We must find the zeros of the following function:
[tex]f(x)=4\cdot(x+3)^2\cdot(x-3)\cdot(x^2-49).[/tex]
Now, the zeros of a function are the values of x when f(x) is equal to 0. So we must find all the values of x that makes f(x) = 0. To do that, it is convenient to rewrite the function f(x) as:
[tex]f(x)=4\cdot(x+3)\cdot(x+3)\cdot(x-3)\cdot(x+7)\cdot(x-7)\text{.}[/tex]
Where we use the fact that:
• (x + 3)² = (x + 3) (x + 3),
,
• (x² + 49) = (x + 7) (x - 7).
From the previous equation, we see that if we replace x with the following values, we get f(x) = 0:
• x = -3,
,
• x = 3,
,
• x = -7,
,
• x = 7.
Answer
All the real zeros of the function are -3, 3, -7, 7.
