How do you factor #x^4 - 13x^2 - 30#?

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Answer 1 · 2015-06-13T22:45:16

#x^4-13x^2-30 = (x^2-15)(x^2+2) = (x-sqrt(15))(x+sqrt(15))(x^2+2)#

#x^4-13x^2-30# is quadratic in #x^2#.

To factor it, first find a pair of factors of #30# whose difference is #13#. Such a pair is #15# and #2#.

That gives us:

#x^4-13x^2-30 = (x^2-15)(x^2+2)#

The first of these quadratic factors can be considered a difference of squares:

#x^2-15 = x^2 - (sqrt(15))^2 = (x-sqrt(15))(x+sqrt(15))#

using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

The second factor #x^2+2# has no linear factors with real coefficients. If you were allowed complex coefficients then you could write:

#x^2+2 = (x-sqrt(2)i)(x+sqrt(2)i)#

Answer 2 · 2015-06-13T22:58:01

#(x+sqrt(15))(x-sqrt(15))(x^2+2)# in #RR#

#(x+sqrt(15))(x-sqrt(15))(x+sqrt(2)i)(x-sqrt(2)i)# in #CC#

(If you have never used #i# or #CC#, just ignore this last answer)

I'll call #P(x)# the polynomial.

We notice that #P(x)=Q(x^2)# for some #Q# quadratic polynomial. (In fact, P(x) has only even powers of #x#)

let #X=x^2#

So the equation becomes:

#X^2 -13X-30#, which I know how to factor:

#a,b=(13+-sqrt(169+120))/2=(13+-17)/2 => a=15,b=-2#

(if you had eyes wide open, you could notice that #-13=-(15-2)# and #-30=(-2)15#, so you could avoid tedious calculations)

i.e. #Q(X)=(X-15)(X+2)#

So we factored the polynomial in

#P(x)=(x^2-15)(x^2+2)#

We notice that #(x^2-15)=(x+sqrt(15))(x-sqrt(15))#

If you work in #RR#, then #(x^2+2)# is irreducible (it has no roots)

So #P(x)=(x+sqrt(15))(x-sqrt(15))(x^2+2)#

If you work in #CC#, #x^2 +2 =(x+sqrt(2)i)(x-sqrt(2)i)#

So #P(x)=(x+sqrt(15))(x-sqrt(15))(x+sqrt(2)i)(x-sqrt(2)i)#

(If you have never used #i# or #CC#, just ignore this last two sentences)