The number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3
The equation of the function is given as:
[tex]f(x) = x^3 + 4x^2 + x - 6[/tex]
Expand the function
[tex]f(x) = x^3 + 5x^2 - x^2 + 6x - 5x - 6[/tex]
Reorder the terms
[tex]f(x) = x^3 + 5x^2 + 6x - x^2 - 5x - 6[/tex]
Factor the expression
[tex]f(x) = x(x^2 + 5x + 6) -1(x^2 + 5x + 6)[/tex]
Factor out x -1
[tex]f(x) = (x^2 + 5x + 6)(x -1)[/tex]
Expand
[tex]f(x) = (x^2 + 3x + 2x + 6)(x -1)[/tex]
Factorize
[tex]f(x) = [x(x + 3) + 2(x + 3)](x - 1)[/tex]
Factor out x + 2
[tex]f(x) = (x + 3)(x + 2)(x- 1)[/tex]
The function has been completely factored and it has 3 linear factors
Hence, the number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3
Read more about functions at:
brainly.com/question/7784687
#SPJ1
