How do you factor #x^4-9x^2+18#?

1 Answer
Jun 4, 2016

Answer:

#(t-6)(t-3)#

Explanation:

Say:
#x^2=t#

Then:
#t^2-9t+18#

After this you can factorise for t:
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So factorisation for t would be: #(t-6)(t-3)#.

Now, we have to find x; in order to do that, we have to find the roots of t's. Since this is a quadratic equation it has to be equal zero, so:
#(t-6)(t-3)=0#
Accordingly: #t-6=0# and #t-3=0#. From here we get: #t1=6# and #t2=3#.

Let's find x1 and x2:
#t1=x^2#
#6=x^2#
#+-sqrt6=x1,x2# (two roots: #x1=sqrt6#, #x2=-sqrt6# )

and x
#t2=x^2#
#3=x^2#
#+-sqrt3=x3,x4# (two roots: #x3=sqrt3#, #x4=-sqrt4# )

Why #+-#? Because any number square-rooted can be negative or positive. RECALL: #(-8)^2=64# as well as #8^2=64# but not #-8^2# is not equal #64#.

To get the correct answers we have to find which of the four roots fits into the equation to satisfy it to be zero.

First, let's try with the first one:
#(sqrt6)^4 - 9(sqrt6)^2+18=0 => 36-9*6+18=0 => 36-54+18=0=> 0=0#

Secondly,
#(-sqrt6)^4-9(-sqrt6)^2+18=0 =>36-9*6+18=>0=0#

Thirdly,
#(sqrt3)^4-9(sqrt3)^2+18=0=>9-9*3+18=0=>9-27+18=0=>0=0#

Lastly,
#(-sqrt3)^4-9(-sqrt3)^2+18=0=>9-9*3+18=0=>0=0#

So the solutions are: #-sqrt3,sqrt3,-sqrt6,sqrt6#