Factoring a cubic equation but I want a way to figure out the factor of the function $(5-x)(x-2)(x-2)-4=0$
I want to expand this, subtract the four from the constant then factor this again. How do I find the linear factor? Are there easier ways to do this?
If it helps, the original question is that we have two equations $y=4/(x-2)^2$ and $y=5-x.$ One intersection is at point $(1,4)$, so I have to find the coordinates of the other points of intersection.
Because we are searching were $y=4/(x−2)^2$ and $y=5−x$ intersects, we have the equation: $5-x- \dfrac{4}{(x−2)^2} =0$, which is equal to $\dfrac{(5-x)(x-2)^2 -4}{(x-2)^2} =0$ expanding this we get that this is also equal to $\dfrac{-(x^3-9x^2+24x-16)}{(x-2)^2}$ as we can see, x cannot be equal to 2, and also, this fraction only will be 0 if the numerator is zero, so, we are searching where $x^3-9x^2+24x-16=0$, as we know, 1 is a solution, so we can factor this as $(x-1)(ax^2+bx+c)$. As we can obtain, $a = 1$, $b = -8$ and $c= 16$ so, finally, we have the equation $(x-1)(x^2-8x+16)$. Now, looking at the roots of the quadratic part, we see that 4 is the only solution, so $y=4/(x−2)^2$ and $y=5−x$ intersects at $(1,4)$ and $(4,1)$.
Expand $(5-x)(x-2)(x-2)-4=0$ and use the fact that $x=1$ is solution.
$$\begin{aligned}(5-x)(x-2)(x-2)-4=&amp;-x^3+9x^2-24x+16\\=&amp;-(x-1)(x^2-8x+16)\\=&amp;-(x-1)(x-4)^2\end{aligned}$$
The second point has coordinates $(4,1)$ and it is a point of tangency due to the square (it is a double point of contact).