Thread: I'm having a hard time factoring this

1. I'm having a hard time factoring this

Hi everyone, I'm trying the find the zeros of the derivative 4x^3 + 9x^2 + 6x +1 so I set it to zero but I am stuck as to how I can factor this. btw the original function is x*(x+1)^2

I've asked a tutor at my school's tutoring center and she said that I should use my calc to find the zeros, which visually look like 1 and ~-2.5.

Any help would be much appreciated!

2. Originally Posted by robo_robb
Hi everyone, I'm trying the find the zeros of the derivative 4x^3 + 9x^2 + 6x +1 so I set it to zero but I am stuck as to how I can factor this.
Rational Root Theorem

Synthetic Division

3. That certainly rings a bell. Thank you very much!

4. Originally Posted by robo_robb
Hi everyone, I'm trying the find the zeros of the derivative 4x^3 + 9x^2 + 6x +1 so I set it to zero but I am stuck as to how I can factor this. btw the original function is x*(x+1)^2

I've asked a tutor at my school's tutoring center and she said that I should use my calc to find the zeros, which visually look like 1 and ~-2.5.

Any help would be much appreciated!
$\displaystyle y = x (x + 1)^2 = x^3 + 2x^2 + x \Rightarrow \frac{dy}{dx} = 3x^2 + 4x + 1$. There should be no trouble solving $\displaystyle \frac{dy}{dx} = 3x^2 + 4x + 1 = 0$.

I don't see where $\displaystyle y = 4x^3 + 9x^2 + 6x + 1$ fits in .... is this another function for which you have to find the zeros of its derivative? Again, the derivative is a quadratic. and there is no trouble.

Or are you trying to solve $\displaystyle 0 = 4x^3 + 9x^2 + 6x + 1$ in which case you note that by inspection $\displaystyle x = -1$ is a solution. Therefore $\displaystyle x + 1$ is a factor. Divide the factor into the cubic to get $\displaystyle (x + 1)(4x^2 + 5x + 1) = 0$. Now solve $\displaystyle 4x^2 + 5x + 1 = 0$.

5. Oh I'm sorry, I meant to say x*(x+1)^3

All of the methods posted thus far are a great help. Thanks everyone.

My roots turned out to be -1, -(1/4)