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Since 4xy2 is a factor of 20x2y4, it is automatically the GCF.

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GCF(4xy^2, 20x^2y^4) = 4xy^2.

Q: What is the greatest common factor of 4xy2 and 20x2y4?

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You can't exactly FOIL it because you don't have a problem in the form of (3x + 2)(4x - 6) Frankly, I don't understand exactly what question you are trying to solve, so I'll give you a couple of examples that might help. (-y32) times (2y - 4xy2) turns into: ((-1)(y32) times 2y) plus ((-1)(y32) times ((-1)(4xy2)) which becomes -2y33 plus 4xy34 which would be more often expressed as 4xy34 - 2y33 Perhaps you meant something more like this: (-y times 32) times (2y - 4xy2) turns into: (-32y) times (2y - 4xy2) which becomes (-y times 2y) + (-y times -4xy2) + (32 times 2y) + (32 times -4xy2) which becomes -2y2 + 4xy3 + 64y - 128xy2 which is 4xy3 -128 xy2 - 2y2 + 64y I'm sure if I'm gotten it wrong, someone will straighten it out. Also, if this doesn't help, please fix up the original question so it can be answered properly.

The GCF is 4.

Two contour lines can intersect. A perfect example is a Lagrange Multiplier which is encountered in Calculus III. We are given a function that has restraints (side conditions). An optimization engineer working for a box factory might be asked to find the maximum volume of a cardboard box given the restraint that it has a surface area of 1500 cm2 and a total edge length of 200 cm.We are seeking the extreme values of f(x,y,z) that lie on the one of the level curves (c) of g(x,y,z) and h(x,y,z). These occur at a point P(x0,y0,z0) where you can find the highest level surfaces (k) of f(x,y,z) that are intersected by the level curves (c) of g(x,y,z) and h(x,y,z). These intersections occur when they just barely touch one another. Meaning they have a common tangent line. Further, their normal lines are the same, implying that their gradient vectors ∇f, ∇g, ∇h are parallel.∇f = λ∇g + μ∇h. This works if ∇g and ∇h ≠ 0.Eq. 1 f: V=xyzEq. 2 g: 1500=2(xy)2+2(xz)2+2(yz)2Eq. 3 h: 200=√x2+y2+z2∇f =(yz,xz,xy)∇g = (4xy2+4xz2,4x2y+4yz2,4x2z+4y2z)∇h = (x/√x2+y2+z2, y/√x2+y2+z2, z/√x2+y2+z2)Eq. 4 yz= λ(4xy2+4xz2) + μ(x/√x2+y2+z2)Eq. 5 xz= λ(4x2y+4yz2) + μ(y/√x2+y2+z2)Eq. 6 xy= λ(4x2z+4y2z) + μ(z/√x2+y2+z2)We have 6 equations and 6 unknowns (x,y,z,λ,μ and V). We will have to use back substitution to solve.