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May 16th, 2018, 03:03 PM   #1
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Factorization

What are the steps to factorize: (m^4)-4(m^3)+8(m^2)-8m+4=0? The answer given is (m-(1+i))^2 (m-(1-i))^2. How?

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May 16th, 2018, 07:27 PM   #2
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Assume it's biquadratic:

$$(m^2+am+b)(m^2+cm+d)$$

Expand, equate coefficients and solve the resulting system:

$$a+c=-4$$

$$bd=4$$

$$ad+bc=-8$$

$$ac+b+d=8$$

$$\Rightarrow a=-2,b=2,c=-2,d=2$$

So it factors as $(m^2-2m+2)^2$, which has roots as given.
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May 16th, 2018, 08:33 PM   #3
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Actually, you do not need to assume that the quartic can be factored into two quadratics with real coefficients: the Fundamental Theorem of Algebra guarantees it. The only guess in greg's method is that the leading coefficients are both 1.

Moreover, the symmetry of the quartic might warrant a more extreme guess along the lines of:

$(m^2 + am + b)^2 =$

$m^4 + am^3 + bm^2 + am^3 + a^2m^2 + abm + bm^2 + abm + b^2 =$

$m^4 + 2am^3 +(a^2 + b)m^2 + 2abm + b^2.$

Now equating coefficients gives $2a = -\ 4 \implies a = -\ 2.$

And $2ab = -\ 8 \implies 2(-\ 2)b = -\ 8 \implies b = 2.$

This is a quick way that might not work.
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May 17th, 2018, 02:46 AM   #4
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Quote:
Originally Posted by JeffM1 View Post
The only guess in greg's method is that the leading coefficients are both 1.
No guess. The given quartic is monic.
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May 17th, 2018, 06:36 AM   #5
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Quote:
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No guess. The given quartic is monic.
Greg

I am not arguing; I am curious.

How do we know in advance that the leading coefficients in the quadratics that factor the quartic are each 1? We know that their product is 1, but that leaves open an infinite number of possible pairs of leading coefficients.

Last edited by JeffM1; May 17th, 2018 at 06:38 AM.
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May 17th, 2018, 07:43 AM   #6
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The factorised expression is a product. It's unique only up to multiples of the factors.

$$(x-1)(x+2) =\tfrac12 (2x-2)(x+2) = (2x-2)(\tfrac12x + 1)$$

So we select a coefficient of 1 for each factor to give us a reference point. The result is the same.
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May 17th, 2018, 10:11 AM   #7
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Originally Posted by v8archie View Post
The factorised expression is a product. It's unique only up to multiples of the factors.

$$(x-1)(x+2) =\tfrac12 (2x-2)(x+2) = (2x-2)(\tfrac12x + 1)$$

So we select a coefficient of 1 for each factor to give us a reference point. The result is the same.
Archie

Thanks I think I have it.

$\text {Given: } a_1 = 1\text { and } x = a_1m^4 + a_2m^3 + a_3m^2 + a_4m + a_5.$

$x = (b_1m^2 + b_2m + b_3)(c_1m^2 + c_2m + c_3) \implies b_1c_1 = a_1 \implies b_1c_1 = 1 \implies$

$x = 1 * x = b_1c_1x = c_1(b_1m^2 + b_2m + b_3) * b_1(c_1m^2 + c_2m + c_3) =$

$x = (b_1c_1m^2 + b_2c_1m + b_3c_1)(b_1c_1m^2 + b_1c_2m + b_1c_3) =$

$x = (m^2 + d_1m + d_2)(m^2 + e_1m + e_2).$

If a polynomial to be factored has a leading coefficient of 1, it can always be factored into two terms with leading coefficients of 1. And then of course those terms can be factored, etc.

Learn something new every day.
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