Thursday, January 22, 2015

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

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)

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