Answer: \(x=1,\ -\frac32,\ \frac43\).

Answer: \(x=1,\ -\frac32,\ \frac43\).

At the points of intersection, the two expressions for \(y\) are equal:

\(7x^3-7x^2-17x-4=x^3-2x^2-4x-16.\)

Rearrange:

\(6x^3-5x^2-13x+12=0.\)

Testing simple roots, \(x=1\) is a root, so \((x-1)\) is a factor. Dividing gives

\(6x^3-5x^2-13x+12=(x-1)(6x^2+x-12).\)

Now factorise the quadratic:

\(6x^2+x-12=(2x+3)(3x-4).\)

Therefore

\((x-1)(2x+3)(3x-4)=0.\)

Hence

\(x=1, \quad x=-\frac32, \quad x=\frac43.\)