How do you factor polynomials of higher powers?

[becomingagodo]

I need to know how to factor equation like these -x^3+5x^2+17x-21 , now how do you do this. I heard you need to use the rational root theorem and something called synthetic division information, however it was explain poorly in my book I am using.

Can someone just factor
-x^3+5x^2+17x-21=0
and show the working and explain why you did it. I can factor equation where the highest power is two and quadratic eqautions.
Note the anwser is this
-(x+3)(x-1)(x-7)=0

[TV's Frank]

Why are we doing your homework?

[JoeTheJuggler]

I need to know how to factor equation like these -x^3+5x^2+17x-21 , now how do you do this. I heard you need to use the rational root theorem and something called synthetic division information, however it was explain poorly in my book I am using.

Can someone just factor
-x^3+5x^2+17x-21=0
and show the working and explain why you did it. I can factor equation where the highest power is two and quadratic eqautions.
Note the anwser is this
-(x+3)(x-1)(x-7)=0

Someone else can probably give you a better answer, but I'd just use trial and error (and some reason).

Edit: I'm assuming the answer is possible with all whole numbers. If not, I'd look to factor out one term (x-5 or something) and be left with an ugly quadratic with fractions.

You know there are three terms with 1x that multiply to the -x^3, so they're either all negatives or one negative and two positives. If they're all negs, I don't think you'd be able to get the middle terms, so I probably would start with one negative and two positives. Three terms that multiply to 21 limits the other value to 7, 3 and 1 or 21, 1 and 1. Again, either all negative or one negative. Then I'd play around with those until I got the middle terms to come out right.

[JoeTheJuggler]

Why are we doing your homework?

He has the answer, so I think this is a legitimate question seeking to understand how you arrive at it.

[becomingagodo]

I just can't understand how (x+3)(x-1)(x-7) can produce 17x. I been trying to work out how they can sum to 17x.
Why are we doing your homework?
It is the summer holidays, how can this be homework?
Three terms that multiply to 21 limits the other value to 7, 3 and 1 or 21, 1 and 1. Again, either all negative or one negative. Then I'd play around with those until I got the middle terms to come out right.
Yeah, I think I understand how to do it know.

[mijopaalmc]

I just can't understand how (x+3)(x-1)(x-7) can produce 17x. I been trying to work out how they can sum to 17x.

It is the summer holidays, how can this be homework?

Yeah, I think I understand how to do it know.

-(x+3)(x-1)(x-7)

Multiply the first two binomials
-(x^{2}-x+3x-3)(x-7)

Collect like terms
-(x^{2}+2x-3)(x-7)

Multiply the trinomial and binomial by distributing the terms of the binomial
-x(x^{2}+2x-3)-7(x^{2}+2x-3)=-(x^{3}+2x^{2}-3x)+(-7x^{2}-14x+21)

Collect like terms
-(x^{3}+2x^{2}-7x^{2}-3x-14x+21)=-(x^{3}-5x^{2}-17x+21)

Distribute the negative sign
-x^{3}+5x^{2}+17x-21

[becomingagodo]

Yeah, I understand it now. For some reason I had the idea that it had to add up to 15x instead of 17x, so when I did my calculation I couldn't add it up to 15. Temporary insanity.

P.S. (3x7)-7+3=17

[Iamme]

I just can't understand how (x+3)(x-1)(x-7) can produce 17x. .

Maybe it's something on the same principle used where you can tell what age you are if you add up the change in your pocket, then add 19 divide by 3, then...I'm joking in a way...but there indeed is this goofy math trick to do something like that, that works for everyone. Maybe the asnwer to YOUR question and the answer to my question have some common denominator.

BTW, I plugged 5 in as X and on the left side I get 32 times negative 2, while on the right side I get 85. I must not follow what your question is.

Oh, I get it now. You said "can", not "will"...for every number that can be X.

[Iamme]

Yeah, I understand it now. For some reason I had the idea that it had to add up to 15x instead of 17x, so when I did my calculation I couldn't add it up to 15. Temporary insanity.

P.S. (3x7)-7+3=17

Can you show this result now plugged into your formula?, as with your formula I only saw multiplications without any minus or plus, like you now show above. In my book (anything)(anything) (anything) = anything times anything times anything. NOT anything minus something plus something else.

[becomingagodo]

Can you show this result now plugged into your formula?, as with your formula I only saw multiplications without any minus or plus, like you now show above. In my book (anything)(anything) (anything) = anything times anything times anything. NOT anything minus something plus something else.
Well, only 1,1,21 or 1,7,3 can sum up to 21.
21-1-1=19x
(-1X-7)+(-1X3)+(-7X3)
-7+3=-4
-4+21=17x

Using the multiplying brackets rule it adds up to 17.
but there indeed is this goofy math trick to do something like that, that works for everyone
I know a way to calculate the day you were born. It something like take 25% of the date you where born and add it to the year, or something like that. See people like Daniel Tammet use that to make them seem super smart or a savant and yet it is a simple trick.

[JoeTheJuggler]

Well, only 1,1,21 or 1,7,3 can sum up to 21.
I think you have the right idea--but "sum" is the wrong word here.



I know a way to calculate the day you were born. It something like take 25% of the date you where born and add it to the year, or something like that. See people like Daniel Tammet use that to make them seem super smart or a savant and yet it is a simple trick.

No way is it that simple. I've seen perpetual calendars, and it's still a substantial memorization feat.

[Complexity]

Summer school?

How about making use of math books and resources on the internet?

[FireGarden]

I heard you need to use the rational root theorem and something called synthetic division information,

Suppose you have to find a factor of f(x)
That means you have to find something like:

f(x)=(x-a)g(x)

Now RHS=0 when x=a,
so surely LHS=0 when x=a

Therefore, to find a factor... it may be useful to find a root. (Not just maybe, take my word for it. :D)

Suppose I have f(x)=x^2-3x+2
then f(1)=1-3+2=0
So try (x-1) as a factor

The division, finding the other factor:
(x-1)(x....

Gives me the x^2,
but I have a "-x" where I need "-3x"
So I take away another 2x

(x-1)(x-2)

That also gives me the +2, so there is no remainder.

Hope that helps

[FireGarden]

Summer school?

How about making use of math books and resources on the internet?

We are a resource on the internet.
:D

[drkitten]

Suppose you have to find a factor of f(x)
That means you have to find something like:

f(x)=(x-a)g(x)

Now RHS=0 when x=a,
so surely LHS=0 when x=a

Therefore, to find a factor... it may be useful to find a root. (Not just maybe, take my word for it. :D)

...

Hope that helps

That may not be as useful as one might think. :p

More seriously, "synthetic division" is simply a process for trying out all possible factors.

Let's pretend that we have f(x) = ax^n + bx^(n-1) ... + p, and we want to find a linear
factor (rx + s). So by hypothesis, we know that f(x) = (rx +s) ( g(x) ).

What do we know about g(x)? Well,

Since f(x) is a polynomial of degree n, g(x) is a polynomial of degree (n-1).
Since the first coefficient of f(x) is a, we know that r times the first coefficient of g must be a.
Since the last coefficient of f(x) is p, we similarly know that s times the last coefficient of g(x) is p.
Therefore, r is a factor of a, and s is a factor of p.


Synthetic division is just trying all possible combinations of "factors of a" and "factors of p" to find all possible rational roots. If a is 1, for example, then the only factors are +/-1.
if p is 6, then the possible factors are +/- 1,2,3,and 6. So by checking

(x-1)
(x+1)
(x-2)
(x+2)
(x-3)
(x+3)
(x-6)
(x+6)


we enumerate all possible rational roots. If a were 2, we would need to check 2x +/ {1,2,3,6} as well. This process is timeconsuming, but guaranteed to be complete.

[elgarak]

That may not be as useful as one might think. :p

More seriously, "synthetic division" is simply a process for trying out all possible factors.

Let's pretend that we have f(x) = ax^n + bx^(n-1) ... + p, and we want to find a linear
factor (rx + s). So by hypothesis, we know that f(x) = (rx +s) ( g(x) ).

What do we know about g(x)? Well,

Since f(x) is a polynomial of degree n, g(x) is a polynomial of degree (n-1).
Since the first coefficient of f(x) is a, we know that r times the first coefficient of g must be a.
Since the last coefficient of f(x) is p, we similarly know that s times the last coefficient of g(x) is p.
Therefore, r is a factor of a, and s is a factor of p.


Synthetic division is just trying all possible combinations of "factors of a" and "factors of p" to find all possible rational roots. If a is 1, for example, then the only factors are +/-1.
if p is 6, then the possible factors are +/- 1,2,3,and 6. So by checking

(x-1)
(x+1)
(x-2)
(x+2)
(x-3)
(x+3)
(x-6)
(x+6)


we enumerate all possible rational roots. If a were 2, we would need to check 2x +/ {1,2,3,6} as well. This process is timeconsuming, but guaranteed to be complete.Another helpful thing is that most school problems have a root in the low single digit integers (1, 2, 3) -- checking if a and p have common factors was simple. Another helpful thing is that the degree is 3 at the most. Higher degrees are only given as school problems if they are simple (like multiple roots at x=0) -- at least during my school time. When I was in training, I could 'see' one possible factor, then divide to reduce the degree of the polynomial by one.

[phildonnia]

The general method for cubic equations:

http://en.wikipedia.org/wiki/Cubic_formula#Cardano.27s_method

[FireGarden]

That may not be as useful as one might think. :p


Hey!
Taking my word for it is always useful! :p

But good answer, Dr K