Answer:
[tex]\left(x-5\right)^6\left(x+3\right)^3\left(11x+1\right)[/tex]
and
[tex]x=5,\:x=-3,\:x=-\frac{1}{11}[/tex]
Step-by-step explanation:
This problem is slightly more difficult.
We start off with the product rule. If y = v*u then in the case:
[tex]u=\left(x+3\right)^4,\:v=\left(x-5\right)^7[/tex]
[tex]=> \frac{d}{dx}\left(\left(x+3\right)^4\right)\left(x-5\right)^7+\frac{d}{dx}\left(\left(x-5\right)^7\right)\left(x+3\right)^4[/tex]
Now we apply the chain rule:
[tex]\frac{d}{dx}\left(\left(x+3\right)^4\right) = 4\left(x+3\right)^3\frac{d}{dx}\left(x+3\right)\\= 4\left(x+3\right)^3\cdot \:1\\= 4\left(x+3\right)^3[/tex]
We can apply the chain rule again for this second bit:
[tex]\frac{d}{dx}\left(\left(x-5\right)^7\right) = 7\left(x-5\right)^6\frac{d}{dx}\left(x-5\right)\\= 7\left(x-5\right)^6\cdot \:1\\= 7\left(x-5\right)^6[/tex]
So now we are left with this long equation:
[tex]4\left(x+3\right)^3\left(x-5\right)^7+7\left(x-5\right)^6\left(x+3\right)^4[/tex]
Which we can actually factor:
[tex]4\left(x+3\right)^3\left(x-5\right)^7+7\left(x-5\right)^6\left(x+3\right)^4 = \left(x-5\right)^6\left(x+3\right)^3\left(4\left(x-5\right)+7\left(x+3\right)\right) \\= \left(x-5\right)^6\left(x+3\right)^3\left(11x+1\right)[/tex]
...And for the second bit, if we equate to 0, we can apply the Zero Product Property to get all the x values:
[tex]\left(x-5\right)^6\left(x+3\right)^3\left(11x+1\right)=0,\\x=5,\:x=-3,\:x=-\frac{1}{11}[/tex]
