<s>[INST] Find the coordinates of the points on the curve $y=2x^4 - 3x^2 + x - 7$ where the gradient is parallel to the line $y=3x$ I got that $\frac{dy}{dx}= 3$ 
(because of the gradient of the parallel line)
and then, I found the derivative of the equation of the curve to be $8x^3 -6x +1= 3$.
This simplifies as $8x^3 - 6x -2 =0$
From here, I simplified the equation and got $4x^3- 3x- 1= 0$
And I don't know how to solve this equation for $x$.
 [/INST] Gradient of  $y=3x$ is $3$.
So by differentiating, we have $\frac{dy}{dx}=8x^3-6x+1$
Now solve $8x^3-6x+1=3$.
Clearly $8x^3-6x+1=3\Rightarrow 8x^3-6x-2=0$.
$\Rightarrow 8x^3-6x-2=(x-1)(8x^2+8x+2)=0\Rightarrow x=1$.
Thus coordinates of the point $=(1,-7)$
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