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October 3rd, 2017, 11:17 PM  #1 
Newbie Joined: Oct 2017 From: Dubai Posts: 2 Thanks: 0  Factoring polynomials
Can someone please explain this to me in detail, esp with the cross multiplication thing. Also, where does the (5y3) go in the 3rd step. Thanksss 
October 4th, 2017, 02:29 PM  #2  
Senior Member Joined: May 2016 From: USA Posts: 802 Thanks: 318  Quote:
Part b seems to be a simple case of substituting a variable. You can see that the expression is similar to a quadratic in form, and it becomes a quadratic with an obvious substitution. $\text {LET: } u = x^2 \implies$ $x^4  10x^2 + 9 = u^2  10u + 9 = (u  1)(u  9) = (x^2  1)(x^2  3) \implies $ $x^4  10x^2 + 9 = (x + 1)(x  1)(x + 3)(x  3).$ The expression in part a is also similar in form to a quadratic, but a substitution is not obvious. The missing text may explain things differently, but what seems to be going on is a method for finding a possible substitution. By the fundamental theorem of algebra, we know that $\exists\ a,\ b,\ c,\ d,\ e,\ k \in \mathbb C \text { such that }\\ 2x^2 + 5xy + 3y^2  3x  5y  2 = \{ax + (by + c)\}\{(dx + (ey + k)\}.$ $\text {A factoring in integers} \implies a,\ b,\ c,\ d,\ e,\ k \in \mathbb Z.$ $a * d = 2 \implies sgn(a) = sgn(d).$ So, without loss of generality, we can say a = 1 and d = 2. $\therefore 2x^2 + 5xy + 3y^2  3x  5y  2 = \{x + (by + c)\}\{(2x + (ey + k)\}.$ Rearranging, $2x^2 + 5xy + 3y^2  3x  5y  2 =\\ 2x^2 + 5xy  3x + (3y^2  5y  2) = 2x^2 + x(5y  3) + (3y + 1)(y  2).$ $\therefore \text {a factoring in integers } \implies 2x^2 + x(5y  3) + (3y + 1)(y  2) = \\ \{x + (3y + 1)\}\{2x + (y  2)\} \text { or } \{x + (y  2)\}\{2x + (3y + 1)\}.$ Now neither will work if there is no factorization in integers, but exactly one will work if there is a factorization in integers. $\{x + (3y + 1)\}\{2x + (y  2)\} =\\ 2x^2 + x(y  2) + 2x(3y + 1) + (3y + 1)(y  2) =\\ 2x^2 + xy  2x + 6xy + 2x + (3y + 1)(y  2) =\\ 2x^2 + 7xy +(3y + 1)(y  2),\text {which is INCORRECT}.$ $\{x + (y  2)\}\{2x + (3y + 1)\} = \\ 2x^2 + x(3y + 1) + 2x(y  2) + (3y + 1)(y  2) =\\ 2x^2 + 5xy  3x + (3y + 1)(y  2), \text { which is CORRECT}.$ $\therefore 2x^2 + 5xy + 3y^2  3x  5y  2 =\\ \{x + (y  2)\}\{2x + (3y + 1)\}.$  
October 4th, 2017, 05:55 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,036 Thanks: 1394 
If p = y  2 and q = 3y + 1, consider 2x² + 2px + qx + pq = 2x² + (2p + q)x + pq = (x + p)(2x + q). 
October 7th, 2017, 03:33 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,719 Thanks: 699 
Going the other way, I presume you know the "distributive laws": a(c+ d)= ac+ ad and (a+ b)c= ac+ bc. To multiply (a+ b)(c+ d), use that twice. First, (a+ b)(c+ d)= a(c+ d)+ b(c+ d) and then ac+ ad+ bc+ bd. The "cross multiplication" is the "ad+ bc" part. (x+ a)(x+ b)= x(x+ b)+ a(x+ b)= x^2+ bx+ ax+ ab= x^2+ (a+ b)x+ ab. 
October 9th, 2017, 07:48 AM  #5 
Newbie Joined: Oct 2017 From: Dubai Posts: 2 Thanks: 0 
Thank you so much for explaining!! Really appreciate it 

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factoring, polynmials, polynomials 
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