Problem -

Find all values of x=c so that the tangent line to the graph of

f(x) = 4x^3(14x^2 +7x-21)^2 at (c,f(c)) will be horizontal.

the the solution the book gives me is 4x^3(14x^2+7x-21)(28x+7) + (12x^2)(14x^2+7x-21)^2 - this part i know how to get to. then 196x^2(x-1)(14x-9)(3+2x)(x+1) - which i don't know how they got to.

can someone please help explain how the book get the last part of the solution? thank you!

Originally Posted by arcticreaver
Problem -
f(x) = 4x^3(14x^2 +7x-21)^2 at (c,f(c)) will be horizontal.
...
the the solution the book gives me is 4x^3(14x^2+7x-21)(28x+7) + (12x^2)(14x^2+7x-21)^2 - this part i know how to get to. then 196x^2(x-1)(14x-9)(3+2x)(x+1) - which i don't know how they got to.
The 1st term of what you quoted as the book's derivative is missing a factor of 2 from the differentiation of (14x^2 +7x-21)^2. Perhaps they corrected that error in a later step?

$\frac{d}{dx}\left(4x^3(14x^2 +7x-21)^2\right) =$

$\left(4x^3\right) \left( 2(14x^2+7x-21)^1(28x+7) \right) + \left( 12x^2 \right)(14x^2+7x-21)^2 \right)$.

By the way, if you intend on working that out, I'd strongly suggest getting those ugly factors of 7 factored out before you even begin:

$4x^3(14x^2 +7x-21)^2$

$= 4x^3 \left[ (7) (2x^2+x-3) \right]^2 = 4x^3 (7)^2 (2x^2+x-3)^2 = cx^3(2x^2+x-3)^2$, where c = (4)(49) = 196.

One more hint - when computing a derivative of a polynomial that you'll eventually want to factor, don't expand the polynomial. Keep it together.

What I'm saying here is do not try to expand $(14x^2 +7x-21)^2$ into a 5 term 4th degree polynomial in x. Keep it factored.

One final hint: You can factor $2x^2+x-3$