Factoring: I can't figure out (x+2)^2 - 3 (x-2) + 2

[Math_Junkie]

How would you factor a question like this?
Much thanks to whoever can help me, obviously you're smarter than I am at this!

(x+2)^2 - 3 (x-2) + 2

 

[Mrspi]

Re: Factoring, took me hours and I can't figure this one out

How would you factor a question like this?
Much thanks to whoever can help me, obviously you're smarter than I am at this!

(x+2)^2 - 3 (x-2) + 2

One way to approach this problem is to let z = (x + 2)

If you do that, you'll have this:

z^2 - 3z + 2

Ok.....factor that.

Then when you're done, "back substitute" replacing z with (x + 2)

 

[arthur ohlsten]

[x+2]^2-3[x-2]+2 expand
x^2+4x+4-3x+6+2
x^2+x+12
can't be factored with real terms

x={-1+/- [1-48]^1/2}
x=-1/2 +/- i sqrt47/2
[x+1/2 +[i sqrt47]/2 ] [x+1/2-[isqrt47]/2]

if the equation was copied in error and should be:
[x+2]^2-3[x+2]+2
[[x+2]-2] [[x+2]-1]
x[x+1] answer

[x+2]^2-3[x+2]+2 expand
x^2+4x+4-3x-6+2
x^2+x
x[x+1] answer

Arthur

 

[Math_Junkie]

Thank you both for your responses!

Both of your posts has helped me alot.

Keep up the great work! :lol:

 

[arthur ohlsten]

glad I was of limited help.
I am afraid MRSPI made a error and misread the equation

[x+2]^2-3[x-2]+2 if you make the substitution [x+2]=z you get
z^2-3[z-2-2]+2
z^2-3z+14
can't be factored with real terms
roots are 3+/-1[9-56]^1/2 all over 2
[z-3/2 +[isqrt47]/2] [ [z-3-isqrt47]/2]

Arthur

 

[stapel]

(x+2)^2 - 3 (x-2) + 2

:?: Tutor arthur ohlsten pointed out that, as posted, this cannot be factored over the reals. Are you working with complex-number factoring, or is there a typo here? :?:

Please reply with clarification. Thank you!

Eliz.

:arrow: P.S. Welcome to FreeMathHelp!

 

[Math_Junkie]

Oh my bad!

There is a typo, it's supposed to be (x+2)^2 - 3(x+2) + 2

I used MSRPI's suggestions and came to the right answer!

Sorry for the confusion guys. :roll: