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Find a pair of factors for each number by using the difference of two squares.
a. 45
b. 77
c. 112
 one year ago
 one year ago
Find a pair of factors for each number by using the difference of two squares. a. 45 b. 77 c. 112
 one year ago
 one year ago

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KingGeorgeBest 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 \(9\cdot 5\)
 one year ago

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

sexy1993Best ResponseYou've already chosen the best response.0
trying to figure it out
 one year ago

KingGeorgeBest 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?
 one year ago

KingGeorgeBest 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 \[7\cdot11\]
 one year ago

KingGeorgeBest 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?
 one year ago

sexy1993Best ResponseYou've already chosen the best response.0
i do not understand how u did this
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
Sorry, I had to go get dinner. What part of this do you now understand?
 one year ago
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