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 May 16th, 2018, 03:03 PM #1 Member   Joined: Apr 2017 From: India Posts: 34 Thanks: 0 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? Last edited by shashank dwivedi; May 16th, 2018 at 03:08 PM.
 May 16th, 2018, 07:27 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond 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.
 May 16th, 2018, 08:33 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,148 Thanks: 479 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.
May 17th, 2018, 02:46 AM   #4
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Quote:
 Originally Posted by JeffM1 The only guess in greg's method is that the leading coefficients are both 1.
No guess. The given quartic is monic.

May 17th, 2018, 06:36 AM   #5
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Quote:
 Originally Posted by greg1313 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.

 May 17th, 2018, 07:43 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra 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. Thanks from JeffM1
May 17th, 2018, 10:11 AM   #7
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Quote:
 Originally Posted by v8archie 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.

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