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October 4th, 2017, 12:17 AM   #1
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Factoring polynomials

Can someone please explain this to me in detail, esp with the cross multiplication thing. Also, where does the (5y-3) go in the 3rd step. Thanksss
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October 4th, 2017, 03:29 PM   #2
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Quote:
Originally Posted by Keikix View Post
Can someone please explain this to me in detail, esp with the cross multiplication thing. Also, where does the (5y-3) go in the 3rd step. Thanksss
You have given no context, and the text is not good English. Consequently, I can not be certain whether this answer is helpful.

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)\}.$
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October 4th, 2017, 06:55 PM   #3
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If p = y - 2 and q = 3y + 1, consider
2x² + 2px + qx + pq
= 2x² + (2p + q)x + pq
= (x + p)(2x + q).
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October 7th, 2017, 04:33 AM   #4
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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.
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October 9th, 2017, 08:48 AM   #5
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Thank you so much for explaining!! Really appreciate it
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