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factor k^2 - 81

 Oct 31, 2019
 #1
avatar+22096 
0

(k-9)(k+9)

 Oct 31, 2019
 #2
avatar+321 
+2

I'm going to explain ElectricPavolv's answer, which is indeed correct. He got the answer using his knowledge of difference of squares, which states that a quadratic equation in the form of a^2-b^2 = (a+b)(a-b). In this case, we substitute k for a and 9 for b, to get a factorization fo (k-9)(k+9).

 Nov 1, 2019
 #3
avatar+7764 
+1

\(\begin{array}{rl} &k^2 - 81\\ =&k^2 \color{red} - 9k + 9k\color{black} -81\\ =&(k^2 - 9k) + (9k - 81)\\ =&\color{red}k\color{black}(k-9)\color{red}+9\color{black}(k-9)\\ =&(k - 9)(\color{red}k + 9\color{black})\\ =&(k -9)(k+9) \end{array}\)

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 Nov 2, 2019

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