The number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3
How to determine the number of real zeros?
The equation of the function is given as:
f(x) = x^3 + 4x^2 + x - 6
Expand the function
f(x) = x^3 + 5x^2 - x^2 + 6x - 5x - 6
Reorder the terms of the function
f(x) = x^3 + 5x^2 + 6x - x^2 - 5x - 6
Factor the expression
f(x) = x(x^2 + 5x + 6) -1(x^2 + 5x + 6)
Factor out x -1
f(x) = (x^2 + 5x + 6)(x -1)
Expand
f(x) = (x^2 + 3x + 2x + 6)(x -1)
Factorize
f(x) = [x(x + 3) + 2(x + 3)](x - 1)
Factor out x + 2
f(x) = (x + 3)(x + 2)(x- 1)
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
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