How do you factor #x^4 - x^2 - 20#?

3 Answers
Jun 25, 2016

Answer:

#(x-2)(x+2)(x-i sqrt(5))(x+i sqrt (5))#

Explanation:

create new variable equal to #x^2#. Call it y.

Your equation now is #y^2 - y - 20# whose roots are #y=4# and #y=-5#
#y=4# is ok. That is #x=pm 2#. #y=-5# leads to imaginary roots #x= pm i sqrt(5)#

Jun 25, 2016

Answer:

#x^4-x^2-20=(x^2+4)(x^2-5)#
= #(x+sqrt5)(x-sqrt5)(x^2+4)#
= #(x+sqrt5)(x-sqrt5)(x+2i)(x-2i)#

Explanation:

#x^4-x^2-20#

= #x^4-5x^2+4x^2-20#

= #x^2(x^2-5)+4(x^2-5)#

= #(x^2+4)(x^2-5)#

These are factors with rational coefficients. If we are allowed to have monomials with irrational coefficients too, #x^2-5=(x+sqrt5)(x-sqrt5)# and

#x^4-x^2-20=(x+sqrt5)(x-sqrt5)(x^2+4)#

If we also allow monomials with complex numbers, then factors of #x^2+4# are #(x+2i)(x-2i)# and

#x^4-x^2-20=(x+sqrt5)(x-sqrt5)(x+2i)(x-2i)#

Jun 25, 2016

Answer:

#Exp.=(x^2-5)(x^2+4).#

Explanation:

Expression #= x^4-x^2-20.#

Note that #5xx4=20, 5-4=1#

#:. Exp.=x^4-5x^2+4x^2-20=x^2(x^2-5)+4(x^2-5)=(x^2-5)(x^2+4).#

If, at the first glance, it may not strike that the given Exp. is a qudr. poly, of #x^2#, then, we may use substitution #x^2=t# in it and see that it is #t^2-t-20 = (t-5)(t+4)=(x^2-5)(x^2+4),# as before!.