Question 725354
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Ok, so far, so good.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x^2\ -\ 4x\ -\ 12x\ +\ 6\ =\ 0]


Group into two binomials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(8x^2\ -\ 4x\right)\ +\ \left(-12x\ +\ 6\right)\ =\ 0]


Go back and review that last step until you completely understand the reasons that the signs are the way they are, especially for the second binomial grouping.


Factor the GCF from each of the groups (4x from the first group, -6 from the second):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\left(2x\ -\ 1\right)\ -\ 6\left(2x\ -\ 1\right)\ =\ 0]


Notice that the binomials in parentheses are now identical.  Factor them out.



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(4x\ -\ 6\right)\left(2x\ -\ 1\right)\ =\ 0]


From there it is just application of the Zero Product Rule.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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