How to factorize x^4 + x^2 + 1 ?

Answer 1

The trick here is to note that it looks like a quadratic formula. Let's do a little manuevering to see this:

x^4 = (x^2)^2

which means that we can re-write the problem as (x^2)^2 + x^2 + 1. Now, substitute another letter, say t for x^2. We get:

t^2 + t + 1

All I've done is replace the two x^2 values with 't'. Now, it's just a quadratic equation of the form:

ax^2 + bx + c

To factor this, you'll need to find two values that multiply to the last term (1, in this case) and add to the coefficient of the second term (which is also 1, in this case).

With a little investigation, you'll find that two integers do not exist with such properties, which for most math books, means that, though the problem has a solution, it can't be factored.

***THE ANSWER STOPS HERE if you haven't learned about imaginary numbers.***

*However*, in case you really want to factor this, you simply use the quadratic formula to find the values of x that solve this equation:

t = (-b +/- sqrt(b^2 - 4ac))/2a

So, we get that t = (-1 + sqrt(-3))/2 and t = (-1 - sqrt(-3))/2, which are imaginary numbers. These two solutions I will call solutionone and solutiontwo.

So the factorization for f(t) is:

(t - solutionone)(t - solutiontwo) where solutionone and solutiontwo are the values that come out of the quadratic formula:

Then, you plug x^2 back in for t and get:

(x^2 - solutionone)(x^2 - solutiontwo)

These need to be factored as well (ugh) because a linear factorization implies that all the factors have x instead of x^2.

Thank goodness, though, that these are really easy.

It is always the case that x^2 - a can be factored like this:

(x - sqrt(a))(x + sqrt(a)).

Always.

So, this means that the complete factorization for the problem is:

(x - sqrt(solutionone))(x + sqrt(solutionone))(x - sqrt(solutiontwo))(x + sqrt(solutiontwo))

Yeow! Good luck!

Answer 2

Notice that x^4 + x^2 + 1 is a factor of x^6 - 1:

x^6 - 1 = (x^2 - 1) (x^4 + x^2 + 1)

So if you can factor x^6 - 1 from first principles, you can pick out the factorization of x^4 + x^2 + 1 !

[By the way, the answer depends on what field you're being asked to factor it over. It's irreducible over the rational numbers, splits into two quadratic factors over the real numbers, and (like every polynomial) factors completely into linear factors over the complex numbers.]

Answer 3

If we treat x^2 as "x", then we get the equation x^2 + x + 1, which cannot be factorized. Therefore, x^4 + x^2 + 1 cannot be factorized either.

Answer 4

x^2(x^2+1)

Answer 5

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

i used www.quickmath.com

Answer 6

x^4 +x^2 + 1

add x^2 and deduct X^2 in order to keep equations same.

hence, new equation will be,

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

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

now,assume x^2 + 1 = A

= A^2 - x^2 = (A - x) * (A + X)

put the actual value of A in the equation

= (x^2 + 1 - x) * (x^2 + 1 +X) = ANSwer.

Answer 7

The simplest way to factor this is simply (x²+x+1)(x²-x+1).

Answer 8

here's an idea............
its a difficult one...not simple..........so some monuverability is needed.........some tricks..........lets solve for you.......

see we know

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

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

=( x^2-x+1)(x^2+x+1) answer



bcoz (a^2-b^2)=(a+b)(a-b).....

here x^2+1=a, x=b

Answer 9

let x^2=y
y^2+y+1=(y+2)(y-1)
y=-2 or 1
therefore x=1 or square root of -2

Answer 10

Josh got the right idea follow his method

Good work