Consider the polynomial f(x) = 3x3 – 2x2 – 7x – 2.
(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.
(a) The Rational Zero Theorem states that if a polynomial f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ has rational zeros, they will be of the form p/q, where p is a factor of the constant term a₀ and q is a factor of the leading coefficient aₙ.
In this case, the constant term is -2 and the leading coefficient is 3. The factors of -2 are ±1 and ±2, and the factors of 3 are ±1 and ±3.
So, the possible rational zeros of the given polynomial are:
±1/1, ±2/1, ±1/3, and ±2/3.
(b) To find the zeros of the polynomial, we can use the Rational Zero Theorem as a starting point and then use synthetic division or factoring to find the actual zeros.
Let's try some of the possible rational zeros from part (a) and use synthetic division to check if they are indeed zeros.
We start with p = 1 and q = 1:
1 | 3 -2 -7 -2
| 3 1 -6
--------------
3 -1 -6 -8
Since the remainder is not 0, 1 is not a zero.
Let's try another possible rational zero:
-1 | 3 -2 -7 -2
| -3 5 2
---------------
0 3 -5
The remainder is 0, so -1 is a zero of the polynomial.
Now that we found one zero, we can use long division or synthetic division again to find the remaining zeros.
Dividing the polynomial by (x + 1):
(x + 1)(3x² + x - 6) = 0
Now we need to solve the quadratic equation:
3x² + x - 6 = 0
This quadratic equation can be factored as:
(3x + 6)(x - 1) = 0
Setting each factor equal to zero:
3x + 6 = 0 --> 3x = -6 --> x = -6/3 --> x = -2
x - 1 = 0 --> x = 1
So, the zeros of the polynomial f(x) = 3x³ - 2x² - 7x - 2 are:
x = -1, x = -2, and x = 1.
To find the possible rational zeros of a polynomial, we can use the Rational Zero Theorem. According to the theorem, if a polynomial has a rational zero, it must be of the form p/q, where p is a factor of the constant term (in this case, -2) and q is a factor of the leading coefficient (in this case, 3).
(a) Using the Rational Zero Theorem, the possible rational zeros of the polynomial f(x) = 3x^3 – 2x^2 – 7x – 2 are the ratios of the factors of -2 (constant term) and 3 (leading coefficient).
The factors of -2 are: ±1, ±2
The factors of 3 are: ±1, ±3
So, the possible rational zeros are:
±(1/1), ±(1/3), ±(2/1), ±(2/3)
Simplifying these rational numbers gives us:
±1, ±1/3, ±2, ±2/3
(b) To find all the zeros of the polynomial, we can use these possible rational zeros and perform synthetic division or check them by substituting in the polynomial equation.
Let's try the possible zeros one by one. Start with the first possible zero, which is 1:
When we substitute x = 1 into the polynomial f(x) = 3x^3 – 2x^2 – 7x – 2 and perform the calculations, we get:
f(1) = 3(1)^3 – 2(1)^2 – 7(1) – 2
= 3 - 2 - 7 - 2
= -8
Since f(1) is not equal to zero, 1 is not a zero of the polynomial.
We repeat this process for all the possible zeros until we either find the zeros or exhaust all possibilities.
After checking all the possible rational zeros, we find that the polynomial does not have any rational zeros.
To find the remaining zeros, we can use numerical methods or graph the polynomial to estimate the locations of the zeros. In this case, the remaining zeros are irrational and need to be approximated using numerical methods like Newton's method or a graphing calculator.
Please use a ^ before exponents.
f(x) = 3x^3 -2x^2 -7x -2
The rational zeros are the rational values of x for which f(x) = 0.
Possible (but not necessarily valid)roots are x = +or- 1/3, 2/3, 1, or 2. One of those values that works is x = -1. That means one of the factors of the polynomial is (x+1).
The other factor is
(3x^3 -2x^2 -7x -2)/(x+1)
Synthetic long division will show that factor to be
3x^2 -5x -2 = 0
Use the quadratic formula to solve for the other two roots.
x = (1/6)[5 +/- sqrt(25 +24)]
= (1/6)(12 or -2)
= 2 or -1/3