<s>[INST] For f(x) = ((3x+7)^8)((4x-5)^3) , find f'(x) and use this answer to find the value(s) of x at which the graph of f(x) has a horizontal tangent line I know that $f'(x) = (24(3+7)^7)*(4x-5)^3+((3x+7)^8)*12(4x-5)^2$, but is there any easier way to find the horizontal tangent line without expanding the terms using pascal's triangle and solving for x by setting f'(x) equal to 0?
 [/INST] $$f(x)=(3x+7)^8(4x-5)^3$$
$$f'(x)=8(3x+7)^7.3(4x-5)^3+(3x+7)^83.(4x-5)^2.4$$
$$f'(x)=(3x+7)^7(4x-5)^2[24(4x-5)+12(3x+7)]$$
$$f'(x)=(3x+7)^7(4x-5)^2[132x-36]$$
$$f'(x)=12(3x+7)^7(4x-5)^2[11x-3]$$
to find the tangent point 
$$(3x+7)^7=0$$
$$(4x-5)^2=0$$
$$(11x-3)=0$$
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