Factoring x^4-11x^2+1 | Two Methods

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[forcelifeforce]

*@ SyberMath* -- Your answer to your second method is *incorrect.* You have (x^2 + 3x - 1)(x^2 - 3x + 1). The actual answer is
*(x^2 + 3x - 1)(x^2 - 3x - 1).*

[adandap]

The first method gives a complete factorisation because you can write the quartic as (x-r)(x-s)(x-t)(x-u) where r, s, t, u are the four roots you found. The facorisation into the product of quadratics is only the last step if you insist on real coefficients. Of course, you can also factor the quadratics in the second method if that's what you want.

[mcwulf]

Ok that second method is pretty slick 😂

[pietergeerkens]

Very nice!

I figured out the first solution - but will work on remembering the second solution for future use.

[paulnokleberg]

You have an error at 8:28. The last 1 in each factor has to be negative to get the original equation back after distribution.

[mbmillermo]

That was another really cool problem. Your second method was nicer, but the first one really taught me something useful.

[andylee]

Both methods were fun in their own ways!

[j.alaniz]

Hey, the second solution the second factor is incorrect....it should be x square minus 3X minus 1

[leif]

Nobody else noticed he wrote the second term wrong at the end..it should be x^2 -3x minus 1 not x^2 - 3x plus 1...anyone else catch it??

[knutthompson]

Love the first method, gives a lot of insights, but you definitely should do the second method if you have a choice. Way easier.

[goldfing]

My mehod: solve
x^4 - 11x^2 + 1 = 0

Add 11x^2 and subtract 1:
x^4 = 11x^2 - 1

Make the Ansatz (x^2 + z)^2 = x^4 + 2zx^2 + z^2.
Thus add 2zx^2 + z^2 on both sides:

x^4 + 2zx^2 + z^2 = 11x^2 - 1 + 2zx^2 + z^2

The left side is a perfect square, reorder the right side:

(x^2 + z)^2 = (2z + 11)x^2 + (z^2 - 1)

In order to make the right side a perfect square, too,
its discriminant must be zero. With

a = 2z + 11
b = 0
c = z^2 - 1

we get D = b^2 - 4ac = 0 and thus
b^2 = 4ac
(0)^2 = 4*(2z + 11)*(z^2 - 1)
Dividing by four:

(2z + 11)(z^2 - 1) = 0

Either 2z + 11 = 0, thus z = -5.5
Or z^2 = 1, thus z = +1 or z = -1.
All three solutions for z are real,
and they lead to three different factorizations of the biquadratic expression.

Let's try z = +1:
(x^2 + 1)^2 = 13x^2
x^2 + 1 = +-sqrt(13)*x
x^2 -+ sqrt(13)*x + 1 = 0
Ugly quadratic equation and factorization.

Let's try z = -1:
(x^2 - 1)^2 = 9x^2
x^2 - 1 = +-3x
x^2 -+ 3x - 1 = 0
Nice factorization:
x^4 - 11x^2 + 1 = (x^2 + 3x - 1)(x^2 - 3x - 1)

Let's try x = -5.5:
(x^2 - 5.5)^2 = (5.5)^2 - 1
(x^2 - 11/2)^2 = (11/2)^2 - 1
((2x^2 - 11)/2)^2 = 121/4 - 4/4 = 117/4
(2x^2 - 11)/2 = +-sqrt(117) / 2
x^2 - 11/2 = +-sqrt(117) / 2
x^2 - (11 -+ sqrt(117))/2 = 0
Ugly factorization.

[mbmillermo]

I just have to point out one thing: The plus-minus sign has the plus on top: ± .

[KnRamesh-wk]

It was great. But Check last part. Both are +1

[ligion]

Why put items together? Isn't this the final answer? (x-x1)(x-x2)(x-x3)(x-x4)

[OrenLikes]

6:43 "put together": difference gives 6x=0 -> x=0 ?!?, did you mean "multiplying"?