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sexy1993
Find a pair of factors for each number by using the difference of two squares. a. 45 b. 77 c. 112
Let's start with a. Using difference of two squares, note that \(45=49-4=7^2-2^2\)This last expression can be factored easily as \[7^2-2^2=(7-2)\cdot(7+2)\]Hence, your factorization is \(9\cdot 5\)
For b, we want to write\[77=81-4=9^2-2^2\]Can you factorize from there?
trying to figure it out
There's an easy way to factorize a difference of two squares. If you have a differnece of two squares \[a^2-b^2\]then this factors into\[a^2-b^2=(a-b)\cdot(a+b)\]Does this help?
So you have an equation of the form \[a^2-b^2\]where \(a=9\) and \(b=2\). Thus, \[a^2-b^2=(a-b)\cdot(a+b)\]can be replaced by \[9^2-2^2=(9-2)\cdot(9+2)\]Which then simplifies to \[7\cdot11\]
Let's move on to c. We want to factor 112. Notice that \[112=121-9=11^2-3^2\]Using the same principles as above, can you try to factor this one?
i do not understand how u did this
Sorry, I had to go get dinner. What part of this do you now understand?