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anonymous

  • one year ago

How do i factor 8x^2y^3+4x^2y^2-2xy?

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  1. butterflydreamer
    • one year ago
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    You want to find the HCF :) So you can break this down by just looking at the numbers first: 8 , 4 and 2 What would the HCF of 2,4 and 8 be?

  2. butterflydreamer
    • one year ago
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    hint: Factors of 2 = 1,2 Factors of 4 = 1, 2, 4 Factors of 8 = 1, 2 , 4 , 8

  3. anonymous
    • one year ago
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    2

  4. butterflydreamer
    • one year ago
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    yess. So factorise out 2 and we get: \[8x^2y^3+4x^2y^2-2xy\] \[= 2 (4x^2 y^3 - 2x^2y^2 -xy) \] Now look at the x's :) We have x, x^2 and x^2 So what would be the HCF of those? Note: factors of x = 1, x Factors of x^2 = 1, x , x^2

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