How do you factor # x^4 + 4x^3 - 22x^2 + 4x + 16#?

1 Answer
George C.
Aug 28, 2016

#x^4+4x^3-22x^2+4x+16#

#=(x+3-sqrt(17))(x+3+sqrt(17))(x-1-sqrt(3))(x-1+sqrt(3))#

Explanation:

#f(x) = x^4+4x^3-22x^2+4x+16#

We can try the rational roots theorem, which tells us that any rational zeros of #f(x)# are expressibl in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #16# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-4, +-8, +-16#

None of these works, so #f(x)# has no rational zeros and therefore no linear factors with integer coefficients.

How about quadratic factors?

Is there a pair of quadratic factors with integer coefficients?

We can simplify the problem by making a linear substitution.

#x^4+4x^3-22x^2+4x+16#

#=(x+1)^4-28(x+1)^2+56(x+1)-13#

#=t^4-28t^2+56t-13#

where #t=x+1#

If this simplified quartic has quadratic factors with integer coefficients then they must take one of the following forms:

  • #(t^2+at+13)(t^2-at-1)#

  • #(t^2+at-13)(t^2-at+1)#

With a little comparing of coefficients we find:

#t^4-28t^2+56t-13 = (t^2+4t-13)(t^2-4t+1)#

Hence:

#x^4+4x^3-22x^2+4x+16#

#=((x+1)^2+4(x+1)-13)((x+1)^2-4(x+1)+1)#

#=(x^2+6x-8)(x^2-2x-2)#

#=((x+3)^2-17)((x-1)^2-3)#

#=(x+3-sqrt(17))(x+3+sqrt(17))(x-1-sqrt(3))(x-1+sqrt(3))#

Related questions
See all questions in Factoring Completely
Impact of this question
1197 views around the world
You can reuse this answer
Creative Commons License