If f, left bracket, x, right bracket, equals, 6, x, cubed, minus, 31, x, squared, plus, 18, x, plus, 40f(x)=6x
3
−31x
2
+18x+40 and x, minus, 4x−4 is a factor of f, left bracket, x, right bracketf(x), then find all of the zeros of f, left bracket, x, right bracketf(x) algebraically.
To find the zeros of f(x), we need to find the values of x for which f(x) equals zero.
Since x - 4 is a factor of f(x), we can set f(x) equal to zero and solve for x:
6x^3 - 31x^2 + 18x + 40 = 0
We can use synthetic division to find the remaining zeros of f(x).
Dividing by x - 4, we get:
4 | 6 -31 18 40
24 -28 -40
_________________
30 -10 0
This tells us that x - 4 is a factor of f(x), and the reduced equation is:
6x^2 - 10x + 0 = 0
Factoring out an x, we have:
x(6x - 10) = 0
Setting each factor equal to zero, we get:
x = 0 or 6x - 10 = 0
For the second equation, solving for x gives:
6x = 10
x = 10/6 = 5/3
So, the zeros of f(x) are x = 0, x = 5/3, and x = 4.