Rewrite 4n^2-49 together \left(2n\right)^2-7^2. The distinction of squares have the right to be factored making use of the rule: a^2-b^2=\left(a-b\right)\left(a+b\right).

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George C. Jun 3, 2015\displaystyle4n^2-49=\left(2n\right)^2-7^2=\left(2n-7\right)\left(2n+7\right) ...using the distinction of squares ...
4r2-64 Final an outcome : 4 • (r + 4) • (r - 4) step by action solution : step 1 :Equation at the end of action 1 : 22r2 - 64 action 2 : step 3 :Pulling out choose terms : 3.1 pull out choose ...
4v2-16 Final result : 4 • (v + 2) • (v - 2) action by step solution : step 1 :Equation at the end of action 1 : 22v2 - 16 step 2 : step 3 :Pulling out prefer terms : 3.1 traction out favor ...
4*(17-24) Final an outcome : 4 Reformatting the entry : changes made to her input have to not impact the solution: (1): "•" was changed by "*". Step by action solution : action 1 : Equation in ~ the end ...
64n2-9 Final result : (8n + 3) • (8n - 3) action by action solution : step 1 :Equation in ~ the finish of action 1 : 26n2 - 9 step 2 :Trying to variable as a difference of Squares : 2.1 Factoring: ...
n2-49=0 Two solutions were found : n = 7 n = -7 action by action solution : action 1 :Trying to element as a difference of Squares : 1.1 Factoring: n2-49 theory : A difference of 2 ...

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Rewrite 4n^2-49 as \left(2n\right)^2-7^2. The distinction of squares have the right to be factored using the rule: a^2-b^2=\left(a-b\right)\left(a+b\right).
\left< \beginarray l l 2 & 3 \\ 5 & 4 \endarray \right> \left< \beginarray l l together 2 & 0 & 3 \\ -1 & 1 & 5 \endarray \right>
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