Factor (1/2)x^4y^2 + (1/3)x^3y^3 + (1/4)x^2y^4 completely.

Step-by-step solution

1. 1

   Handle the variables and the coefficients as two separate jobs. For a GCF you take the lowest power of each variable that appears in every term, and separately the largest number that divides all coefficients. Doing them together is what makes fractional coefficients confusing.

2. 2

   Take the lowest power of each variable. The $x$ exponents are $4,3,2$ so the smallest is $x^2$; the $y$ exponents are $2,3,4$ so the smallest is $y^2$. The variable part of the GCF is

   $x^2y^2$

3. 3

   Choose the fractional coefficient using the LCM of the denominators. The coefficients are $\tfrac12,\tfrac13,\tfrac14$. Since $\operatorname{lcm}(2,3,4)=12$, the useful common factor is $\tfrac{1}{12}$: dividing each coefficient by $\tfrac{1}{12}$ means multiplying by $12$, and $\tfrac{12}{2}=6$, $\tfrac{12}{3}=4$, $\tfrac{12}{4}=3$ are all integers. No larger fraction would clear all three denominators, so the GCF is

   $\frac{1}{12}x^2y^2$

4. 4

   Divide each term by the GCF. Dividing by $\tfrac{1}{12}$ multiplies by $12$, and the variable exponents subtract:

   $\frac{\tfrac12 x^4y^2}{\tfrac1{12}x^2y^2}=6x^2,\qquad \frac{\tfrac13 x^3y^3}{\tfrac1{12}x^2y^2}=4xy,\qquad \frac{\tfrac14 x^2y^4}{\tfrac1{12}x^2y^2}=3y^2$

5. 5

   Write the factored form and test whether the bracket factors further.

   $\frac{1}{12}x^2y^2\left(6x^2+4xy+3y^2\right)$

   As a quadratic in $x$ the bracket has discriminant $(4y)^2-4(6)(3y^2)=16y^2-72y^2=-56y^2<0$ for $y\neq 0$, so it is irreducible over the reals and the factorisation is complete.

6. 6

   Check by expanding back. $\tfrac1{12}x^2y^2\cdot 6x^2=\tfrac12x^4y^2$, $\tfrac1{12}x^2y^2\cdot 4xy=\tfrac13x^3y^3$, $\tfrac1{12}x^2y^2\cdot 3y^2=\tfrac14x^2y^4$. All three original terms are reproduced exactly.