Finding horizontal tangents to a function. 
Find the points at which the line tangent to the following function is horizontal $$q(x)=(x+3)^4(2x-1)^7$$

Every time I've gotten to the point of finding $x$ the numbers are all irrationally too large. I've gotten $2(2x-1)^6(x+3)^3(11x+19).$ But I do not see a way to solve from these numbers.
Thank you for any help.
 A: Hint, use product rule ( and chaining rule) for derivative:
$$
q'(x)=4(x+3)^3(2x-1)^7+14(2x-1)^6(x+3)^4= 2(x+3)^3(2x-1)^6(11x+19)
$$
So you have: $ q'(x)=0 \iff (x+3)^3(2x-1)^6(11x+19)=0$ that is factorized.
A: You correctly found that $q'(x) = 2(2x-1)^6(x+3)^3(11x+19).$
Now you just need to find where $q'(x) = 0.$
The formula for $q'(x)$ may look complicated but you only need some very
simple things from it. It is the product of four terms:
$$2 \times (2x-1)^6 \times (x+3)^3 \times (11x+19).$$
Then $q'(x)=0$ exactly when at least one of those terms is zero.
Obviously $2 \neq 0$ always,
so you need to find
$$\begin{align}
x_1 \quad &\mbox{such that}\quad (2x_1-1)^6 = 0,\\
x_2 \quad &\mbox{such that}\quad (x_2+3)^3 = 0, \quad \mbox{and}\\
x_3 \quad &\mbox{such that}\quad 11x_3+19 = 0.
\end{align}$$
That is, you may have up to three different values of $x$ 
that will make $q'(x)=0.$
Notice that the only way in which you need to use the exponents here
is to confirm that they are all positive; for example,
either  $2x-1 = 0$ and therefore $(2x-1)^6 = 0$,
or  $2x-1 \neq 0$ and therefore $(2x-1)^6 \neq 0$.
A: We need $q'(x)=\left((x+3)^4(2x-1)^7\right)'=0$.
$$(\ln\, q(x))'=\frac{q'(x)}{q(x)} \Rightarrow q'(x) = q(x)\cdot(\ln\, q(x))'$$
$$q'(x)=(x+3)^4(2x-1)^7\left(\frac{4}{x+3}+\frac{7\cdot 2}{2x-1}\right)=$$
$$(x+3)^3(2x-1)^6\left(4(2x-1)+14(x+3)\right)=(x+3)^3(2x-1)^6(22x+38).$$
So, there are the roots of q'(x): $-3,\frac{1}{2},-\frac{19}{11}$.
A: Use factorization to find the roots:
\begin{align}
\frac{d q}{dx}
&= 4(x+3)^3 (2x-1)^7+14 (x+3)^4 (2x-1)^6
\\
&=(x+3)^3(2 x-1)^6 (4 (2x-1)+14(x+3))
\\
&=
2(11 x+19)(x+3)^3 (2x-1)^6
=0.
\end{align}
