How do you factor completely $a^2 - 1 - 4b^2 - 4b$ ?

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How do you factor completely $a^2 - 1 - 4b^2 - 4b$ ?

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Answer 1 · 2016-12-18T00:00:12

$a^2-1-4b^2-4b = (a-2b-1)(a+2b+1)$

Explanation:

The difference of squares identity can be written:

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

Use this with $A=a$ and $B=2b+1$ as follows:

$a^2-1-4b^2-4b = a^2-(4b^2+4b+1)$

$color(white)(a^2-1-4b^2-4b) = a^2-(2b+1)^2$

$color(white)(a^2-1-4b^2-4b) = (a-(2b+1))(a+(2b+1))$

$color(white)(a^2-1-4b^2-4b) = (a-2b-1)(a+2b+1)$

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How do you factor completely $a^2 - 1 - 4b^2 - 4b$ ?

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How do you factor completely a^2 - 1 - 4b^2 - 4ba^2 - 1 - 4b^2 - 4b?

1 Answer

Explanation:

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How do you factor completely $a^2 - 1 - 4b^2 - 4b$ ?