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Factor completely p4 – 81
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I dont even get it, what times 81 twice then adds to p^4?
write it as (p^2)^2 - (9)^2
now apply rule for difference of 2 squares a^2 - b^2 = (a - b)(a + b)
do you follow that?
(p – 3)^2(p + 3)^2
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yeah, thats my answer than right
a = p^2 and b = 9
(p^2)^2 - (9)^2 = (p^2 - 9)(p^2 + 9) - now you apply rule to p^2 - 9
You can also brute force this if you like: |dw:1342821494997:dw|
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