how do you factorise polynomials like this?

Question

I have to factorise the following:
X3-4X2-X+4
(Xcubed-4Xsquared-X+4
Could Someone please explain what steps I need to do to factorise the expression, and other similar ones and don't know the method.

Answer 1

Factor each polynomial by grouping the first two terms and the last two terms. x^3-4x^2-x+4

*You don't always have to group the very 1st terms. In this case, you could also group the 1st & 3rd term together.

First: group "like" terms.

(x^3 -x) - (4x^2 - 4)

Sec: factor both sets of parnethesis.

x(x^2 - 1) - 4(x^2 - 1)

Third: make sure the terms in parenthesis are the same - combine the inner term (once) with the outer term.

(x - 4)(x^2 - 1) or, (x^2 - 1)(x - 4)

Answer 2

this is going to be quite long..

first, i want you to know that this method is for getting rational roots only i.e. real factors (usually, this is the kind of factors that you would like to have), i wouldnt want to go to extracting irrational factors..

Now, for the method.. first, take the coefficient of the first term, which in your example, is 1. Take then, the coefficient of the last term which is 4. Now, take all the possible factors of 1 and 4..

for 1 --> +/- 1 (let this equal p)
for 4 --> +/- 1, +/-2, and +/-4 (let this equal q)

what you then do is get all the possible numbers that will arise from dividing p over q.. Thus, the possibilities are: +/- 1, +/- 1/2 and +/- 1/4

Now, using synthetic division, choose one of the p/q 's (trial and error, but i'll choose the right one for brevity), say you choose +1, which means you chose the root (x-1).. Divide the original polynomial by +1, and you will get, x^2 -3x -4 (x squared minus three x minus four)..

So now you have two factors, (x-1) and (x^2-3x-4) Now, factor out the trinomial expression.. you will get (x-4)(x+1)

Thus, combining all the factors you have obtained the roots are (x-4)(x+1) and (x-1)

***Note: This implies that had you chosen -1 or +4 in the synthetic division process, you would still arrive at the correct answer

Answer 3

Check by putting x =1, x = -1, x = 2, x = -2, x = 3, x = -3.
If any of these values of x makes the value of the function zero, that gives you a clue to one of the possible factors.

The value of x^3 - 4x^2 - x + 4 becomes zero with x = 1. This means x - 1 is one of the factors.

Now, rearrange the terms keeping in mind that x - 1 is one factor as under.

x^3 - 4x^2 - x + 4
= x^3 - x^2 - 3x^2 + 3x - 4x + 4
= x^2 ( x - 1 ) - 3x ( x - 1 ) - 4 ( x - 1 )
= ( x - 1 ) ( x^2 - 3x - 4 )
= ( x - 1 ) ( x - 4 ) ( x + 1 )

Answer 4

not sure this will be helpful, but if you look at tthe equation you can see that if you set your polynomial to zero, x=1 is a solution, so you can factor (x-1) out of the equation. you can do that with polynomial long division or with synthetic division, but i prefer long division, and with it i got this:

x^3 - 4x^2 + 4 = (x - 1)(x^2 - 3x - 4)

the second part of the right side factors to (x - 4)(x + 1), so

x^3 - 4x^2 + 4 = (x - 1)(x + 1)(x - 4)

hopefully someone will come forward with a better way than "just seeing" one solution, but keep in mind, if the assignment is to factor the polynomial to simple roots, there will always be an (x - a) factor. just look for it.

Answer 5

x^3-4*x^2-x+4....... Start with the given cubic polynomial


(x^3-4*x^2)+(-x+4)....... Group like terms into two groups


x^2(x-4)-1(x-4)....... Factor out x^2 out of the first group. Factor out -1 out of the second group.


Notice we now have the common term x-4. This means we can combine like terms


(x^2-1)(x-4)....... Combine like terms


(x+1)(x-1)(x-4)....... Factor x^2-1 to get (x+1)(x-1) (note: this is a difference of squares)



So x^3-4*x^2-x+4 factors to (x+1)(x-1)(x-4)

Answer 6

4x^3 - 8x^2 -x +2 =0 from observation, you can easily factorise by grouping instead of wasting time on trial methods of factorising polynomials i.e (4x^3 -8x^2) -x +2 =0 4x^2(x-2)-1(x-2) = 0 (4x^2-1)(x-2) = 0 [(2x)^2 -1^2](x-2) = 0 use dif .of 2 squares (2x - 1)(2x + 1) (x - 2) =0 ==> x = 1/2, x= -1/2, x=2

Answer 7

x^3-4x^2-x+4
=x^2(x-4)-1(x-4)
=(x-4)(x^2-1)
=(x-4){(x)^2-(1)^2}
=(x-4)(x+1)(x-1)
[This particular factoring was done by grouping and then taking common factor out.Alongwith that factorization of identity a^2-b^2=(a+b)(a-b) was also followed in the later stage.It is very difficult to explain in details everything about factoring in this short spae and time.i would advise you to take the help of a experienced teacher]

Answer 8

X^3-4x^2-x+4
=x^2(x-4)-1(x-4)
=(x-4)(x^2-1)
=(x-4){(x)^2-(1)^2}
=(x-4)(x+1)(x-3)

Answer 9

The previous answers are good if you can easily find one of the roots. But if you run into a more complicated problem with non-integer coefficients, you might have to resort to the more general solution of the cubic equation. You can find Cardano's method described in Wikipedia.

Answer 10

f (x) = x ³ - 4 x ² - x + 4
f(1) = 1 - 4 - 1 + 4
f(1) = 0 thus (x - 1) is a factor.
To find other factors , use synthetic division:-
"""|1"""- 4"""- 1"""4
"1|""""""1""""-3"""-4
------------------------
""|1""""-3""""-4"""0

( x - 1 ) ( x ² - 3 x - 4 )
( x - 1 ) ( x - 4 ) ( x + 1 )