2x^2 + x -3 = 0 Facotr each polynomill?

Question

I have the answer but I need help in understanding the steps to get the answer.

Answer 1

The rigorous way is to use the quadratic equation, but that may not be what you need for this problem.

Consider if you have (ax+b)(cx+d)...
... you get (acx^2 + (ad+bc)x + bd)

The coefficient for x^2 is ac, which is 2 in your example. That means that one of a or c is almost certainly 2 and the other is 1. Factored, your equation will almost certainly be:

(2x+b)(x+d)

The constant value is bd, which is -3 in your example. Factored, that will probably be (3)*(-1) or (-3)*(1).

Then turn your attention to the middle term, which is (ad+bc), or (2d+b)

Then you know that 2d+b = 1 (the coefficient of x). What values of +/-3 and +/-1 work with that? It works when d=-1 and b=3 (2d+b = -2+3 = 1).

And that leaves you with:

(2x+3)(x-1)

Answer 2

2x^2 + x -3
= 2( x^2 + x/2 - 3/2)
= 2(x^2 + 2 * 1/4*x + 1/4^2 - 3/2 -1/4^2)
= 2[ (x + 1/4)^2 - 25/16]
= 2 [ x + 1/4 - 5/4] [ x + 1/4 + 5/4]
= 2[ x - 1][x+6/4]
= [x - 1][2x - 2 * 6/4]
= [x-1] [2x + 3]

Answer 3

2x^2 + x -3=2x^2+3x-2x-3 =x(2x+3)-(2x+3 =(2x+3(x-1)
So the equation is;

=(2x+3(x-1)=0. this leads to:
x=1 or x=-(3/2) answer.

Answer 4

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

Answer 5

(2x+3)(x-1)

Answer 6

nvm