Question 333056
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This process is good for quadratic trinomials where the lead coefficient is 1.  Given a quadratic trinomial of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ bx\ +\ c]


Step 1:  Write *[tex \Large (x\ \ \ )(x\ \ \ )].


Step 2:  Put in the signs.  If the sign on the constant, that is the *[tex \Large c] term, is positive, then the signs in the factors are the same as the sign on the first degree, that is the *[tex \Large bx] term.  If the sign on the  constant term is negative then one of the binomials has a *[tex \Large +] and the other has a *[tex \Large ^-].


Step 3: Find all factors *[tex \Large p] and *[tex \Large q] such that *[tex \Large pq\ =\ c]


Step 4:


If the sign on the constant term was *[tex \Large +], then select a pair of factors from the group of factor pairs developed in Step 3 such that *[tex \Large p\ +\ q\ =\ b].  Insert *[tex \Large p] and *[tex \Large q] into *[tex \Large (x\ +\ p)(x\ +\ q)] or *[tex \Large (x\ -\ p)(x\ -\ q)] depending on the sign on the *[tex \Large bx] term.


If the sign on the constant term was *[tex \Large -], then select a pair of factors from the group of factor pairs developed in Step 3 such that *[tex \Large p\ -\ q\ =\ b].   Insert *[tex \Large p] and *[tex \Large q] into *[tex \Large (x\ -\ p)(x\ +\ q)] or *[tex \Large (x\ +\ p)(x\ -\ q)] such that whichever of *[tex \Large p] or *[tex \Large q] has the larger absolute value goes with the sign that is on the original *[tex \Large bx] term.


Your example:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ +\ 3]


Step 1:


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


Step 2: Both signs *[tex \Large +], so:


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


Step 3: Factors of 3 are 1 and 3.


Step 4: 3 plus 1 is 4, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 3)(x\ +\ 1)]


Use FOIL to multiply the two binomials to check your work.  The product should be the original trinomial.


Trinomials with a lead coefficient other than 1 are a little trickier.  Basically you are looking for numbers *[tex \Large p], *[tex \Large q], *[tex \Large r], and *[tex \Large s], such that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ pq\ =\ c]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ rs\ =\ a]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ rq\ +\ sp\ =\ b]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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