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01 Feb 2020, 11:11
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Factor the following quadratic expressions: 4x^2+8xy+4y^2.
Please, I need to understand why one of the following answers is wrong.
1. 4x^2+8xy+4y^2
= 4x^2+4xy+4xy+4y^2
= 4x(x+y)+4y(x+y)
so possible solutions are (4x+4y)(x+y)
2. 4x^2+8xy+4y^2
=(2x+2y)(2x+2y) ------ possible solution: (2x+2y)^2
---------------------------------------------
MY QUESTION: why is solution 2 correct and solution 1 incorrect?
Please, I need to understand why one of the following answers is wrong.
1. 4x^2+8xy+4y^2
= 4x^2+4xy+4xy+4y^2
= 4x(x+y)+4y(x+y)
so possible solutions are (4x+4y)(x+y)
2. 4x^2+8xy+4y^2
=(2x+2y)(2x+2y) ------ possible solution: (2x+2y)^2
---------------------------------------------
MY QUESTION: why is solution 2 correct and solution 1 incorrect?
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Additi
Joined: 29 May 2019
Last visit: 22 Jul 2021
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01 Feb 2020, 11:21
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Soln 2 is correct PFB the solution
4x^2 + 4y^2 + 8xy
This is (a+b)^2 = a^2+ b^2 +2ab
Replace a = 2x and b = 2y
(2x + 2y)^2 = 4x^2 + 4y^2 + 8 xy
Posted from my mobile device
4x^2 + 4y^2 + 8xy
This is (a+b)^2 = a^2+ b^2 +2ab
Replace a = 2x and b = 2y
(2x + 2y)^2 = 4x^2 + 4y^2 + 8 xy
Posted from my mobile device
01 Feb 2020, 12:11
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phuntie
Hello
Both are correct!
(4x+4y)(x+y) ... your first solution
= 4 * (x+y) * (x+y)
= 2 * 2 * (x+y) * (x+y)
= 2(x+y) * 2(x+y)
= (2x+2y)(2x+2y) ... your other solution
