Enter your answer and show all the steps that you use to solve this problem in the space provided. Find all the zeroes of the polynomial function f(x)=x3−5x2+6x−30. If you use synthetic division, show all three lines of numbers.
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To find the zeroes of the polynomial function f(x) = x^3 - 5x^2 + 6x - 30, we can use synthetic division as well as factoring.
To begin with synthetic division, we set up the division as follows:
5 | 1 -5 6 -30
The first step is to bring down the coefficient of the highest degree term, which is 1 in this case.
5 | 1 -5 6 -30
| 5
Next, we multiply the divisor (5) by the number at the bottom (5) and write the result above the next coefficient.
5 | 1 -5 6 -30
-25 | 5
We then add the values vertically:
5 | 1 -5 6 -30
-25 | 5
______________
1 -20 -19 -5
The numbers on the bottom row represent the coefficients of the resulting polynomial, which is 1x² - 20x - 19, with a remainder of -5.
Now, we have a quadratic equation. To find the remaining zeroes, we need to factor the quadratic equation or solve it using the quadratic formula.
Factorizing this equation, we find:
(x - 5)(x + 1)
Setting each factor equal to zero, we have:
x - 5 = 0 or x + 1 = 0
Solving for x in each equation, we find:
x = 5 or x = -1
Therefore, the zeroes of the polynomial function f(x) = x^3 - 5x^2 + 6x - 30 are x = 5, x = -1.
To find the zeroes of the polynomial function f(x) = x^3 - 5x^2 + 6x - 30, we can use synthetic division.
Let's start by finding the possible rational zeroes. They can be determined by taking the factors of the constant term (-30) divided by the factors of the leading coefficient (1).
The factors of -30: ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30
The factors of 1: ±1
So, the possible rational zeroes are: ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30
Now, let's try these values using synthetic division to check if any of them are zeroes:
1 | 1 -5 6 -30
1 -4 2 -28
----------------------
1 -4 2 -28
The remainder is -28, which means x = 1 is not a zero.
-1 | 1 -5 6 -30
-1 6 -12 6
--------------------
1 -6 -6 -24
The remainder is -24, so x = -1 is not a zero either.
2 | 1 -5 6 -30
2 -6 4 -52
--------------------
1 -3 10 -22
The remainder is -22, indicating that x = 2 is not a zero.
-2 | 1 -5 6 -30
-2 14 -20 28
--------------------
1 9 -14 -2
The remainder is -2, so x = -2 is not a zero.
3 | 1 -5 6 -30
3 -6 3 -81
--------------------
1 -2 9 -51
The remainder is -51, which means x = 3 is not a zero either.
-3 | 1 -5 6 -30
-3 24 -78 168
--------------------
1 -8 -72 138
The remainder is 138, indicating that x = -3 is not a zero.
5 | 1 -5 6 -30
5 0 30 180
--------------------
1 0 36 150
The remainder is 150, meaning x = 5 is not a zero.
-5 | 1 -5 6 -30
-5 50 -280 1350
---------------------
1 -10 -274 1320
The remainder is 1320, so x = -5 is also not a zero.
Thus, there are no rational zeroes for the polynomial function f(x) = x^3 - 5x^2 + 6x - 30.
To find the zeroes of a polynomial function, we need to find the values of x for which f(x) is equal to zero. In this case, we are given the polynomial function f(x) = x^3 - 5x^2 + 6x - 30.
One way to find the zeroes is to use synthetic division. To perform synthetic division, we need to choose a test zero value. We can then use the remainder theorem to determine if the test zero is a factor of the polynomial.
Let's start by testing a few values for x to find a test zero:
When x = 1:
f(1) = (1)^3 - 5(1)^2 + 6(1) - 30
= 1 - 5 + 6 - 30
= -28
When x = -1:
f(-1) = (-1)^3 - 5(-1)^2 + 6(-1) - 30
= -1 - 5 - 6 - 30
= -42
When x = 2:
f(2) = (2)^3 - 5(2)^2 + 6(2) - 30
= 8 - 20 + 12 - 30
= -30
When x = -2:
f(-2) = (-2)^3 - 5(-2)^2 + 6(-2) - 30
= -8 - 20 - 12 - 30
= -70
Based on these test values, we can see that f(2) = -30, which means x = 2 is a zero of the polynomial. Next, we will perform synthetic division using x = 2 as the test zero:
2 | 1 -5 6 -30
| 2 -6 24
|____________________
1 -3 0 -6
The result of this synthetic division is a polynomial of degree 2: 1x^2 - 3x + 0. Now, we can proceed to solve this second-degree equation.
Setting the quadratic equation equal to zero:
1x^2 - 3x + 0 = 0
Factoring out an x:
x(x - 3) = 0
Setting each factor equal to zero:
x = 0, x - 3 = 0
Solving for x in each equation:
x = 0, x = 3
Therefore, the zeros of the polynomial function f(x) = x^3 - 5x^2 + 6x - 30 are: x = 2, x = 0, and x = 3.