factoring polynomials?

Question

Could someone please guide me through factoring the following:

x^3-6x^2+11x-6

The answer should come out to (x-1)(x-2)(x-3)
Thank you!

Answer 1

give you a little hint

look up the factor theorem... it's quite handy... essentially it says that for any polynomial ax^n + ax^n-1 + ... axn^0 + c = 0 , if x = b and f(b) = 0, then x - b is a factor

all this is saying if there is a value b for x such that f(b) = 0 - this is a root to the equation - then x - b is a factor of f(x).

so you have 2 things to do...

1. find a root of your equation (try 1, -1, 2, -2) etc.

2. once you find that root (hint it's 1), then x - 1 is your factor...
do long division/synthetic division of the polynominal using that factor

set it up this way...
........_________________
x - 1 | x^3 - 6x^2 + 11x - 6

I'll do the first step...

........______x^2________
x - 1 | x^3 - 6x^2 + 11x - 6
x^3 - x^2
-----------

Subtract and bring down the 11x

.........______x^2________
x - 1 | x^3 - 6x^2 + 11x - 6
- x^3 + x^2
-----------
- 5x^2 + 11x


Eliminate the -5x^2

........______x^2_- 5x___
x - 1 | x^3 - 6x^2 + 11x - 6
- x^3 + x^2
-----------
- 5x^2 + 11x
- 5x^2 + 5x
---------------

Subtract and bring down the -6

........______x^2_- 5x_+ 6__
x - 1 | x^3 - 6x^2 + 11x - 6
- x^3 + x^2
-----------
- 5x^2 + 11x
+ 5x^2 - 5x
---------------
6x - 6
6x - 6


Subtract and you're done... no remainders...

Your quotient will be a quadratic and if you're working on polynomials greater that 2 - degree you know how to factor a quadratic...

So you've got (x-1) and (x^2 - 5x + 6) as factors to your equation...


Good luck, hope this helps...

Answer 2

One way is to find a root for the equation x^3-6x^2+11x-6 = 0. Then the first factor would be known. This means, if you put 1 for x and the answer is zero, then (x-1) is a factor. Divide by this factor to get a quadratic, and factor it by FOIL.

If you know synthetic division, that's even easier. Divide til it comes out with 0 remainder, then do it again.

Answer 3

If you are trying to factor the cubic when you already know the answer, synthetic division works fine. Although , if you want to prove that the three roots you have are correct, just multiply the three terms together to see if you get the right cubic.

Without knowing the answers in advance, it is a very difficult problem. For the equation:

x^3 + a*x^2 + b*x + c = 0

You can determine the three roots as follows:

define f = squareroot(12*b^3 - 3*b^2*a^2 - 54*a*b*c + 81*c^2 + 12*a^3*c)

define g = cuberoot((a*b)/6 - c/2 - (a^3)/27 + f/18)

define h = (b/3 - (a^2)/9)/f

The three solutions are:

x1 = f - h - a/3
x2 = -f/2 + h/2 - a/3 + squareroot(-3)*(f + h)/2
x3 = -f/2 + h/2 - a/3 - squareroot(-3)*(f + h)/2

Answer 4

first, think of the most simple factor that is the most probably contained in the equation..

for example, choose (x-1), so that x = 1(*)

then x^3 x^2 x^1 X^0
1(*) 1(#) -6 11 -6

1(@) -5 6 +
-------------------------------------------------------
1 (#) -5 6 0

- directly drop coefficient of x^3 into the bottom line
- multiply 1(#) with 1 (*), put the result in the second line. the result is 1 (@)
- sum coefficient of x^2 and previous result, and so on, UNTIL YOU GET THE FINAL RESULT OF ZERO. If not, try other possibility
- use the coefficeint at the last line as follows

x^3-6x^2+11-6=0
(x-1) (this is our guess) (x^2-5x+6) this is our calculation
(x-1) (x-2) (x-3)

Answer 5

tri using the ... i forgot what it's called, but it's like this

write the coefficient of the equation:

1 -6 11 6

then write a one in front (oh man, i won't be able to explain it to u but i know how to do it, but i need to show u on paper)

u write 1 in front of the above numbers and thin u add

first, u bring down the 1 from (1 -6 11 6)
then u multiply 1 times 1 = 1 and u add that to -6
1 + -6 = -5
so u bring that one down and multiply it by one
-5 times 1 = -5

so u again add: 11+ -5: which again, is the same as 11-5 which is 6

then you multiply 6 by 1 to equal 6

and in the end the last two sixes are supposed to give u zero so u are left with the queastion of only to the second power

something like
x squared minus five x plus six. and then you can easily factor that.

it's super hard to do it like this. i need paper. but look in your math book aBOUT what i showed u

Answer 6

okay. here we go.

you have to start off by taking two sets of two like terms.

so this is the problem (x^3)-(6x^2)+(11x)-(6)

x^3 and 11x are alike [both have an x] and 6x^2 and 6 are alike [both have a 6]

so you put (x^3 + 11x) together and (-6x^2 - 6) together

factor out the common factors.

x(x^2 + 11) + -6(x^2 - 1)

ehhh..nevermind. that didnt work how it was suppose to... hmm.. oh well. im sorry. maybe if you email me i could help you . sorry man. but anyway, thats a way you could solve it. if i did it right...

Answer 7

This Site Might Help You.

RE:
factoring polynomials?
Could someone please guide me through factoring the following:

x^3-6x^2+11x-6

The answer should come out to (x-1)(x-2)(x-3)
Thank you!

Answer 8

Factoring polynomials completely: x^3y + 2x^2y^2 + xy^3 Factor out the Greatest Common Factor (GCF) = xy xy(x^2 + 2xy +y^2) Factor a trinomial xy(y + x) (y +x) Final result: xy(y + x) (y + x)

Answer 9

X 3-6x 2 11x-6

Answer 10

I don't think you can factor that by hand. Use a graphing calculator: graph y = x³-6²+11x-6, and see where y = 0, which is where the graph intersects the x-axis. Or use the solver on your graphing calculator.