Compute p '(x) and use it to find all values of x for which p '(x) = 0. (could you correct my logic please?)

Question

Let p(x) = x^3 + 10x^2 − 7x + 58. Compute p '(x) and use it to find all values of x for which p '(x) = 0.

 

So I took the derivative which would be

p'(x)=3x^2  +20x-7

 

Then I tried to factor it to find where p'(x)=zero using the diamond method and I got -1/3 and 7 but it says I got it wrong. Could you please explain where the error is in my thought process? please?

Answer 1

Your derivative is correct: 3x² + 20x -7

 

The problem is asking you to solve for x when this is zero:

 

3x² + 20x -7 = 0

 

This can be factored into the following form:

(3x + a)(x + b) = 0

 

We know that a * b = -7, and since the only factors of -7 are 7 and 1, then a and b MUST be -7 and 1 or 1 and -7.

 

We know that 3b + a = 20. Since this number is positive, we know that b must be 7 and a must be -1.

 

Therefore, we have this:

(3x - 1)(x + 7) = 0

 

Therefore 3x-1 = 0, 3x = 1, x = 1/3

AND x + 7 = 0, x = -7

 

Therefore the derivative equals zero for x = 1/3 and x = -7

Answer 2

3x² + 20x - 7 = 0

 

 

Factor using FOIL.  Where did you get diamond method from?  I never heard of this method.

 

(3x - 1)(x + 7) = 0

 

 

Set the factors equal to zero and solve for x to find your critical points.