Question

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

Answer 1

We will find the rational zeros of f(x) = 3x^3 -5x^2 - 12x + 20

1) Arrange in descending order: f(x) = 3x^3 -5x^2 - 12x + 20

2) write down the factors of the constant term, 20. 
1,2, 4, 5, 10, 20 - these could all be the possible values of p.

3) write down the factors of the leading coefficient, 3.
1,3 - these are the possible values of q.
.
4) write down all the possible values of p/q. both positive and negative values must be written down. Simplify
1/1 ; 1/3 ; 2/1 ; 2/3 ; 4/1 ; 4/3 ; 5/1 ; 5/3 ; 10/1 ; 10/3 ; 20/1 ; 20/3
1 ; 0.33 ; 2 ; 0.67 ; 4 ;  1.33 ; 5 ; 1.67 ; 10 ; 3.33 ; 20 ; 6.67

5) use the simplified p/q in both positive and negative values in the synthetic division.  Pls. see attachment for my synthetic division.

Based on my computations, a factor of the polynomial f(x) = 3x^3 -5x^2 - 12x + 20 is:  x = -1.67 or x = -5/3

IF THESE ARE THE MISSING CHOICES:
A)2x + 1 B)2x - 1 C)3x+5 D)3x - 5.

MY ANSWER IS: D) 3x - 5.

Answer 2

In algebra, the rational root theorem states a constraint on rational solutions of a polynomial equation

$a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x+a_0=0$

with integer coefficients. Solutions of the equation are roots of the polynomial on the left side of the equation.

If $a_0$ and $a_n$ are nonzero, then each rational solution x, when written as a fraction $x=\dfrac{p}{q}$ in lowest terms satisfies

• p is an integer factor of the constant term $a_0,$
• q is an integer factor of the leading coefficient $a_n.$

For the polynomial $f(x) = 3x^3 -5x^2 - 12x + 20$, $a_3=3,\ a_0=20.$ Find factors of these coefficients:

• factors of $a_3=3$ are $\pm 1,\ \pm 3;$
• factors of $a_0=20$ are $\pm 1,\ \pm 2,\ \pm 4,\ \pm 5,\ \pm 10,\ \pm 20.$

Therefore, possible roots are

$\pm 1,\ \pm 2,\ \pm 4,\ \pm 5,\ \pm 10,\ \pm 20,\ \pm \dfrac{1}{3}, \ \pm \dfrac{2}{3},\ \pm \dfrac{4}{3},\ \pm \dfrac{5}{3},\ \pm \dfrac{10}{3},\ \pm \dfrac{20}{3}.$

Check all that apply:

$f(2)=3\cdot 2^3 -5\cdot 2^2 - 12\cdot 2 + 20=3\cdot 8-5\cdot 4-24+20=24-20-24+20=0;$

$\\f(-2)=3\cdot (-2)^3 -5\cdot (-2)^2 - 12\cdot (-2) + 20=3\cdot (-8)-5\cdot 4+24+20=-24-20-24+20=0;$

$\\f\left(\dfrac{5}{3}\right)=3\cdot \left(\dfrac{5}{3}\right)^3 -5\cdot \left(\dfrac{5}{3}\right)^2 - 12\cdot \left(\dfrac{5}{3}\right) + 20=3\cdot \left(\dfrac{125}{27}\right)-5\cdot \left(\dfrac{25}{9}\right)-20+20=\dfrac{125}{9}-\dfrac{125}{9}=0.$

Then

$f(x) = 3x^3 -5x^2 - 12x + 20=3(x-2)(x+2)\left(x-\dfrac{5}{3}\right).$

Answer: there are three possible factors

$(x-2),\ (x+2),\ \left(x-\dfrac{5}{3}\right).$

The last one can be rewritten as 3x-5, then $f(x) =(x-2)(x+2)\left(3x-5\right).$