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anonymous

  • 5 years ago

(4X)^2 - (4X)^3 A: X^-1 B:12X^-1 C:16x^2 - 4X^3 D:16x^2 - 64X^3

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  1. anonymous
    • 5 years ago
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    \[(4x)^2 - (4x)^3 = (4x^2)(1-4x) = 16x^2(1-4x) = 16x^2 - (16\times 4)x^3 = 16x^2 - 64x^3\]

  2. anonymous
    • 5 years ago
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    sorry there is a little issue\[ (4x)^2−(4x)^3=(4x)^2(1−4x)\] the rest of what i wrote is right

  3. anonymous
    • 5 years ago
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    Ha Can you explain that a little bit like how you got the one and started simplifying ? im kind of confused

  4. anonymous
    • 5 years ago
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    \[(4x)^2 - (4x)^3 = (4x)^2(1-4x)\] because (4x)^2 is common to both (4x)^2 and -(4x)^3, since \[1\times (4x)^2 = (4x)^2,\] and \[-(4x)\times (4x)^2 = (4x)^3.\] \[(4x)^2(1-4x) = 16x^2(1-4x)\] because (4x)^2 = (4^2)(x^2) = 16x^2. Now just multiply out.

  5. anonymous
    • 5 years ago
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    \[\left( 4x ^{2} \right)=4^{2}x ^{2}=16x ^{2} \] in the same way \[\left( 4x ^{3} \right)=4^{3}x ^{3}=64x ^{3}\] substitute both of them u will get\[ 16x ^{2}-64x ^{3}\] 16x square -64x cube

  6. anonymous
    • 5 years ago
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    That makes sense guys thanks :)

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