How do you factor $18 b^{2} + 24 b t + 8 t^{2}$ $18b^2+24bt+8t^2$ ?

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How do you factor $18 b^{2} + 24 b t + 8 t^{2}$ $18b^2+24bt+8t^2$ ?

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Answer 1 · 2015-06-03T10:57:58

Given $18 b^{2} + 24 b t + 8 t^{2}$ $18b^2+24bt+8t^2$

If you extract the obvious common factor of $2$ $2$
$XXXX$ $color(white)("XXXX")$ $2 (9 b^{2} + 12 b t + 4 t^{2})$ $2(9b^2+12bt+4t^2)$
$XXXX$ $color(white)("XXXX")$ $= 2 ({(3 b)}^{2} + 2 (3 b) (2 t) + {(2 t)}^{2})$ $=2((3b)^2 + 2(3b)(2t) + (2t)^2)$
we end up with some obvious squares for the first and last terms and using their roots can see that together they will generate the middle term.
$XXXX$ $color(white)("XXXX")$ $2 {(3 b + 2 t)}^{2}$ $2(3b+2t)^2$

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How do you factor $18 b^{2} + 24 b t + 8 t^{2}$ $18b^2+24bt+8t^2$ ?

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How do you factor $18 b^{2} + 24 b t + 8 t^{2}$ $18b^2+24bt+8t^2$ ?