7x^3-24x^2-19x+12
-----------------=f(x)
2x^3-4x^2-40x+80
Determine all values of x for which f is not defined.
thats supposed to be a fraction all equel to f(x).
A function is not defined if the denominator is zero
So we just have to look at the bottom which I will assume is
2x^3 - 4x^2 - 40x + 80
by grouping
= 2x^2(x-2) - 40(x-2)
= (x-2)(2x^2 - 40)
so x-2 cannot be zero, or x ≠ 2
or
2x^ - 40 ≠ 0
x^2 ≠ 20
x ≠ ±√20 or ±2√5
so x ≠ 2, 2√5, -2√5
To determine the values of x for which f(x) is not defined, we need to find the values of x that would make the denominator of the fraction equal to 0. In this case, the denominator is the polynomial 2x^3 - 4x^2 - 40x + 80.
To find the values of x for which the denominator is equal to 0, we can use the factoring method or synthetic division. Let's use factoring to solve it:
2x^3 - 4x^2 - 40x + 80 = 0
First, we can factor out a common factor of 2 from each term:
2(x^3 - 2x^2 - 20x + 40) = 0
Next, we can try to factor the remaining part of the polynomial:
(x^3 - 2x^2 - 20x + 40) = 0
To simplify the process, we can try substituting different integers as roots and check if the equation becomes zero. By trial and error, you can find that x = 2 is a root:
(2^3 - 2(2)^2 - 20(2) + 40) = 0
(8 - 2(4) - 40 + 40) = 0
(8 - 8 - 40 + 40) = 0
0 = 0
Since the equation becomes 0 with x = 2, we know that (x - 2) is a factor of the polynomial. We can perform polynomial long division or synthetic division to find the resulting quadratic equation.
Using polynomial long division or synthetic division, we get:
2x^3 - 4x^2 - 40x + 80 = (x - 2)(2x^2 + 0x - 40)
Now, we can solve the quadratic equation:
2x^2 + 0x - 40 = 0
To solve the quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 2, b = 0, and c = -40.
x = (0 ± √(0^2 - 4(2)(-40)))/(2(2))
x = (0 ± √(0 + 320))/(4)
x = (± √(320))/(4)
x = ± √(80)/(4)
x = ± (4√(5))/(4)
x = ± √(5)
Therefore, the values of x for which f(x) is not defined are x = +√(5) and x = -√(5).