+0

# help

0
123
3

factor k^2 - 81

Oct 31, 2019

#1
+19914
0

(k-9)(k+9)

Oct 31, 2019
#2
+308
+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
+7763
+1

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

.
Nov 2, 2019