May 16th, 2018, 04:03 PM  #1 
Member Joined: Apr 2017 From: India Posts: 39 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(1i))^2. How?
Last edited by shashank dwivedi; May 16th, 2018 at 04:08 PM. 
May 16th, 2018, 08:27 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,898 Thanks: 1093 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 $(m^22m+2)^2$, which has roots as given. 
May 16th, 2018, 09:33 PM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,249 Thanks: 515 
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, 03:46 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,898 Thanks: 1093 Math Focus: Elementary mathematics and beyond  
May 17th, 2018, 07:36 AM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,249 Thanks: 515  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 07:38 AM. 
May 17th, 2018, 08:43 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,551 Thanks: 2554 Math Focus: Mainly analysis and algebra 
The factorised expression is a product. It's unique only up to multiples of the factors. $$(x1)(x+2) =\tfrac12 (2x2)(x+2) = (2x2)(\tfrac12x + 1)$$ So we select a coefficient of 1 for each factor to give us a reference point. The result is the same. 
May 17th, 2018, 11:11 AM  #7  
Senior Member Joined: May 2016 From: USA Posts: 1,249 Thanks: 515  Quote:
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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