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

Oct 21, 2016

$49 \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + \frac{\sqrt{273}}{7} x + \frac{13}{7}\right) \left({x}^{2} - \frac{\sqrt{273}}{7} x + \frac{13}{7}\right)$

#### Explanation:

The sum of the coefficients in

$f \left(x\right) = 49 {x}^{6} - 140 {x}^{4} + 260 {x}^{2} - 169$ is 0.

So, $x - 1$ is a factor.

As$f \left(- x\right) - f \left(x\right)$, x+1 is also a factor.

Let $f \left(x\right) = 49 \left({x}^{2} - 1\right) \left({x}^{2} + a x + c\right) \left({x}^{2} + b x + d\right)$

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

$49 \left(a + b\right) = 0 \to b = - a$.

$49 \left(c + d + a b\right) = - 140 \to c + d - {a}^{2} = - \frac{20}{7}$

$49 \left(a d + b c - a - b\right) = 0 \to c = d$

$49 \left(c d - a b - c - d\right) = 260 \to {a}^{2} + {c}^{2} - 2 c = \frac{260}{49} \to a = \pm \frac{\sqrt{273}}{7}$

$49 \left(- a d - b c\right) = 0 \to c = d$

$49 c d = - 169 \to {c}^{2} = \frac{169}{49} \to c = \pm \frac{13}{7.}$

The negative c would make ${a}^{2}$ negative.

So, $c = \frac{13}{7}$.

Now, $f \left(x\right) =$

$49 \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + \frac{\sqrt{273}}{7} x + \frac{13}{7}\right) \left({x}^{2} - \frac{\sqrt{273}}{7} x + \frac{13}{7}\right)$

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

Note that I have not used $i = \sqrt{- 1}$, in this method.

Oct 29, 2016

$49 {x}^{6} - 140 {x}^{4} + 260 {x}^{2} - 169$

$= \left(x - 1\right) \left(x + 1\right) \left(7 {x}^{2} - \sqrt{273} x + 13\right) \left(7 {x}^{2} + \sqrt{273} x + 13\right)$

#### Explanation:

$f \left(x\right) = 49 {x}^{6} - 140 {x}^{4} + 260 {x}^{2} - 169$

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

$49 - 140 + 260 - 169 = 0$

and hence $f \left(1\right) = 0$, $x = 1$ is a zero and $\left(x - 1\right)$ a factor.

Also $f \left(- 1\right) = f \left(1\right) = 0$, so $\left(x + 1\right)$ is a factor too:

$\left(x - 1\right) \left(x + 1\right) = {x}^{2} - 1$

and we find:

$49 {x}^{6} - 140 {x}^{4} + 260 {x}^{2} - 169 = \left({x}^{2} - 1\right) \left(49 {x}^{4} - 91 {x}^{2} + 169\right)$

We can factor the remaining quartic using the identity:

$\left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right) = {a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}$

Put $a = \sqrt{7} x$, $b = \sqrt{13}$

Then ${a}^{2} {b}^{2} = 7 {x}^{2} \cdot 13 = 91 {x}^{2}$

So we want $2 - {k}^{2} = - 1$ and we can choose $k = \sqrt{3}$

Then:

$49 {x}^{4} - 91 {x}^{2} + 169 = \left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right)$

$\textcolor{w h i t e}{49 {x}^{4} - 91 {x}^{2} + 169} = \left(7 {x}^{2} - \sqrt{3 \cdot 7 \cdot 13} x + 13\right) \left(7 {x}^{2} + \sqrt{3 \cdot 7 \cdot 13} x + 13\right)$

$\textcolor{w h i t e}{49 {x}^{4} - 91 {x}^{2} + 169} = \left(7 {x}^{2} - \sqrt{273} x + 13\right) \left(7 {x}^{2} + \sqrt{273} x + 13\right)$

Putting it all together:

$49 {x}^{6} - 140 {x}^{4} + 260 {x}^{2} - 169$

$= \left(x - 1\right) \left(x + 1\right) \left(7 {x}^{2} - \sqrt{273} x + 13\right) \left(7 {x}^{2} + \sqrt{273} x + 13\right)$