Math problem, please help, I'm stuck

[Nocturnal]

Please help me with this problem. It is bothering me and making me frustrated.

I'm following this formula: ax^2 + bx + c

#50 from 7.4

28x^2 - 28x + 7

This is what I did so far:

I use the grouping method because the trial and error method is too confusing for me.

So I found the GCF of 7.

Turned the problem into:

7(4x^2 - 4x + 1)

Next I start to do the grouping method.

I see that the constant is a positive or plus so I know that both signs of the two factors will be the same.

So I write out ac = 4(1) and that that 4(1) has to add up to -4.

2(2) will do it, well -2x(2x) = -4x.

Next step is to add in the -2x + 2x into the problem.

7(4x^2 - 2x + 2x + 1)

Now I find the GCF for the first two terms. The GCF is 2x.

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

Then you find the GCF of the second term which is 1.

1(2x + 1)

So

(7)2x(2x - 1) 1(2x + 1)

THIS IS WHERE I GET EXTREMELY CONFUSED.

2x - 1 does not equal 2x + 1

 

[Nocturnal]

Wait, I am looking at the problem now and I'm guessing it can't be factored?

It can't be factored because ax^2 - bx + c would mean that both signs would NEED to be -, -. Otherwise the problem can't work.

 

[rocadelpunk]

just glancing at it

i think you set it to 0, factor out by 7..divide it off, and use the quadratic formula

 

[dighn]

-2 and -2 multiple to 4 and add to -4

 

[Luagsch]

7*(2x-1)^2

 

[kaizersose]

its a repeated root, the answer is x=1/2 and x = 1/2.

if you divide the entire thing by 28, you get x^2 - x + 1/4. easy to factor from there.

edit: spelling

 

[Nocturnal]

I'm not even doing roots or anything. What I'm trying to do is called "Factoring Trinomials of the form ax^2 + bx + c, a is not equal to 1.

 

[Atlechnik]

This is easy, you are just factoring incorrectly

you could factor out a 7 and get 7(2x-1)^2=0
divide both sides by 7, then you get (2x-1)^2=0
now squar root both sides sqrt of 0=0; thus 2x-1=0
add 1 to both sides; 2x=1, now just divide both sides by 2
and you get x=1/2

 

[Nocturnal]

I already got it. Thanks guys for the help. Even reading a message board is hard. I have to physically do the problem to get it down. Thanks again though.

 

[CrazyPerson]

only if someone will help me with this stupid bio...