How do you find the real factors of #49x^6-140x^4+260x^2-169?#

The sum of the coefficients in

#f(x)=49x^6-140x^4+260x^2-169# is 0.

So, #x-1# is a factor.

As# f(-x) -f(x)#, x+1 is also a factor.

Let #f(x)=49(x^2-1)(x^2+ax+c)(x^2+bx+d)#

Equating coefficients of #x^k, k =5, 4, 3, 2, 1, 0#,, in order,

#49(a+b)=0 to b=-a#.

#49(c+d+ab)=-140 to c+d-a^2=-20/7 #

#49(ad+bc-a-b)=0 to c=d#

#49(cd-ab-c-d)=260 to a^2+c^2-2c=260/49 to a=+-sqrt273/7#

#49(-ad-bc)=0 to c=d#

#49cd=-169 to c^2=169/49 to c = +-13/7.#

The negative c would make #a^2# negative.

So, #c = 13/7#.

Now, #f(x)=#

#49(x-1)(x+1)(x^2+sqrt273/7x+13/7)(x^2-sqrt273/7x+13/7)#

Sign test shows that the number of real factors is restricted to the

maximum of 2. There fore, there are four complex linear factors

occurring in conjugate pairs that combine to form the real qadratic

factors in the answer.,

Note that I have not used #i=sqrt(-1)#, in this method.

#f(x) = 49x^6-140x^4+260x^2-169#

Note that the sum of the coefficients is #0#, that is:

#49-140+260-169 = 0#

and hence #f(1) = 0#, #x = 1# is a zero and #(x-1)# a factor.

Also #f(-1) = f(1) = 0#, so #(x+1)# is a factor too:

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

and we find:

#49x^6-140x^4+260x^2-169 = (x^2-1)(49x^4-91x^2+169)#

We can factor the remaining quartic using the identity:

#(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#

Put #a = sqrt(7)x#, #b = sqrt(13)#

Then #a^2b^2 = 7x^2*13 = 91x^2#

So we want #2-k^2 = -1# and we can choose #k=sqrt(3)#

Then:

#49x^4-91x^2+169 = (a^2-kab+b^2)(a^2+kab+b^2)#

#color(white)(49x^4-91x^2+169) = (7x^2-sqrt(3*7*13)x+13)(7x^2+sqrt(3*7*13)x+13)#

#color(white)(49x^4-91x^2+169) = (7x^2-sqrt(273)x+13)(7x^2+sqrt(273)x+13)#

Putting it all together:

#49x^6-140x^4+260x^2-169#

#=(x-1)(x+1)(7x^2-sqrt(273)x+13)(7x^2+sqrt(273)x+13)#