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sexy1993
Group Title
Find a pair of factors for each number by using the difference of two squares.
a. 45
b. 77
c. 112
 2 years ago
 2 years ago
sexy1993 Group Title
Find a pair of factors for each number by using the difference of two squares. a. 45 b. 77 c. 112
 2 years ago
 2 years ago

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KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
Let's start with a. Using difference of two squares, note that \(45=494=7^22^2\)This last expression can be factored easily as \[7^22^2=(72)\cdot(7+2)\]Hence, your factorization is \(9\cdot 5\)
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
For b, we want to write\[77=814=9^22^2\]Can you factorize from there?
 2 years ago

sexy1993 Group TitleBest ResponseYou've already chosen the best response.0
trying to figure it out
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
There's an easy way to factorize a difference of two squares. If you have a differnece of two squares \[a^2b^2\]then this factors into\[a^2b^2=(ab)\cdot(a+b)\]Does this help?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
So you have an equation of the form \[a^2b^2\]where \(a=9\) and \(b=2\). Thus, \[a^2b^2=(ab)\cdot(a+b)\]can be replaced by \[9^22^2=(92)\cdot(9+2)\]Which then simplifies to \[7\cdot11\]
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
Let's move on to c. We want to factor 112. Notice that \[112=1219=11^23^2\]Using the same principles as above, can you try to factor this one?
 2 years ago

sexy1993 Group TitleBest ResponseYou've already chosen the best response.0
i do not understand how u did this
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
Sorry, I had to go get dinner. What part of this do you now understand?
 2 years ago
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