Question

for this exercise, consider the quintic polynomial f(x) shown below which has the complex root -2 + i: f(x)=2x^5 + 5x^4 - 5x^3 - 25x^2 - 7x + 10 x_1=-2 + i using the complex conjugate root theorem, find an irreducible quadratic factor of f(x): irreducible quadratic factor: divide the original function f(x) by the quadratic factor to find a cubic quotient: reducible cubic factor: list all possible rational roots of the cubic factor above, using integers or exact fractions: p/q = f(x) is a 5^th order polynomial, with five actual roots. list them all, including complex roots, using integers or exact fractions: x = the fully factored form of the polynomial f(x) will include three linear factors and one irreducible quadratic factor that came from complex roots. write the fully factored form: f(x)=

Explanation:

Step1: Find the conjugate root

If \(x_1=-2 + i\) is a root, by the Complex - Conjugate Root Theorem, its conjugate \(x_2=-2 - i\) is also a root. Then the irreducible quadratic factor is \((x-(-2 + i))(x-(-2 - i))\).
\[

$$\begin{align*} &(x + 2 - i)(x + 2 + i)\\ =&(x + 2)^2-i^2\\ =&x^{2}+4x + 4+1\\ =&x^{2}+4x + 5 \end{align*}$$

\]

Step2: Divide the polynomial

Divide \(f(x)=2x^{5}+5x^{4}-5x^{3}-25x^{2}-7x + 10\) by \(x^{2}+4x + 5\) using polynomial long - division or synthetic division. The quotient is \(2x^{3}-3x^{2}+7x + 2\).

Step3: Find possible rational roots

For a polynomial \(a_nx^n+\cdots+a_1x + a_0\), the possible rational roots are of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading - coefficient \(a_n\). For the polynomial \(2x^{3}-3x^{2}+7x + 2\), \(a_0 = 2\) and \(a_n=2\). The factors of \(2\) are \(\pm1,\pm2\). So the possible rational roots \(\frac{p}{q}\) are \(\pm1,\pm2,\pm\frac{1}{2}\).

Step4: Find all roots

By testing the possible rational roots \(\frac{p}{q}\) of \(2x^{3}-3x^{2}+7x + 2\), we find that \(x = - \frac{1}{2}\) is a root. Then we divide \(2x^{3}-3x^{2}+7x + 2\) by \(2x + 1\) to get \(x^{2}-2x + 2\). Solving \(x^{2}-2x + 2 = 0\) using the quadratic formula \(x=\frac{2\pm\sqrt{4 - 8}}{2}=1\pm i\). The roots of \(f(x)\) are \(-2 + i,-2 - i,-\frac{1}{2},1 + i,1 - i\).

Step5: Write the fully - factored form

\(f(x)=2(x + 2 - i)(x + 2 + i)(2x + 1)(x-(1 + i))(x-(1 - i))=(x^{2}+4x + 5)(2x + 1)(x^{2}-2x + 2)\)

Answer:

Irreducible Quadratic Factor: \(x^{2}+4x + 5\)
Reducible Cubic Factor: \(2x^{3}-3x^{2}+7x + 2\)
Possible rational roots \(\frac{p}{q}\): \(\pm1,\pm2,\pm\frac{1}{2}\)
Roots of \(f(x)\): \(-2 + i,-2 - i,-\frac{1}{2},1 + i,1 - i\)
Fully - factored form of \(f(x)\): \((x^{2}+4x + 5)(2x + 1)(x^{2}-2x + 2)\)