How do you solve
y=15x^2+x-2

Question

How do you solve
y=15x^2+x-2

jack1 · Answer

depends what you mean by solve?

jack1 · Answer

your unable to find a numerical solution to x and y with only one equation.
you could factor this or find the zeroes if you like?

anonymous · Answer

i don't know exactly what it means by solve. what i typed is all i have from the question. i guess we just factor this?

jack1 · Answer

ok, so these equations are in the form:
$$\large y = ax^2 + bx + c$$
so what you need to factor out are a, b and c into the form:
$$\large y = (--x + -- )(--x + --)$$
so u need to solve for the dashes of the above equation

jack1 · Answer

so start with a simple one: c
what are the factors of -2?

jack1 · Answer

@mef1717 ?

anonymous · Answer

i honestly don't know.

jack1 · Answer

ok, so another way to put it is factors of 16 = "what times what = 16"

anonymous · Answer

4x4

jack1 · Answer

so factors of 16 are 
1 and 16,
2 and 8, 
4 and 4

miracrown · Answer

Well if your looking to factor what you have to do is look at the two coefficients at the x^2 and the constant term. Then you need to take factors for those terms  and those factors should foil into what we need...

So for the first term we have 15 which has the factor of 1,3,5,15

and then there is also the constant term 2 with factors 1 and 2

miracrown · Answer

so there are a couple possible ways we can factor this...

jack1 · Answer

so there's really only one factor pair of 2, and thats 1 and 2
so factors of -2 are either (-1) x 2     or 1 x  (-2)

miracrown · Answer

there is only 1 right way to factor but the trick is to find the right combination

miracrown · Answer

do you know how to do it from here or do you need more help?

@mef1717

anonymous · Answer

i still need help, i don't understand what to do with the 15, we factor it but how do we know what numbers to use?!

miracrown · Answer

Well we need to find a combination of those number so we can foil it out and get 15* x ^2+x-2

the +x is really important because those other factors have to (in a sense) combine to become that...

miracrown · Answer

now because the coefficient on the x is 1, which is relatively small for the numbers we have, we can assume that the numbers are going to be fairly close together to one another

so there is a good chance we would actually use the 5 and 3 factors as opposed to the 15 and 1 factor

anonymous · Answer

okay so what do we do next? like how do i write the answer?

miracrown · Answer

well assuming that is what they meant by solve the answer would look something like...

miracrown · Answer

4= (5x + 2)( 3x -1) 

like that!

anonymous · Answer

im assuming too thats what my teacher meant by solve, he wasn't really clear on that, thanks for the help

miracrown · Answer

No worries! :)

jack1 · Answer

y = (qqx+ss)(ppx+tt)
now to cull our selection down, we need to find the correct numbers that
1.     for ss times tt (above): Multiply to = -2    (our c value)
2.     for qq times pp (above):  Multiply to = 15  (our a value)
3.     add to equal our b value using FOIL - so (qq times tt)   +   (ss times pp)

now just find the right combo
y = 15x^2 +    x    -  2
y = ax^2    +  bx  +  c
to
y = (qqx+ss)(ppx+tt)

1.  factors of c (-2) are -1, 2    or    -2, 1     so lets pick -1 and 2
2.  factors of a (15) are 1,15    or       3,5     so lets pick 3 and 5
3.  so b = 1
     so (3   times   2)  +  (5   times   -1) = 1

therefore put qq as 3, pp as 5, ss as -1 and tt as 2

so all up
 y = (qqx+ss)(ppx+tt)
y = (3x-1)(5x+2)

final equation:
$$\large y = (3x-1)(5x+2) $$

does that make sense @mef1717 ?