# random kumon homework

1. 3x²(x-2y)²+ 3y²(x+2y)**²**

2. 4x(x+2y)**²** + 3y²(x+2y)

Are their solving processes different because of the exponent thingy (the ones in red because when i was doing my kumon i thought something was up with it and might be why i got it wrong)

but thats all i dont even know what this is i think its algebra 1 but idk.

Ruklvr

2

## 1 Answer

1. Are you sure there is no typo? I don't think it can be factored. The correct form might be $3x²(x+2y)²+ 3y²(x+2y)²$. in that case

\[3x²(x+2y)²+ 3y²(x+2y)²=3(x+2y)^2(x^2+y^2).\]

2. The second expression can be factored as follows:

\[4x(x+2y)^2+ 3y^2(x+2y)=(x+2y)[4x(x+2y)+3y^2]=(x+2y)(4x^2+8xy+3y^2)\]

\[=(x+2y)(4x^2+8xy+4y^2-y^2)=(x+2y)[(2x+2y)^2-y^2]\]

\[=(x+2y) (2x+2y -y)(2x+2y+y)\]

\[=(x+2y)(2x+y)(2x+3y.)\]

Savionf

575

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What are the questions?

1. 3x²(x-2y)²+ 3y²(x+2y)² 2. 4x(x+2y)² + 3y²(x+2y) how do you like factorize them out thats what I'm doing in kumon i think