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Commentary for "Factoring Quadratics"

agentmulder

Active member
Feb 9, 2012
33
Moderator edit: This commentary topic pertains to the following tutorial:

http://mathhelpboards.com/math-notes-49/factoring-quadratics-3396.html
I agree with everything MarkFL has posted so i'll just give a quick 'scheme' leaving out the details.

Another approach is to factor by grouping, consider,

$6x^2 - 17x - 45 $

$(6)(-45) = -270$

Now find 2 integers that multiply to -270 and sum to -17,

-27 with 10 work

rewrite $-17x$ as $-27x + 10x $

$6x^2 - 27x + 10x - 45 $

use grouping, paying mind to the signs,

$ (6x^2 - 27x) + (10x -45)$

factor,

$3x(2x - 9) + 5(2x - 9)$

$(2x - 9)(3x + 5)$

You can multiply this out using FOIL to confirm the factoring is correct.

:)
 
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mathmaniac

Active member
Mar 4, 2013
188
Re: Factoring Quadratics

$6x^2 - 17x - 45 $

$(6)(-45) = -270$

Now find 2 integers that multiply to -270 and sum to -17,

-27 with 10 work
How to do it:

We have to find p and q such that pq=6 x (-45)= 2 x 3 x 3 x 3 x 5

and p+q=-17

We observe that 17 is a prime.So p and q cannot have common factors (otherwise p+q will be have a common factor).

So either p or q will have all the 3 "3"s.Let it be p.
We have a 2 and 5 remaining which can go into either p or q.Let us first try both of them in q.
So we have p=27 and q=10

One and only one of p and q is negative and also the greater of p and q has to have the sign "-" (because the sum is negative).

So we have p + q = -27 + 10 = -17 and it works.If it hadn't worked then we had to exchange the 2 and 5 a couple of times until we get it right.

If p + q had a common factor,then we had to allow it to be the GCD of p and q and continue this.
 

agentmulder

Active member
Feb 9, 2012
33
Re: Factoring Quadratics

How to do it:

We have to find p and q such that pq=6 x (-45)= 2 x 3 x 3 x 3 x 5

and p+q=-17

We observe that 17 is a prime.So p and q cannot have common factors (otherwise p+q will be have a common factor).

So either p or q will have all the 3 "3"s.Let it be p.
We have a 2 and 5 remaining which can go into either p or q.Let us first try both of them in q.
So we have p=27 and q=10

One and only one of p and q is negative and also the greater of p and q has to have the sign "-" (because the sum is negative).

So we have p + q = -27 + 10 = -17 and it works.If it hadn't worked then we had to exchange the 2 and 5 a couple of times until we get it right.

If p + q had a common factor,then we had to allow it to be the GCD of p and q and continue this.
These are nice observations, i especially like the part of all 3's must belong to p or q. As the product AC gets large your observations can in fact reduce the work involved trying to find p and q.

:)
 

Petrus

Well-known member
Feb 21, 2013
739
Re: Factoring Quadratics

Hello,
I can add a method :)
Let's say we want to factor \(\displaystyle p(x)=-4x^2+24x-32\) we can use this formula \(\displaystyle p(x)=k*(x-a)(x-b)\)
we want to calculate crittical point (without derivate). In Sweden we learned a formula caled "pq-formula"(too calculate critical point in second polynom) as you can see there cant be any number or negative number infront of \(\displaystyle x^2\) so we start to break out -4 then we got \(\displaystyle -4(x^2-6x+8)\)
now we calculate the critical point \(\displaystyle x^2-6x+8=0\) and we get \(\displaystyle x_1=4\) and \(\displaystyle x_2=2\) now we use the formula \(\displaystyle p(x)=k*(x-a)(x-b)\) where k is our constant and a,b our crit point so we got \(\displaystyle p(x)=-4*(x-4)(x-2)\)
here is pq formula:
 
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