find the real zeros of each polynomial
f(x)=5x^3-40
Please show work I want to learn how to did it
well, note that
f(x) = 5(x^3-2^3) = 5(x-2)(x^2+2x+4)
That help?
so what's the correct answer Steve?
To find the real zeros of the polynomial f(x) = 5x^3 - 40, we need to set f(x) equal to zero and solve for x.
Step 1: Set f(x) = 0
0 = 5x^3 - 40
Step 2: Solve for x
To solve this equation, we will use a method called factoring. The first step is to factor out the greatest common factor (GCF).
GCF of 5x^3 and -40:
The GCF of 5 and -40 is 5.
5(x^3 - 8) = 0
Now, we have a difference of cubes, which can be factored further using the formula: a^3 - b^3 = (a - b)(a^2 + ab + b^2).
(x - 2)(x^2 + 2x + 4) = 0
Step 3: Solve each factor separately:
Setting (x - 2) = 0:
x - 2 = 0
x = 2
Setting (x^2 + 2x + 4) = 0:
This is a quadratic equation, and it can be solved using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 1, b = 2, and c = 4.
x = (-2 ± √(2^2 - 4(1)(4))) / (2(1))
x = (-2 ± √(4 - 16)) / 2
x = (-2 ± √(-12)) / 2
Since the square root of a negative number (in this case -12) is not a real number, there are no real solutions for this quadratic factor.
Therefore, the real zeros of the polynomial f(x) = 5x^3 - 40 are x = 2.