Consider the polynomial:
f(x) = 2x^3 – 3x^2 – 8x – 3.
(a) By using the Rational Zero Theorem, list all possible rational zeros of the given polynomial.
(b) Find all of the zeros of the given polynomial. Be sure to show work, explaining how you have found them.
values I would try are
±1, ±3, ±1/2, and ±3/2
b)
on the second try
f(-1) = -2-3+8-3 = 0
so x+1 is a factor.
by synthetic division I got
2x^3 – 3x^2 – 8x – 3 = (x+1)(2x^2 - 5x -3)
with a couple trial and error stabs, I factored the quadratic into (x-3)(2x+1)
so the factors are
(x+1)(x-3)(2x+1)
of course I could have continued with the above values of a) and tried
f(3), f(-3), f(1/2) etc
and would have found
f(3) and f(-1/2) also to result in zero.
f(x)=0
F(x) = (2x+1) (x+1) (x-3)
x=-1/2
x=-1
x=3
F(x) =0
F(x) = (2x+1) (x+1) (x-3)
(2x+1) (x+1) (x-3) = 0
Is this a good way to show the procedure?
(a) To find all the possible rational zeros of a polynomial, we use the Rational Zero Theorem. According to the theorem, if a rational number p/q is a zero of a polynomial, then p must be a factor of the constant term and q must be a factor of the leading coefficient.
In this case, the constant term is -3 and the leading coefficient is 2. So, the possible rational zeros can be obtained by taking the factors of -3 (±1, ±3) and dividing them by the factors of 2 (±1, ±2).
Therefore, the possible rational zeros of the polynomial f(x) = 2x^3 – 3x^2 – 8x – 3 are: ±1/2, ±1, ±3/2, ±3.
(b) To find the zeros of the polynomial, we need to solve the equation f(x) = 2x^3 – 3x^2 – 8x – 3 = 0.
One way to find the zeros is to use synthetic division or long division to divide the polynomial by one of the possible rational zeros. I will choose x = -1 as a potential zero and use synthetic division to test it:
-1 | 2 -3 -8 -3
| -2 5 3
____________________
2 -5 -3 0
The remainder is 0, which means -1 is a zero of the polynomial. The resulting quotient is 2x^2 - 5x - 3.
Now, we have a quadratic equation 2x^2 - 5x - 3 = 0 to solve. We can solve it by factoring, completing the square, or using the quadratic formula.
By factoring, we can write the equation as (2x + 1)(x - 3) = 0. Setting each factor equal to zero, we get 2x + 1 = 0 and x - 3 = 0.
Solving these equations, we find x = -1/2 and x = 3.
So, the zeros of the polynomial f(x) = 2x^3 – 3x^2 – 8x – 3 are x = -1/2, x = -1, and x = 3.