How do you factor $a^{2} + 4 a b - 3 b^{2}$ $a^2+4ab-3b^2$ ?

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Answer 1 · 2015-06-28T17:04:25

Complete the square to find:

$a^{2} + 4 a b - 3 b^{2} = (a + (2 - \sqrt{7}) b) (a + (2 + \sqrt{7}) b)$ $a^2+4ab-3b^2 = (a+(2-sqrt(7))b)(a+(2+sqrt(7))b)$

${(a + 2 b)}^{2} = a^{2} + 4 a b + 4 b^{2}$ $(a+2b)^2 = a^2+4ab+4b^2$

So

$a^{2} + 4 a b - 3 b^{2}$ $a^2 + 4ab - 3b^2$

$= {(a + 2 b)}^{2} - 4 b^{2} - 3 b^{2}$ $= (a+2b)^2 - 4b^2 - 3b^2$

$= {(a + 2 b)}^{2} - 7 b^{2}$ $= (a+2b)^2 - 7b^2$

$= {(a + 2 b)}^{2} - {(\sqrt{7} b)}^{2}$ $= (a+2b)^2 - (sqrt(7)b)^2$

$= ((a + 2 b) - \sqrt{7} b) ((a + 2 b) + \sqrt{7} b)$ $= ((a+2b)-sqrt(7)b)((a+2b)+sqrt(7)b)$

$= (a + (2 - \sqrt{7}) b) (a + (2 + \sqrt{7}) b)$ $= (a+(2-sqrt(7))b)(a+(2+sqrt(7))b)$

...using the difference of squares identity

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

with $A = a + 2 b$ $A=a+2b$ and $B = \sqrt{7} b$ $B=sqrt(7)b$

How do you factor a2+4ab−3b2a^2+4ab-3b^2 ?