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blagie [28]
3 years ago
15

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

Mathematics
2 answers:
kodGreya [7K]3 years ago
5 0
We will find the rational zeros of <span>f(x) = 3x^3 -5x^2 - 12x + 20

1) Arrange in descending order: </span><span>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, </span>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:
<span>A)2x + 1 B)2x - 1 C)3x+5 D)3x - 5.
</span>
MY ANSWER IS: D) 3x - 5.

charle [14.2K]3 years ago
4 0

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).


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