Factoring a polynomial

[J.Ole]

I'm trying to brush up on the maths I took 11 years ago so that I can test into trig. when I go back to college in Fall, but am really rusty. I'm working on factoring polynomials presently, but am baffled by this problem because I haven't been able to find an example like it online to try to follow.

The instructions simply say "Factor Completely."

a^2 -ac + ab - bc

I have found that (a-c) can be factored out, so my answer at present looks rather like: (a-c)(a+c)+(a-c)(b+c), but that doesn't seem to fit into the form of the examples I've found of other factored polynomials. Thanks for looking and helping if you can.

 

[Deleted member 4993]

I'm trying to brush up on the maths I took 11 years ago so that I can test into trig. when I go back to college in Fall, but am really rusty. I'm working on factoring polynomials presently, but am baffled by this problem because I haven't been able to find an example like it online to try to follow.

The instructions simply say "Factor Completely."

a^2 -ac + ab - bc

I have found that (a-c) can be factored out, so my answer at present looks rather like:

a² - ac + ab - bc

= a * (a - c) + b * (a - c) ← factor out (a-c)

= (a - c) * (a + b)

and we are done.....

but that doesn't seem to fit into the form of the examples I've found of other factored polynomials. Thanks for looking and helping if you can.

.

 

[mmm4444bot]

trying to brush up on the maths I took 11 years ago so that I can test into trig. when I go back to college in Fall, but am really rusty.

MY OPINION: No real math work in 11 years? Bad idea to try to test into trig. Start with an algebra course, instead.

 

[lookagain]

Here are a couple of other ways (not exhaustive) to factor it
(where I opted at the outset to keep the squared term first):


a^2 - ac + ab - bc =

a^2 + ab - ac - bc =

a(a + b) - c(a + b) =

(a + b)(a - c)



---------------------------------- OR


a^2 - ac + ab - bc =

a^2 - bc - ac + ab =

a^2 - bc - a(c - b) =

a^2 - a(c - b) - bc =

a[a - (c - b)] - bc =

a[(a - c) + b] - bc =

a(a - c) + ab - bc =

a(a - c) + b(a - c) =

(a - c)(a + b)

 

[Willie]

MY OPINION: No real math work in 11 years? Bad idea to try to test into trig. Start with an algebra course, instead.

It depends on learning style and motivation. It's certainly possible to study math on your own. I'm having pretty good success with it myself. You just have to be honest with yourself about how much study and practice time it will really take (lots), and whether you are motivated enough to do that without the structure and schedule of a class.

 

[mmm4444bot]

It's certainly possible to study math on your own.

Agreed. And, in the OP's case, I believe that it is a bad idea to try to test out of some prerequisites because (based on what I sense from this thread and my experience) the trig course will likely be tantamount to taking two or more courses simultaneously after a decade of no real math work. The OP may check out the prerequisites unit in Mrs. Cronquist's class, for a good idea -- watch the video lectures and slideshows, read the notes, try some worksheets. That experience will be real.

I'm not trying to be nasty or insensitive, when I say this, but one who is competent at self-study with Internet access hardly needs to ask us to help with exercise in this thread.

If the OP is planning on taking an on-line trigonometry course, then the difficulties will be compounded. If the OP is in some type of vocational program, then the difficulties may be reduced.

All of these comments are unsolicited opinion, of course. I suggest an honest placement test, and allow academic advising to do their thing.

I wish the OP good fortune, regardless of how full their plate becomes. We'll be here. :cool: (I wish Mrs. Pi were, right now.)