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 July 27th, 2017, 10:32 AM #1 Member   Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 Curious about high school factoring When we have a quadratic polynomial of the form: $\displaystyle ax^2 + bx + c$ WHY is it factored in such a way that the newly split $\displaystyle b$ term MUST MULTIPLY to make $\displaystyle ax^2*c$? For example, if we have $\displaystyle 4x^2+4x+1$, $\displaystyle 4x^2 +2x+2x+1$ $\displaystyle =2x(2x+1)+ 1(2x+1)$ $\displaystyle =(2x+1)^2$
 July 27th, 2017, 10:56 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,595 Thanks: 1493 (ax + b)(cx + d) = acx² + adx + bcx + bd implies (adx)(bcx) = (ac)x²(bd). Thanks from Antoniomathgini
July 27th, 2017, 02:25 PM   #3
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Quote:
 Originally Posted by Antoniomathgini When we have a quadratic polynomial of the form: $\displaystyle ax^2 + bx + c$ WHY is it factored in such a way that the newly split $\displaystyle b$ term MUST MULTIPLY to make $\displaystyle ax^2*c$? For example, if we have $\displaystyle 4x^2+4x+1$, $\displaystyle 4x^2 +2x+2x+1$ $\displaystyle =2x(2x+1)+ 1(2x+1)$ $\displaystyle =(2x+1)^2$
We cannot GENERALLY factor a quadratic that way.

The general way to factor a quadratic is

$ax^2 + bx + c = a * \left ( x + \dfrac{b - d}{2a} \right ) * \left ( x + \dfrac{b + d}{2a} \right ), \text { where } d = \sqrt{b^2 - 4ac}.$

Let's PROVE that first.

$a * \left ( x + \dfrac{b - d}{2a} \right ) * \left ( x + \dfrac{b + d}{2a} \right ) =$

$a * \left ( x * \left \{x + \dfrac{b + d}{2a} \right \} + \dfrac{b - d}{2a} * \left \{ x + \dfrac{b + d}{2a} \right \} \right ) =$

$a * \left ( x^2 + \dfrac{bx + dx}{2a} + \dfrac{bx - dx}{2a} + \dfrac{b^2 - d^2}{4a^2} \right ) =$

$a * \left ( x^2 + \dfrac{bx + dx + bx - dx}{2a} + \dfrac{b^2 - \left ( \sqrt{b^2 - 4ac} \right )^2}{4a^2} \right ) =$

$a \left ( x^2 + \dfrac{\cancel 2 bx}{\cancel 2 a} + \dfrac{b^2 - (b^2 - 4ac)}{4a^2} \right ) =ax^2 + \dfrac{\cancel a bx}{\cancel a} + \dfrac{\cancel{4a^2}c}{\cancel{4a^2}} = ax^2 + bx + c.$

There is a very important special case.

$b^2 - 4ac = 0 \implies d = 0 \implies$

$ax^2 + bx + c = a * \left (x + \dfrac{b + 0}{2a} \right ) * \left ( x + \dfrac{b - 0}{2a} \right ) =$

$a * \left (x + \dfrac{b}{2a} \right )^2 = \left \{ \sqrt{a} * \left ( x + \dfrac{b}{2a} \right ) \right \}^2 = \left ( x\sqrt{a} + \dfrac{b\sqrt{a}}{2a} \right )^2.$

Your example is a very special case of the special case.

$a = 4,\ b = 4,\ c = 1 \implies b^2 - 4ac = 4^2 - 4 * 4 * 1 = 16 - 16 = 0.$

$So\ 4x^2 + 4x + 1 = \left ( x * \sqrt{4} + \dfrac{4 * \sqrt{4}}{2 * 4} \right )^2 = \left ( 2x + \dfrac{4 * 2}{8} \right )^2 = (2x + 1)^2.$

 July 27th, 2017, 02:32 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,595 Thanks: 1493 It was not suggested that the term in x has to be split into equal halves.
July 27th, 2017, 03:37 PM   #5
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Quote:
 Originally Posted by skipjack It was not suggested that the term in x has to be split into equal halves.
That was exactly what the example in the original post did. I was addressing that post, not being contrary with you. My fault: I should have quoted the original post to clarify what I was responding to.

 July 31st, 2017, 06:44 PM #6 Newbie   Joined: Jul 2017 From: New Zealand Posts: 7 Thanks: 0 Just wondering what is the easiest method to factorise ax^2 +bx + c at high school level... many thanks
 July 31st, 2017, 06:55 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,745 Thanks: 1001 Math Focus: Elementary mathematics and beyond There are many methods that can be applied. What the "easiest" is would depend on what values we have for a, b and c.
 July 31st, 2017, 07:33 PM #8 Newbie   Joined: Jul 2017 From: New Zealand Posts: 7 Thanks: 0 For example something like 3x^2 + 19x + 6 Thanks
 August 2nd, 2017, 09:12 AM #9 Member   Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 $\displaystyle 3x^2 + 19x + 6$ $\displaystyle = 3x^2+x+18x+6$ $\displaystyle = x(3x+1)+6(3x+1)$ $\displaystyle =(x+6)(3x+1)$
August 2nd, 2017, 10:17 AM   #10
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Quote:
 Originally Posted by JeffM1 That was exactly what the example in the original post did.
That post didn't state that splitting exactly in half was part of the method used, as distinct from being just an insignificant (for the purposes of the question) consequence of the fact that the example used happened to be a perfect square. The post had emphasized that the product of the terms resulting from the split must be $ax^2*c$, thus implying that achieving this should be part of a general method, or was already part of the method taught.

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