1, \pm2\)). Possible roots: \(\pm1, \pm7, \pm\frac{1}{2}, \pm\frac{7}{2}\).

Test \(x = -1\): \(f(-1)=2(-1)^3 + 2(-1)^2 - 7(-1) - 7=-2 + 2 + 7 - 7 = 0\). 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. Using synthetic division with root \(-1\):

Coefficients: \(2\) (x³), \(2\) (x²), \(-7\) (x), \(-7\) (constant)

Bring down \(2\). Multiply by \(-1\): \(-2\). Add to next coefficient: \(2 + (-2)=0\). Multiply by \(-1\): \(0\). Add to next coefficient: \(-7 + 0=-7\). Multiply by \(-1\): \(7\). Add to last coefficient: \(-7 + 7 = 0\).

The quotient is \(2x^2 + 0x - 7=2x^2 - 7\). So \(f(x)=(x + 1)(2x^2 - 7)\).

Step3: Find Remaining Roots

Solve \(2x^2 - 7 = 0\). \(2x^2=7\) → \(x^2=\frac{7}{2}\) → \(x=\pm\sqrt{\frac{7}{2}}=\pm\frac{\sqrt{14}}{2}\).

Answer:

The rational root is \(-1\), and the complex (including real and irrational) roots are \(-1\), \(\frac{\sqrt{14}}{2}\), \(-\frac{\sqrt{14}}{2}\). The rational root found via Rational Root Theorem is \(\boxed{-1}\).