How do you factorise 2x^4 + x^3 - 6x^2 - x - 2?

Question

I was set this by a maths teacher and I can't get how to factorise it. Help!

I'm after a run through of how to do it. My tutor hinted at it being something to do with substitution. E.g. x^4 = y^2 when y = x^2 making it a quadratic equation, etc.

btw this isn't homework its more of a challenge as he gave us the answer which is (x-1)(x-1)(2x+1)(x+2)

i think people are going a little too deep with the use of functions etc. this is part of a section of my course concerning quadratics and factorisation so it'll be something far more simple that the suggestions so far (am grateful for these however)

Thanks for pointing that out regarding the symbols being wrong. I copied it down right but typed it wrong?! Sorry! It should say 2x^4 + x^3 - 6x^2 + x + 2 Ooops

Any ideas?

Answer 1

By trial and error f(-2) = 2*16-8-24+2-2 =0
so x+2 is a factor

now
2x^4+x^3-6x^2-(x+2)
= 2x^4+4x^3-3x^3-6x^2-(x+2) spliting x^3 = 4x^3-3x^3
= 2x^3(x+2)-3x^2(x+2)-(x+2)
=(x+2)(2x^3-3x^2-1)

Now I take the correct form and see

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

now coffecient of x/ coefficient of x^4 = 1
try 1 and -1
with 1 f(1) = 2+1-6+1+2 =0
so (x-1) is a factor
2x^4 -2x^3 +3x^3-3x^2-3x^2+3x-2x+2 ( split the item so that it is = of the previous coffecient + some thing)
= 2x^3(x-1) +3x^2(x-1)-3x(x-1)-2(x-1)
=(x-1)(2x^3+3x^2-3x-2)
again try 1 and -1 and for 1 it is zero so x-1 comes again
2x^3-2x^2+5x^2-5x+2x-2
= 2x^2(x-1)+5x(x-1) +2(x-1)
= (x-1)(2x^2+5x+2)
=(x-1)(
2x+1)(2x+2) last being quaddratic can be factored
hence sloution = (x-1)^2(2x+1)(x+2)

Answer 2

E = 2x^4 + x^3 - 6x^2 - x - 2.

If you are not supposed to know the general formula, then it means that there is a trick. I tried to see if the formula has some obvious roots [because, if r is a root, then (x - r) is a factor].

I tried 0, 1, -1, 2 : No.
-2 : Bingo: 32 + (-8) - 24 - (-2) - 2 = 0.

So we know that (x - (-2)) = (x+2) is a factor.
Now we have to calculate the other factors.
You can do it progressively starting at the highest power of x.
You pose the factor that you want (x+2) and you see what you should add / remove to obtain the term.

2x^4 = 2x^3 (x + 2) - 4 x^3.
[because 2x^3 (x+2) = 2x^4 [what you need] + 4x^3]

So: E = 2x^3 (x + 2) - 4 x^3 + x^3 - 6x^2 - x - 2
= 2x^3 (x+2) - 3 x^3 - 6 x^2 - x - 2.

Do the same with next term (3 x^3), replace and simplify.
-3x^3 = -3x^2 (x+2) + 6x^2

E = 2x^3 (x+2) -3x^2 (x+2) + 6x^2 - 6 x^2 - x - 2.
= 2x^3 (x+2) -3x^2 (x+2) - x - 2

This one is obvious:
-x - 2 = - (x+2)

So:

E = 2x^3 (x+2) -3x^2 (x+2) - (x+2)

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

Edit: Ok, other people were faster.

However you can still factorize it, there is another real root, but it is not obvious and you need the formula to factorize a 3rd degree formula. I guess that the teacher is not asking for that. The root is: R2 = 1/2 + 1/6 V3(81 - 54 . V2(2)) + 1/2 . V3(3 + 2 V2(2))

Where V2(x) is square root of x and V3 is cubic root.
So (x - R2) is another factor... but I guess this is not wanted.

Answer 3

Let f(x) = 2x^4 + x^3 - 6x^2 - x - 2

Using Remainder Theorem,
f(-2) = 32 - 8 - 24 + 2 - 2 = 0
So (x+2) must be a factor of f(x).

Let f(x) = (x+2)(ax^3 + bx^2 + cx + d)
By comparing coefficients,
x^4: a = 2
constant: d = -1
x^3: 1 = 2a + b => 1 = 2(2) + b => b = -3
x^2: -6 = 2b + c => -6 = 2(-3) + c => c = 0

So f(x) = (x+2)(2x^3 - 3x^2 - 1)

The function 2x^3 - 3x^2 - 1 is difficult to solve and when I used an online graphic calculator, I discovered that the value for x is 1.688.

Answer 4

hi.

2x^4 + x^3 - 6x^2 - x - 2= 0

using trial and error method u'll get the first root, x= -2
since x= -2 is a root of the eq., therefore, (x+2) is one of the factor,


let P(X) = 2x^4 + x^3 - 6x^2 - x - 2
P(-2) = 0
therefore,
P(X) = 2x^4 + x^3 - 6x^2 - x - 2 = (x+2)(2x^3 + Ax^2 + Bx -1)

comparing the coefficient of x
2B - 1 = -1
2B = 0
B = 0
therefore,

P(X) = 2x^4 + x^3 - 6x^2 - x - 2 = (x+2)(2x^3 + Ax^2 -1)

comparing the coefficient x^2,
2A = -6
A = -3

therefore,

P(X) = 2x^4 + x^3 - 6x^2 - x - 2 = (x+2)(2x^3 - 3x^2-1)

u can't factorise further as u'll get imaginary roots.

hope i helped a little. Good Luck !!!

Answer 5

For one thing your teacher has given you the wrong answer. If you expand (x-1)(x-1)(x+2)(2x+1) you actually get 2x^4+x^3-6x2+x+2.

The equation you have given can only be factorised as:

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

Which I think has already been stated above.

You need to double check with your teacher that have copied out the assignment correctly.

Answer 6

lol ok anyway this equals
x^2(2x^2+ x-6) - x-2 so this equals
x^2(2x-3)(x+2)-(x+2) =(x+2)(x^2 ( 2x-3) -1 )
so this equals (x+2) (2x^3-3x^2-1)
just simple math

Answer 7

When doing factorization - use this acronym:

Please Excuse My Dear Aunt Sally

Please = parenthesis
Excuse = exponents
My & Dear = multiplication & division
Aunt & Sally = addition and subtraction

If you use this order of operations, then you will get the correct answer.

Good luck!

Answer 8

Gas mark 5 for 2 1/2 hours....then baste and cook for a further 3 hours on gas mark 3....
As you can see , i have no idea , but i bet you smiled at my answer.? :)

Answer 9

Go here. Job done!
http://www.1728.com/quartic.htm

Answer 10

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

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