# Number of real roots of the following polynomial

I have a cubic polynomial:

$f(x)=x^3+3x^2+3x+7$

I wanted to obtain number of real roots of the provided polynomial.

What i did: I took the derivative of the function which turned out to be $3x^2+6x+3$ which is a perfect square after taking $3$ common. Hence, the function is strictly increasing. Also, it has two negative roots. Now, it's function can have at the most three roots using this monotonicity concept. Since, the polynomial is of order three, it can have at the most three roots. Using the rule of sign, we can also say that this polynomial will have either three negative roots or one negative roots. The three negative root situation can be ignored using the monotonicity. Only plausible situation is one negative roots. Hence, this polynomial will have one real root.

I think the number of steps that I took can be reduced. I am also using some redundant things please guide me. Are my steps correct to obtain the number of roots. If not, please add some point that I can think of while obtaining number of roots of a polynomial or of any other function whatsoever. Thanks in advance.

• If you can deduce that the function is monotonic, then there is at most one real root immediately.
– xbh
Jan 5 '18 at 2:27
• You say "the function is strictly increasing" and you say that "it has two negative roots" but these two statements are incompatible.
– lulu
Jan 5 '18 at 2:32
• @lulu He means that $f'$ has two negative roots (both $x=-1$), Jan 5 '18 at 2:33
• @ThePhenotype The derivative $f'$ is not strictly increasing. Maybe when he wrote "Hence, the function is strictly increasing. Also, it has two negative roots" the word "it" does not refer to "the function" but that's seriously confusing.
– lulu
Jan 5 '18 at 2:41
• @lulu Yes you're right about the referencing being confusing. I just eliminated the wrong interpretations myself to find out what he exactly meant, fortunately ending with a unique interpretation. Jan 5 '18 at 2:56

$$(x+1)^3 = -6$$