Hello Kitty's MM Question Thread - Cubics!

[hello_kitty]

Hey guys! i have some troubles getting the answer after doing the working!
Any help would be awesome!!!


1)   Factorise each of the following:
      2x^3 + x^2 - 2x -1           

I used the factor theorem, trial and error
therefore got P(1) = 2 * (1)^3 + (1)^2 - 2 (1) - 1
= 0
Therefore the Factor is (x-1)
But then the answer is (x-1)(x+1)(2x+1) and im not sure how to get to this???



2) Same as No.1        x^3 + 3x^2 + 3x + 1
                              Factor is (x + 1)
                               Answer is (x+1)^3  - not sure how to get to this??


3) Same sort           x^3 + 21x + 20   
                            Factor is (x-1)
                            Answer is (x-1)(x+5)(x-4) not sure again??

4)  Same sort         2x^3 + 3x^2 -1       Factor is (x + 1)           
                           Answer is (x+1)^2 (2x - 1) - not sure again?

5)  Same sort           4x^3 + 3x - 38       Factor is (x-2)
                             Answer is (x-2)(4x^2 + 8x + 19)  - not sure??


6) Same Sort        4x^3 + 4X^2 - 11X - 6          Factor is (x+2)
                          Answer is (x+2)(2x+1)(2x-3)   - not sure??



For these ones it's asking to factorise:

1) 64x^3 - 125
    (4x)^3 - (5)^3
    I know that the first part is (4x-5)
    The answer is (4x-5)(16x^2 + 20x + 25)  and im not sure how to get to this??

2) 8 + 27x^3
    (2)^3 + (5x)^3
    First part (2 + 3x)
    Answer is (2 + 3x) (9x^2 - 6x + 4)  not sure how?

3) 64m^3 - 27n^3
    (4m)^3 - (3n)^3
     First part being (4m - 3n)
    Answer is (4m - 3n) (16m^2 + 12mn + 9n^2)   not sure!?

4)  27b^3 +  8a^3
     (3b)^3 + (2a)^3
     First part (3b + 2a)
     Answer is (3b + 2a)(9b^2 - 6ab + 4a^2)   not sure how?


With these I have no idea how to do??It's asking to factorise. Do i have to make it equal 0? Because i tried and i couldnt find anythign to work

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

Answer is (x-3)(x+1)(x-2)

2) 6x^3 + 17x^2 - 4x - 3

Answer is (3x+1)(x+3)(2x-1)


3a) Show that, for any constant 'a' and any natural number 'n',    x-a is a factor of x^n  - a^n
       


b) Find the conditions if any on 'n' that required in order that:
i) x + a is a factor of x^n + a^n


Answer -  'n' odd   (not sure how)

ii) x + a is a factor of x^n - a^n

Answer -  'n' even    (not sure how)
 

THANKS!!

[pi]

For the first few, read up on 'long-division of polynomials'. There should be a heading like this in a yr11 or yr12 MM textbook.

Or if you want to be really tricky, google up 'synthetic division', there is a good module here though.


The general methods (for beginners) to factorising a cubic polynomial:
1. Use factor theorem (P(x)...) with numbers 1, -1, 2, -2. etc.
2. Once found, form a bracket (x-a), where 'a' was the found factor
3. Use long division, where the divisor is (x-a)
4. A quadratic factor will hence be found, this can factorised in a number of ways (just sight, completing square, etc.), ONLY IF REAL FACTORS EXIST TO IT!
5. Write final answer in three sets of brackets (Step 4. could be final step if real factors to quadratic do not exist)


For others, remember formulas like this:

[/0]

If you've found 1 factor with the factor theorem, then you can find the other factors pretty easily.

One way is by Polynomial long division to get the other factor, just like if you had to factor a number you could divide it by one factor to find another factor.

For the second way (of which I'm quite fond), let's take your first example. You already know one factor, (x-1). Write the other factor like this: (the other factor must be quadratic because a degree 1 poly times a degree 2 poly is a degree 1+2 = 3 poly).

Then you have:





Now equating coefficients you can solve for a,b,c.


For your last question, simply use the factor theorem. You need to use the fact that if is odd and if n is even. You can then derive Rohitpi's above formulas for general 'n' by using one of the above two methods.

[luffy]

If you've found 1 factor with the factor theorem, then you can find the other factors pretty easily.

One way is by Polynomial long division to get the other factor, just like if you had to factor a number you could divide it by one factor to find another factor.

For the second way (of which I'm quite fond), let's take your first example. You already know one factor, (x-1). Write the other factor like this: (the other factor must be quadratic because a degree 1 poly times a degree 2 poly is a degree 1+2 = 3 poly).

Then you have:





Now equating coefficients you can solve for a,b,c.


For your last question, simply use the factor theorem. You need to use the fact that if is odd and if n is even. You can then derive Rohitpi's above formulas for general 'n' by using one of the above two methods.

Agree with you completely. But, I think you meant:
if is odd
if is even

[hello_kitty]

sorry i'm a bit confused on what you mean :/
my teacher said not to use long division of polynomials :/ ??

[pi]

sorry i'm a bit confused on what you mean :/
my teacher said not to use long division of polynomials :/ ??

Personally, I use /0's method of factorising, as I find it much faster. I'm not sure why you can't use long division though, its the most basic approach, other than plugging random numbers into factor theorem, which could take a while.

I'd just go and learn it anyway

[/0]

If you've found 1 factor with the factor theorem, then you can find the other factors pretty easily.

One way is by Polynomial long division to get the other factor, just like if you had to factor a number you could divide it by one factor to find another factor.

For the second way (of which I'm quite fond), let's take your first example. You already know one factor, (x-1). Write the other factor like this: (the other factor must be quadratic because a degree 1 poly times a degree 2 poly is a degree 1+2 = 3 poly).

Then you have:





Now equating coefficients you can solve for a,b,c.


For your last question, simply use the factor theorem. You need to use the fact that if is odd and if n is even. You can then derive Rohitpi's above formulas for general 'n' by using one of the above two methods.

Agree with you completely. But, I think you meant:
if is odd
if is even

lol yep, my bad :/