# Solve by factoring 2x^2+3x=2

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### 1 Answer

Solve `2x^2+3x=2` :

First write in standard form:

`2x^2+3x-2=0`

I. You can use a guess and check strategy: you know the factored form will be of the form (a+b)(c+d). So ac=`2x^2` ; the only way to get that is `(2x+b)(x+d)` . Now bd=-2 and there are four ways to do that:

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

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

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

(2x-2)(x+1)

Checking using the distributive property (FOIL) we see that the second is thecorrect one:

`(2x-1)(x+2)=0` Using the zero product property we get:

`2x-1=0` or `x+2=0` So

`x=1/2` or -2

(2) Another method: `2x^2+3x-2=0`

Let M=2(-2)=-4 (Mulitply the leading coefficient and the constant term)

We need a p and q such that pq=-4 and p+q=3 (The product of p and q is M, while the sum of p and q is the linear term)

We find p=4 and q=-1to work. (You could also use p=-1,q=4)

Rewrite `2x^2+3x-2` as `2x^2+4x-1x-2` ; now factor 2 terms at a time, taking out the common factor:

`2x^2+4x-1x-2=2x(x+2)-1(x+2)=(2x-1)(x+2)` so:

`2x^2+3x-2=0==>(2x-1)(x+2)=0==>x=1/2,-2`

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**The solution is `x=1/2` or `x=-2` :**

`2(1/2)^2+3(1/2)=1/2+3/2=2`

`2(-2)^2+3(-2)=8-6=2`

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