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

##### 2 Answers

#### Explanation:

The sum of the coefficients in

So,

As

Let

Equating coefficients of

The negative c would make

So,

Now,

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

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

#### Explanation:

Note that the sum of the coefficients is

#49-140+260-169 = 0#

and hence

Also

#(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

Then

So we want

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)#