Its a trinomial with the Degree 2
6x^2+74x+44
The zeros of the function f(x) = x^3 + 3x^2 + 2x are x = 0, x = -1 and x = -2
<h3>How to determine the zeros of the function?</h3>
The function is given as:
f(x) = x^3 + 3x^2 + 2x
Factor out x in the above function
f(x) = x(x^2 + 3x + 2)
Set the function to 0
x(x^2 + 3x + 2) = 0
Factorize the expression in the bracket
x(x + 1)(x + 2) = 0
Split the expression
x = 0, x + 1 = 0 and x + 2 = 0
Solve for x
x = 0, x = -1 and x = -2
Hence, the zeros of the function f(x) = x^3 + 3x^2 + 2x are x = 0, x = -1 and x = -2
Read more about zeros of function at
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Step-by-step explanation:
u just have to draw a graph nd start at -2 which goes down -2 twice then go up 4 over 1 nd u just connect the lines nd find the slope of the lines that's how I learned it
Answer:
the logarithmic solution of log₅(25) is equal to 2
Step-by-step explanation:
given,
log₅(25)
now using logarithmic identity
log ₓ (x) = 1 and
log (aˣ) = x log (a)
now,
log₅(25)
=log₅(5²)
=2 log₅(5)
=2
hence, the logarithmic solution of log₅(25) is equal to 2
