Factor (y')^2 - 4x^2 y' + 4x^4 - x^2, treating y' as the variable.

Step-by-step solution

1. 1

   Treat the derivative symbol as the variable. Written in descending powers of $y^{\prime}$, the expression is an ordinary quadratic:

   $\left(y^{\prime}\right)^2-\left(4x^2\right)y^{\prime}+\left(4x^4-x^2\right)$

   Recognising which symbol plays the role of the unknown is the first real decision here; the powers of $x$ are just coefficients.

2. 2

   Spot the perfect square hiding in the first three terms. The pieces $\left(y^{\prime}\right)^2$, $-4x^2y^{\prime}$ and $4x^4$ fit the pattern $a^2-2ab+b^2$ with $a=y^{\prime}$ and $b=2x^2$:

   $\left(y^{\prime}\right)^2-4x^2y^{\prime}+4x^4=\left(y^{\prime}-2x^2\right)^2$

   The constant term $4x^4-x^2$ was deliberately split as $4x^4$ plus a leftover $-x^2$ to make this grouping possible.

3. 3

   Rewrite the whole expression as a difference of two squares.

   $\left(y^{\prime}-2x^2\right)^2-x^2=\left(y^{\prime}-2x^2\right)^2-(x)^2$

4. 4

   Apply $A^2-B^2=(A-B)(A+B)$. With $A=y^{\prime}-2x^2$ and $B=x$:

   $\left(y^{\prime}-2x^2-x\right)\left(y^{\prime}-2x^2+x\right)$

5. 5

   Verify by expanding back. The product of the two factors is $\left(y^{\prime}-2x^2\right)^2-x^2$, which expands to $\left(y^{\prime}\right)^2-4x^2y^{\prime}+4x^4-x^2\;\checkmark$. A numeric spot check with $y^{\prime}=5$ and $x=1$: the original is $25-20+4-1=8$, and the factors give $(5-2-1)(5-2+1)=2\cdot 4=8\;\checkmark$.