I have an equation 2x^2+x-1
How do I rewrite in parenthesis? Like (x+ )(x- )
Etc.
I got x = .5 and -1 using quadratic formula
Thanks
Just a heads up, this isn't a calculus problem.
Anyways, I believe you can only factor a quadratic $\displaystyle ax^2 +bx+c$ strictly into the form $\displaystyle (x + h)(x+k)$ when $\displaystyle a=1$
You can still factor by grouping though.
In general for a quadratic equation ax^2 + bx + c = 0 for factorization just write the product of the coefficient of x^2 and constant, in this case ac. Thereafter find such factors of the product ac such that their sum is equal to the coefficient of x. there after split the middle term and group into factors.
The reason I put it in calculus is because im doing a partial fraction integral and need to seperate the denomenamtor.
I'm still confused how to use the quadratic formula to factor into (x )(x )
Because I get .5 and -1 from the quadratic formula
However graphing (x-.5)(x+1) does not give the same as the original equation :S
Perhaps it is that "2" that is bothering you knowing that .5 and -1 are roots of the polynomial tells you that you want $\displaystyle (x- .5)(x- (-1))= (x- .5)(x+ 1)$ However, with that "2" coefficient of $\displaystyle x^2$ you need 2(x- .5)(x+ 1).
in this case your equation is 2x^2 + x - 1
step 1: Product of coefficient of x square 9 in this case = 2 0 and constant term 9 in this case = -1 ) is = -2
step 2. factors of -2 are
-2 = 1x -2 = -1 x 2
step 3. select the factors whose sum is equal to the coefficient of x. in this case 2 + ( -1 ) = 1
step 4. split the middle term with these factors.
2x^2 + 2x - x - 1 = 2x ( x + 1 ) -1 ( x + 1 ) = ( x+1 ) ( 2x - 1)