Question

How do you factor `2x^2+3x-2`?

Answer 1

the answer is `(x+2)(2x-1)`

Explanation:

since you can't take anything out first you have to multiply the first and last number. you do this to find two numbers that multiply to give you that number but add to give you the middle number. those numbers are `-1,4` and ow you replace it in equation `2x^2-1x+4x-2` and now group and factor. first one `x(2x-1)` second `2(2x-1)` the outside number is what you took out or what they had in common. make sure both parenthesis are the same or you did it wrong. and there you got your answer

Answer 2

`=>(2x-1)(x+2)`

Explanation:

`2x^2+3x-2`

We want to achieve a form: `(ax+b)(cx+d)`

We need to satisfy:

`(1) =>ax*cx = acx^2 = 2x^2 -> color(blue)(ac = 2)`

`(2) => bcx + adx = (bc+ad)x =3x -> color(blue)(bc+ad=3)`

`(3) =>color(blue)(bd = -2)`

We can see that we need to find an `ac = 2` and `bd = -2` that satisfies `bc+ad=3`.

Factors of `ac = 2`:
`(i)=>(a,c) = (1,2)`
`(ii) =>(a,c) = (2,1)`
`(iii) =>(a,c) = (-1,-2)`
`(iv) => (a,c) = (-2,-1)`

Factors of `bd = -2`:
`(v)=>(b,d) = (-1,2)`
`(vi) =>(b,d) = (1,-2)`

We find the combination that gives us `bc+ad=3`:

`(i),(v)=>bc+ad = (-1)(2)+(1)(2) = 0`
`(i),(vi)=>bc+ad = (1)(2)+(1)(-2) = 0`
`(ii),(v)=>bc+ad = (-1)(2)+(1)(2) = 0`
`(ii),(vi)=>bc+ad = (1)(2)+(1)(-2) = 0`
`(iii),(v)=>bc+ad = (-1)(-2)+(-1)(2) = 0`
`(iii),(vi)=>bc+ad = (1)(-2)+(-1)(-2) = 0`
`(iv),(v)=>bc+ad = (-1)(-1)+(-2)(2) = -3`
`color(red)((iv),(vi)=>bc+ad = (1)(-1)+(-2)(-2) = 3)`

So we found that `a = -2, b = 1, c = -1, d = -2`.

Hence the factored form `(ax+b)(cx+d)` is:

`=>(-2x+1)(-x-2)`

Simplifying

`=>(-2x+1)(-1)(x+2)`

`=>(2x-1)(x+2)`