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

Question

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

2 Answers

Impact of this question

2166 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-21T16:38:19

$49 (x - 1) (x + 1) (x^{2} + \frac{\sqrt{273}}{7} x + \frac{13}{7}) (x^{2} - \frac{\sqrt{273}}{7} x + \frac{13}{7})$ $49(x-1)(x+1)(x^2+sqrt273/7x+13/7)(x^2-sqrt273/7x+13/7)$

Explanation:

The sum of the coefficients in

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

So, $x - 1$ $x-1$ is a factor.

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

Let $f (x) = 49 (x^{2} - 1) (x^{2} + a x + c) (x^{2} + b x + d)$ $f(x)=49(x^2-1)(x^2+ax+c)(x^2+bx+d)$

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

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

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

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

$49 (c d - a b - c - d) = 260 \to a^{2} + c^{2} - 2 c = \frac{260}{49} \to a = \pm \frac{\sqrt{273}}{7}$ $49(cd-ab-c-d)=260 to a^2+c^2-2c=260/49 to a=+-sqrt273/7$

$49 (- a d - b c) = 0 \to c = d$ $49(-ad-bc)=0 to c=d$

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

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

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

Now, $f (x) =$ $f(x)=$

$49 (x - 1) (x + 1) (x^{2} + \frac{\sqrt{273}}{7} x + \frac{13}{7}) (x^{2} - \frac{\sqrt{273}}{7} x + \frac{13}{7})$ $49(x-1)(x+1)(x^2+sqrt273/7x+13/7)(x^2-sqrt273/7x+13/7)$

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 $i = \sqrt{- 1}$ $i=sqrt(-1)$ , in this method.

Answer 2 · 2016-10-29T22:11:44

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

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

Explanation:

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

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

$49 - 140 + 260 - 169 = 0$ $49-140+260-169 = 0$

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

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

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

and we find:

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

We can factor the remaining quartic using the identity:

$(a^{2} - k a b + b^{2}) (a^{2} + k a b + b^{2}) = a^{4} + (2 - k^{2}) a^{2} b^{2} + b^{4}$ $(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4$

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

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

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

Then:

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

$49 x^{4} - 91 x^{2} + 169 = (7 x^{2} - \sqrt{3 \cdot 7 \cdot 13} x + 13) (7 x^{2} + \sqrt{3 \cdot 7 \cdot 13} x + 13)$ $color(white)(49x^4-91x^2+169) = (7x^2-sqrt(3*7*13)x+13)(7x^2+sqrt(3*7*13)x+13)$

$49 x^{4} - 91 x^{2} + 169 = (7 x^{2} - \sqrt{273} x + 13) (7 x^{2} + \sqrt{273} x + 13)$ $color(white)(49x^4-91x^2+169) = (7x^2-sqrt(273)x+13)(7x^2+sqrt(273)x+13)$

Putting it all together:

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

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

Science

Math

Humanities

... and beyond

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

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the real factors of 49x^6-140x^4+260x^2-169?49x6−140x4+260x2−169?49x^6-140x^4+260x^2-169?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

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