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.

1 answer

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.