# Give two examples how to factor x^6-4x^4-3x^2+12.

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### 1 Answer

One way to factor completely the given expression is to group the first 2 terms and the last 2 terms:

(x^6 - 4x^4) + (- 3x^2+ 12)

We'll factorize by x^4 the first group and by -3 the second group:

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

We'll factorize by (x^2 - 4):

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

We'll recognize that the 1st factor is the difference of 2 squares:

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

So, the result of factorization is:

**x^6-4x^4-3x^2+12 = (x - 2)*(x + 2)*(x^4 - 3) **

The other way factor completely is to group the first and the 3rd terms together and the 2nd and the last terms together.

(x^6 - 3x^2) + (- 4x^4 + 12)

We'll factorize by x^2 the first group and by -4 the second group:

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

We'll factorize by (x^4 - 3):

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

We notice that the result of factorization is:

**x^6-4x^4-3x^2+12 = (x^4 - 3)*(x - 2)*(x + 2)**