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anonymous
 4 years ago
can some one tell me how to get factors i mean method 3b^220b12
anonymous
 4 years ago
can some one tell me how to get factors i mean method 3b^220b12

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0aaaaa what's that ? i don't think that's the way

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this will give u two value for x..........& say it x1,x2...... now factor is (xx1)(xx2)...ok

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1330616930509:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(b6) (3b2) am i right ?

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1we have \[3b^220b12\] standard quadratic equation \[ax^2+bx+c=0\] we know its solution is given as \[x=\frac{b\pm \sqrt{b^24ac}}{2a}\] here a=3 b=20 and c=12 so \[x=\frac{(20)\pm \sqrt{(20)^24\times 3\times 12}}{2\times 3}\] we get \[x=\frac{20\pm \sqrt{400+144}}{6}\] now we get \[x=\frac{20\pm \sqrt{400+144}}{6}\] we get \[x=\frac{20\pm \sqrt{544}}{6}\] \[544= 17\times 16\times 2\] so \[x=\frac{20\pm \sqrt{16\times 34}}{6}\] we get \[x=\frac{20\pm 4\sqrt{ 34}}{6}\] let's divide by 2 \[x=\frac{10\pm \sqrt{34}}{3}\] so the two factors are \[(x(\frac{10+ \sqrt{34}}{3}))(x(\frac{10 \sqrt{34}}{3}))\]

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1Sorry it'd be \[(x(\frac{10+ 2\sqrt{34}}{3}))(x(\frac{10 2\sqrt{34}}{3}))\] I missed the 2:(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah that's too advance i guess but right factors (b6)(3b2)

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1No, these are not the factors \[ (b6)(3b2)= 3b^220b+12\] but in your question you have \[3b^220b12\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yah :) do u know how to get (b6)(3b2)
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