How do you factor #2n^4 - 9n^2 +4 = 0#?

2 Answers
Jun 30, 2018

Answer:

#(2n^2-1)(n-2)(n+2)=0#

Explanation:

Start by noticing that there are no #n^3# or #n# terms in the expression on the LHS to be factored. Thus we can attempt factorisation as a quadratic in #n^2# of the form #(an^2+b)(cn^2+d)#=0

If we are looking for integer factorisations, we immediately know #a# and #c# - one must be 2, the other 1 in order to get the correct #n^4# coefficient. So our desired form is #(2n^2+b)(n^2+d)#=0

Multiply this out:
#2n^4+(b+2d)n^2+bd=0#
where
#b+2d=-9# and #bd=4#.

We now known that #b# and #d# must have the same sign and that sign must be negative. So, assuming integer solutions, the second equation tells us that #b# and #d# must either be #(-1,-4)#, #(-4,-1)# or #(-2,-2)#. Trialling these in the first equation, we see that #b=-1#, #d=-4#.

So our quadratic factorisation is
#(2n^2-1)(n^2-4)=0#

Can we factor this further? Yes. The second bracket is a difference of two squares, #n^2-4=(n-2)(n+2)#. The first requires a square root to express the same, so we leave it as a quadratic. Thus our full integer factorisation is

#(2n^2-1)(n-2)(n+2)=0#

The question doesn't ask it, but if we wish to solve this equation, then the four quartic roots are immediately available:
#n=(-1/sqrt(2),+1/sqrt(2),-2,+2)#
It is good practice to verify that each of these is in fact a solution to the original equation.

Jun 30, 2018

Answer:

#(n^2-4)(2n^2-1)#

Explanation:

Re-Write the given equation as follows:

#(2.(n^4)) - 3^2n^2)+4=0#

#(2n^4 - 3^2.n^2)+4=0#

Step 1: Trying to factor by splitting the middle term

Find two factors of #8# whose sum equals the coefficient of the middle term, which is # -9# .

#-8 + -1= -9# and #-8 xx -1 = 8#

Step 2: Rewrite the polynomial splitting the middle term using the two factors found in step 1 above, #-8# and #-1#

#2n^4-8n^2-n^2+4#

Pulling out the common factors

#2n^2(n^2-4) - 1(n^2-4)#

#(n^2-4)(2n^2-1)# ----> these are the factors!