Question 202586
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2 -17xy+24y^2]


First thing: There is no equation here.  That is because there is no equals sign anywhere to be seen.  What you want to do is to factor the polynomial expression.


You have a trinomial which suggests that it is the product of two binomials.  In order for the first term to be *[tex \Large 3x^2], the first terms of the two binomials must be *[tex \Large 3x] and *[tex \Large x], so start with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3x\ \ \ )(x\ \ \ )]


The next thing to notice is that the last term in your polynomial has a positive sign.  That means that the two signs in the binomials must be the same sign.  Since the sign on the second term in your polynomial is negative, those two signs must be negative, so now you have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3x\ -\ \ )(x\ -\ \ )]


The second terms of the binomials must be of the form *[tex \Large ay] and *[tex \Large by] where *[tex \Large a\cdot b = 24].  Therefore, possibilities for *[tex \Large a] and *[tex \Large b] are 1 and 24, 2 and 12, 3 and 8, and 4 and 6 and we know that 3 times one of the pair of numbers plus 1 times the other of the pair of numbers must equal 17, so:


3 times 1 = 3 plus 24 is 27, not 17
3 times 24...well never mind
3 times 2 = 6 plus 12 is 18, closer but no cigar
3 times 12...again no good
3 times 3 = 9 plus 8 is 17:  Ah Ha!


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3x\ -\ 8y)(x\ -\ 3y)]


Check the work using FOIL:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\cdot x = 3x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\cdot-3y = -9xy]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -8y\cdot x = -8xy]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -8y\cdot-3y=24y^2]


Put 'em together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2 - 9xy - 8xy + 24y^2\ =\ 3x^2 -17xy+24y^2]


Answer checks.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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