How do you factor completely $a^{2} - 121 b^{2}$ $a^2 - 121b^2$ ?

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Answer 1 · 2018-06-13T06:07:21

$a^{2} - 121 b^{2} = (a - 11 b) (a + 11 b)$ $a^2-121b^2 = (a-11b)(a+11b)$

The difference of sqares identity can be written:

$A^{2} - B^{2} = (A - B) (A + B)$ $A^2-B^2 = (A-B)(A+B)$

Given:

$a^{2} - 121 b^{2}$ $a^2-121b^2$

Note that both $a^{2}$ $a^2$ and $121 b^{2} = {(11 b)}^{2}$ $121b^2 = (11b)^2$ are perfect sqares.

So we can use the difference of squares identity with $A = a$ $A=a$ and $B = 11 b$ $B=11b$ to find:

$a^{2} - 121 b^{2} = a^{2} - {(11 b)}^{2} = (a - 11 b) (a + 11 b)$ $a^2-121b^2 = a^2-(11b)^2 = (a-11b)(a+11b)$

Answer 2 · 2018-06-13T06:08:36

Shown below...

Use difference of two squares:

$(A^{2} - B^{2}) = (A + B) (A - B)$ $(A^2-B^2) = (A+B)(A-B)$

$\Rightarrow ({(a)}^{2} - {(11 b)}^{2})$ $=> ( (a)^2 - (11b)^2 )$

$\Rightarrow (a + 11 b) (a - 11 b)$ $=> (a+11b)(a-11b)$

How do you factor completely a2−121b2a^2 - 121b^2?