factoring

[sysD]

how would one get from:

 

(2a+3)^2 - 8(2a+3) + 12

 

to:

 

[math]

[(2a + 3 ) -2] [ (2a+3) - 6]

[/math]

 

i expanded the first part and got:

 

4a^2 - 4a - 3

[mooeypoo]

Start by opening it up, sysD.

[DrRocket]

how would one get from:

 

(2a+3)^2 - 8(2a+3) + 12

 

to:

 

[math][(2a + 3 ) -2] [ (2a+3) - 6][/math]

 

i expanded the first part and got:

 

4a^2 - 4a - 3

 

Think of "2a+3" as "x".

[sysD]

d'oh. thanks, drrock, that really helped.

[sysD]

Factoring pt2.

 

rather than start a new topic i thought i'd ask my next question here.

 

 

The expression is:

 

6x^2 + 11x - 10

 

 

How would I factor this?

[the asinine cretin]

I just imagine the FOIL process in my mind and try factors that make sense given the numerical coefficients. For encouragement I'll say that after practicing enough you can usually glance at a polynomial like that one and figure it out.

 

 

This page might be helpful. http://www.jamesbren...polynomials.htm

Edited September 12, 2011 by Ceti Alpha V

[sysD]

dag

I'll read that over when I get a few minutes. I'm slowly becoming faster with practice.

For the moment, though... I'm also having troulbe factoring:

 

2x^2 + 4x -3

 

 

 

there has to be a better way than trial and error though

Edited September 12, 2011 by sysD

[metalilty]

Factoring pt2.

 

rather than start a new topic i thought i'd ask my next question here.

 

 

The expression is:

 

6x^2 + 11x - 10

 

 

How would I factor this?

 

dag

I'll read that over when I get a few minutes. I'm slowly becoming faster with practice.

For the moment, though... I'm also having troulbe factoring:

 

2x^2 + 4x -3

 

 

 

there has to be a better way than trial and error though

 

It's a quadratic expression ax^2+bx+c where a=/=1.

 

This is what you do:

 

1. Take the product of ac. Ignore the signs.

2. Write down the possible pairs of factors of absolute value ac.

3. - If c is positive select the two factors of ac whose sum is equal to b.

- If c is negative select the two factors of ac whose difference is equal to b.

4. Write down bx as the sum(or difference) of two factors obtained above.

5. perform grouping

6. write down the factors.

 

6x^2 + 11x - 10

 

ac= 60

 

Factor pairs of ac: 12 and 5; 6 and 10; 1 and 60; 20 and 3; 30 and 2; 15 and 4

 

Since c is negative we use 15x-4x to replace 11x.

 

6x^2 + (15x-4x) - 10

 

= 6x^2 + 15x-4x - 10

 

Group it:

 

(6x^2 + 15x)-(4x +10)

 

Factor out the common factor:

 

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

 

Factor out the common factor again:

 

2x+5(3x-2)

 

usually written as:

 

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

 

As for:

 

2x^2 + 4x -3

 

It's unfactorable.

 

This is how you check "factorability"

 

1.b^2- 4(a) [c]

2. If the result is a perfect square then it's factorable.

 

1.16- 4(2) (-3)=40

2. √40=2√10

 

unfactorable.

 

Lets check the previous example:

 

6x^2 + 11x - 10

 

1. 121- 4(6) (-10)=361

2. √361=19

 

factorable.

Edited September 13, 2011 by metalilty

[sysD]

damn, thanks. that was an awesome response.

[the asinine cretin]

+1