Is this factorable: 7x^2+8x-12

And in general, can a trinomial with a number in front of the first coefficient (i.e. 7x^2 in this case) ever be factored (please refresh my memory)..., and if so, how?

There are two ways , I know because I take this in school this year .
1st way : you write (7x )(x ) , because difinitely 7x² equals 7x*x , then you start trying numbers until you get the right ones , but this way is boring and dificult
2nd way (which is more favorable ) :
7x² + 8x -12
x² + 8x - 84 (you remove the coefficient and multiply it with the last term )
then you look for two numbers if multiplied by each other give you (-84) , and since the number is negative then the two mulriplied numbers must contain a negative and a psitive one . and in the same time since since the second term is (+ 8 ) , a positive number , the the positive number should be bigger tha the negative one .
the numbers will be (+14) and (-6)
x² + 8x - 84
(x + 14 ) (x - 6 ) and to finish we have to divide both second terms by the coefficient number from the beginning which was 7
(x + 14/7 ) (x - 6/7)
(x + 2 ) (x - 6/7)

* (x - 6/7) = (7x - 6 ) because it equals (x * 7 - 6/7 * 7 )
the final answer is (x+2)(7x - 6)

every polynomial expression in one variable can be factored into linear factors.


what most teachers are asking is "can you factor using just integers?"

One way to determine if you can factor using integers (i.e. counting numbers) is to ask:

are there two counting numbers that multiply to the first # times last # AND add to the middle #?

for this example, first # times last # = -84, so you want 2 numbers that multiply to -84 and add to 8 (8 being the middle number)

in this case, YES. -6 times 14 is -84 and -6 + 14 is 8. so the expression factors using counting numbers.


use the numbers (-6 and 14) to rewrite middle number