The number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3
<h3>How to determine the number of real zeros?</h3>The equation of the function is given as:
Expand the function
Reorder the terms
Factor the expression
Factor out x -1
Expand
Factorize
Factor out x + 2
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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<u>Answer:</u>
The value of f(x) = when x=3 is 19, hence the correct option is ‘d’.
<u>Solution:</u>
A two degree polynomial equation is given in the question.
The equation is:
We need <em>to find the value of f(x) when the value of x is 3. </em>
To solve a question like this we have to substitute the given value of x in the given equation and then simplify it.
Now let's substitute the value of x in the given equation.
First we square the number 3 and then multiply it with 2 and then add 1.
This is done because of the BODMAS rule, in which multiplication is given higher preference as compared to addition.
On solving the equation we get,
F(x)=2×9+1
F(x)=18+1
F(x)=19.
Therefore the value of f(x) when x=3 is 19.
Hence the correct option is ‘d’.