How can i find the roots of this polynomial equation? #(x^4)(a^2)-(x^2)(a^4)-(x^2)+(a^2)#

1 Answer
Sep 7, 2017

Factor by grouping as:

#x^4a^2-x^2a^4-x^2+a^2 = (ax-1)(ax+1)(x-a)(x+a)#

which has zeros:

#x = 1/a#, #" "x=-1/a#, #" "x=a#, #" "x=-a#

Explanation:

The given expression is not an equation, but we can factor it and find its zeros.

The difference of squares identity can be written:

#A^2-B^2 = (A-B)(A+B)#

We can use this with #A=ax, B=1# and with #A=x, B=a# as follows:

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

#color(white)(x^4a^2-x^2a^4-x^2+a^2) = (a^2x^2-1)(x^2-a^2)#

#color(white)(x^4a^2-x^2a^4-x^2+a^2) = ((ax)^2-1^2)(x^2-a^2)#

#color(white)(x^4a^2-x^2a^4-x^2+a^2) = (ax-1)(ax+1)(x-a)(x+a)#

which has zeros:

#x = 1/a#, #" "x=-1/a#, #" "x=a#, #" "x=-a#