Simple Algrebra: Factoring quadrinomials

[floriflee]

I feel silly for having to post this, but for the life of me I can't remember how to factor quadrinomials. My mother is taking an algebra class and was asking for my help. Shamefully, I couldn't figure out how to get the answer. I'd like to blame it on the fact that I haven't done this kind of math in well over 10 years, and that it's so late so my brain isn't working. So here I am posting to find out how to solve this problem because it's going to drive me crazy until I figure out what it is that I'm missing:

x^3 + 4x^2+x -6

The answer is (x-1)(x+2)(x+3), but how do you get to that?
I've tried various forms of grouping, but so far nothing has come up as a GCF for both groups. What am I forgetting?

 

[siurpeeman]

you could try synthetic division. looking at the coefficient of x^3 and the last term, the only possible factors are ±1, ±2, ±3 and ±6**. once you've found one factor, the other two shouldn't be hard.

**this is because of the rational root theorem.

*edit*
you can make your guessing and checking a little easier by knowing there's only one positive root (descarte's rule of signs), as there is only one sign change. testing +1 will give you your first root on the first go.

 

[swiftaw]

Consider a factored cubic (x-a)(x-b)(x-c)

Expanding this you get x^3 - (a+b+c)x^2 + (ab + bc + ac)x - abc

thus, if you are trying to find the factors of x^3 + 4x^2+x -6 you know that

(a+b+c) = -4
ab+bc+ac = 1
abc = 6

3 equations for 3 unknowns can be solved for the desired result.

Of course, another method is trial and error to find one of the roots, then that will leave you with a quadratic which is easy to factorize.

 

[floriflee]

Ah, yes... synthetic division. I'd completely forgotten about that. Hopefully, they've covered that in her class already--I can't remember which order I learned what in in my classes....

Thanks, again!

 

[ethernet76]

You also should be able to graph the equation to find the roots.

Anywhere the graph crosses the line is a factor.

Grapher shows it crosses it at -3, -2, and 1.

Taking the negative of each results in (x+3), (X+2) and (x-1)

 

[floriflee]

Thanks, all! I can finally go to bed.

 

[siurpeeman]

(a+b+c) = -4
ab+bc+ac = 1
abc = 6

3 equations for 3 unknowns can be solved for the desired result.

no offense but i think you made the problem worse by doing what you did. just because you get three equations with three unknowns doesn't mean the solution is going to be a walk in the park. normally, you solve systems of equations by isolating and eliminating one variable from two equations. then you do the same with two other equations to get two equations with two unknowns. this simply isn't practical, especially with your second and third equations. i mean, could you solve it?

this is the best i could do:
ab + bc + ac = 1
a + b + c = -4
abc = 6

b (a + c) + ac = 1
-b (b + 4) + ac = 1
-b (b + 4) + 6/b = 1
-b ^2 - 4b + 6/b = 1
b^2 + 4b + 1 - 6/b = 0
b^3 + 4b^2 + b - 6 = 0

working with all three equations got me back to the original problem, which i would solve using synthetic division.

 

[swiftaw]

no offense but i think you made the problem worse by doing what you did. just because you get three equations with three unknowns doesn't mean the solution is going to be a walk in the park. normally, you solve systems of equations by isolating and eliminating one variable from two equations. then you do the same with two other equations to get two equations with two unknowns. this simply isn't practical, especially with your second and third equations. i mean, could you solve it?

this is the best i could do:
ab + bc + ac = 1
a + b + c = -4
abc = 6

b (a + c) + ac = 1
-b (b + 4) + ac = 1
-b (b + 4) + 6/b = 1
-b ^2 - 4b + 6/b = 1
b^2 + 4b + 1 - 6/b = 0
b^3 + 4b^2 + b - 6 = 0

working with all three equations got me back to the original problem, which i would solve using synthetic division.

My Bad, they are solvable, however to solve them means solving a cubic, which was the original problem so in this case not worth the effort. I must admit I did set up the equations without trying to solve them, all I know is that they could be solved - but it makes no sense to introduce a method which is just a complex as the original question. Oh well, serve me right for giving a rushed solution and not finishing my work.

 

[cruxed]

I typed up the working pretty briefly in Microsoft Word.
The good thing about long division is that it can be used not only for quadronomials, but probably ever other polynomial.

Hope it helps

 

[polevault139]

It's funny, since I have been taking claculus I sometimes forget some of the easier algebra. Luckily the AP Test is over and I can start to regain my sanity.

 

[notjustjay]

It's funny, since I have been taking claculus I sometimes forget some of the easier algebra. Luckily the AP Test is over and I can start to regain my sanity.

No kidding! I have TWO engineering degrees, and I've done YEARS of calculus (and other courses which were really just calculus in disguise, like physics) and yet I can't even remember any of this stuff anymore.

 

[siurpeeman]

well, there is a way, but it sure isn't pretty.

 

[jonasfan393]

Help with this problems!

My problem has two different variables and I can't figure it out. Here it is: 15a^2-3x-3+7a