Question

According to the rational root theorem which is a factor of the polynomial f(x)=3x^3-5x^2-12x+20

Answer 1

Answer:

$f(x)=3(x+2)(x-2)(x-\frac{5}{3})$

Step-by-step explanation:

The rational roots theorem tells you that given a polynomial function with integer or whole number coefficients, a list of possible solutions can be found by listing the factors of the constant, or last term, over the factors of the coefficient of the leading term.

In your case, for the polynomial $f(x)=3x^3-5x^2-12x+20:$

• the last term is 20;
• the leading coeeficient is 3.

So, possible rational roots can be among:

$\pm1,\pm2,\pm4,\pm5,\pm10,\pm20,\pm\dfrac{1}{3},\pm\dfrac{2}{3},\pm\dfrac{4}{3},\pm\dfrac{5}{3},\pm\dfrac{10}{3},\pm\dfrac{20}{3}.$

Note that

$f(-2)=3\cdot (-2)^3-5\cdot (-2)^2-12\cdot (-2)+20=-24-20+24+20=0.$

This means that $x=-2$ is a root of the polynomial and $x-(-2)=x+2$ is the factor. Also

$f(2)=3\cdot 2^3-5\cdot 2^2-12\cdot 2+20=24-20-24+20=0.$

This means that $x=2$ is a root of the polynomial and $x-2$ is the factor. Also

$f(\frac{5}{3})=3\cdot (\frac{5}{3})^3-5\cdot (\frac{5}{3})^2-12\cdot \frac{5}{3}+20=\frac{125}{9}-\frac{125}{9}-20+20=0.$

This means that $x=\frac{5}{3}$ is a root of the polynomial and $x-\frac{5}{3}$ is the factor.

Then

$f(x)=3(x+2)(x-2)(x-\frac{5}{3}).$