How to

[naved_s9994]

How to factorise

2x^2 - 3x - 14 = 0              solve for x

arghh...step by step procedure please!!

[Flaming_Arrow]

(2x-7) (x+2)

just do it using trial and error

[naved_s9994]

as in.... can you explain the procedure though (hand - wise),, like not using calc

[TrueTears]

as in.... can you explain the procedure though (hand - wise),, like not using calc

He just did it by inspection.

If you want to get a more 'algebraic' way, then use the quadratic formula.

[naved_s9994]

...okay then by inspection,,,, like WHAT im saying is,,, what do you look out for?? and then conclude

[TrueTears]

I do the cross method

2x           -7

x             +2

in other words (this is the a term)

(this is the c term)

hence factors are

[naved_s9994]

thnx true tears

[shinny]

How to factorise

2x^2 - 3x - 14 = 0              solve for x

arghh...step by step procedure please!!

This takes a bit more training, but...




As for knowing what numbers to use in the second step, one way is to list all the factors of the constant and find the pair which has a number divisible by the coefficient of the and work from there perhaps. I never really did much working out for these types of questions so it's kind of hard to teach what I do... =\

EDIT: Better explanation for a systematic almost fool-proof approach to this question;


Since the coefficient of is a prime, it must multiply with B to equal -14 (you might work out why later once you use this technique a few more times and understand how it works, but I cbf explaining), so B is -7. Since B+A=-3, A is +4, and you can work from there as above.

[Mao]

To add to shinny's method:

--> multiply A and C (-28)

Find a pair of numbers that multiply to -28 and add to -3 (-7,4)

Split the x term according to these numbers, and write it out as shinny has.

[shinny]

To add to shinny's method:

--> multiply A and C (-28)

Find a pair of numbers that multiply to -28 and add to -3 (-7,4)

Split the x term according to these numbers, and write it out as shinny has.

Ah yes, that was the proper way I think. Was kind of making up a systematic way myself just before for explanatory purposes. Well you'll eventually get to the stage where you just see it though anyway.

[naved_s9994]

shinny, thanks....that AWESOME.....even all the rest -- i knew about cross multiply method, like used above, but all i was intending to see was a quicker method or (easier) in  a way to do this type of question