Factor these (polynomials) please
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Factor these (polynomials) please

[From: ] [author: ] [Date: 13-08-18] [Hit: ]
......
like in rewrite them like (3x+4)(1x-1) or whatever, thank YOU :D
36x^2 + 12x + 1
x^2 +12x +36
36x^2 - 1
x^2 - 36

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(6x+1)
(x+6)^2
(6x+1)(6x-1)
(x+6)(x-6)

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NOTE: The expressions below are perfect squares.

36x^2 + 12x + 1 = (36/36)x^2 + (12/36)x + 1/36 =

x^2 + (1/3)x + 1/36 = (x + 1/6)(x + 1/6)

x^2 + 12x + 36 = (x + 6)(x + 6)

NOTE: The expressions are the difference between two squares.

36x^2 - 1 = (6x - 1)(6x + 1)

x^2 - 36 = (x - 6)(x + 6)

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36x^2 + 12x + 1 => using (a + b)^2 = a^2 + 2ab + b^2:
= (6x + 1)^2

x^2 + 12x + 36
= (x + 6)^2

36x^2 - 1 => difference of squares:
= (6x - 1)(6x + 1)

x^2 - 36
= (x - 6)(x + 6)

-
So for part 1 (ax+b)(cx+d) = ac x^2 + (bc+ad)x + cd. What can c and d be to multiply to give 1?
1
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