Question

for this exercise, consider the polynomial shown below: f(x)=3x^4 + 7x^3 - 8x^2 - 20x list all possible non - zero rational roots for f(x). no decimals! p/q = the fully factored form of the polynomial is: f(x)= now list the actual roots for f(x). no decimals! x=

Explanation:

Step1: Identify leading - coefficient and constant term

For \(f(x)=3x^{4}+7x^{3}-8x^{2}-20x = 3x^{4}+7x^{3}-8x^{2}-20x+0\), the leading - coefficient \(a_{n}=3\) and the constant term \(a_{0} = 0\). But we can factor out an \(x\) first: \(f(x)=x(3x^{3}+7x^{2}-8x - 20)\). For the cubic polynomial \(g(x)=3x^{3}+7x^{2}-8x - 20\), \(a_{n}=3\) and \(a_{0}=-20\).

Step2: Find factors of \(a_{n}\) and \(a_{0}\)

The factors of \(a_{n}=3\) are \(\pm1,\pm3\), and the factors of \(a_{0}=-20\) are \(\pm1,\pm2,\pm4,\pm5,\pm10,\pm20\).

Step3: Determine possible non - zero rational roots

By the Rational Root Theorem, the possible non - zero rational roots \(\frac{p}{q}\) of \(g(x)\) are \(\pm1,\pm2,\pm4,\pm5,\pm10,\pm20,\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{5}{3},\pm\frac{10}{3},\pm\frac{20}{3}\).

Step4: Factor the polynomial

We can test the possible roots. By testing \(x = 2\): \(g(2)=3\times2^{3}+7\times2^{2}-8\times2 - 20=24 + 28-16 - 20 = 16
eq0\). By testing \(x=-2\): \(g(-2)=3\times(-2)^{3}+7\times(-2)^{2}-8\times(-2)-20=-24 + 28 + 16-20 = 0\). So \((x + 2)\) is a factor of \(g(x)\). Using long - division or synthetic division: \(3x^{3}+7x^{2}-8x - 20=(x + 2)(3x^{2}+x - 10)\). Then factor \(3x^{2}+x - 10=(3x - 5)(x+2)\). So \(f(x)=x(x + 2)^{2}(3x - 5)\).

Step5: Find the actual roots

Set \(f(x)=0\), then \(x(x + 2)^{2}(3x - 5)=0\). The roots are \(x = 0,x=-2,x=\frac{5}{3}\).

Answer:

Possible non - zero rational roots: \(\pm1,\pm2,\pm4,\pm5,\pm10,\pm20,\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{5}{3},\pm\frac{10}{3},\pm\frac{20}{3}\)
Fully factored form: \(f(x)=x(x + 2)^{2}(3x - 5)\)
Actual roots: \(x = 0,x=-2,x=\frac{5}{3}\)