Asked by meow meow
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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Answers
Answered by
GPT 3.5
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.
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.
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