$(5-x)(x-2)(x-2)-4=0$
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.
On
This equation with its unique feature provides us a quick solution by factorizing 4.
Given:
\begin{align}(5-x)(x-2)(x-2)-4=0 , (5-x)(x-2)(x-2)=4 \\(5-x)(x-2)^2=4\cdot(\pm1)^2=1\cdot(\pm2)^2\end{align}
Try:
\begin{align}5-x=4,x-2=\pm1 \to x=1, y=4\end{align} \begin{align}5-x=1,x-2=\pm2 \to x=4, y=1\end{align}
Answer: \begin{align}(1,4) , (4,1)\end{align}
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 $(4,1)$.