(2x+1)(x-3)=0
x= -1/2, or x=3
Using factoring to solve each quadriatic equation. 2x^2-5x-3=0.
This is what I have so far but don't know what to do from here.
GCF:1
A:2
B:-5
C:-3
AC:-6
Goal is the find two numbers that add to -5 and multiply to -6
Factors of 6: 1,6
-2,3 (That's the one)
I don't know what to do from here thanks for your help!!
2 answers
You must be learning a method called "decomposition"
So you want two numbers
whose sum is -5
and whose product is -6
your choice of -2,3 does not work,
but -6,1 does.
So we are going to replace the middle term -5x with
-6x + x
2x^2-5x-3=0
2x^2 - 6x + x - 3 = 0
now group it using common factor in pairs
2x(x-3) + 1(x-3)
common factor again
(x-3)(2x+1) = 0
I am sure you can finish the solving part ...
If you can find a pair of numbers which satisfy the sum and product condition, this method will always work.
So you want two numbers
whose sum is -5
and whose product is -6
your choice of -2,3 does not work,
but -6,1 does.
So we are going to replace the middle term -5x with
-6x + x
2x^2-5x-3=0
2x^2 - 6x + x - 3 = 0
now group it using common factor in pairs
2x(x-3) + 1(x-3)
common factor again
(x-3)(2x+1) = 0
I am sure you can finish the solving part ...
If you can find a pair of numbers which satisfy the sum and product condition, this method will always work.