• Jan 8th 2011, 11:47 AM
Natasha1
Could someone explain to me as simply as possible how to factorise

2x^2 + x - 3

I know the answer is (2x + 3) (x - 1) from my textbook but I don't understand the method used.
• Jan 8th 2011, 11:51 AM
snowtea
Quote:

Originally Posted by Natasha1
Could someone explain to me as simply how to simplify

2x^2 + x -3

I know the answer is (2x + 3) (x - 1) from my textbook but I don't understand the method used.

I think the method is called guess and check :)

A more procedural method would be solve for the roots r1,r2 of Ax^2 + Bx + C = 0.
Then the equation factors into A(x - r1)(x - r2).
• Jan 8th 2011, 11:53 AM
Natasha1
I saw on the net a method called the AC method but that is also confusing
• Jan 8th 2011, 11:55 AM
Quote:

Originally Posted by Natasha1
Could someone explain to me as simply as possible how to factorise

2x^2 + x - 3

I know the answer is (2x + 3) (x - 1) from my textbook but I don't understand the method used.

It comes from an understanding that

$(2x+a)(x+b)=2x(x+b)+a(x+b)=2x^2+2xb+ax+ab=2x^2+(a+ 2b)x+ab$

so you are looking for the factors of $-3$ for which one of them plus twice the other sum to 1.

They will be 3 and twice $-1.$
• Jan 8th 2011, 11:56 AM
Also sprach Zarathustra
Quote:

Originally Posted by snowtea
I think the method is called guess and check :)

A more procedural method would be solve for the roots r1,r2 of Ax^2 + Bx + C = 0.
Then the equation factors into A(x - r1)(x - r2).

snotew mentioned "guess"...

First guess that one root is x=1(it is an easy guess) then do long division: {2x^2 + x -3}/{x-1} you will get x+3/2
• Jan 8th 2011, 11:58 AM
Wilmer
Nawwww....I say if not apparent, use the quadratic formula...but that's me!
• Jan 8th 2011, 11:59 AM
dwsmith
$(\pm\alpha x\pm\beta)(\pm\lambda x\pm\phi)$

$\beta*\phi=-3\Rightarrow 3*1=3$ Now need to figure out which one is negative.

$\alpha*\lambda=2\Rightarrow 2*1=2$

$\alpha*\phi+\beta*\lambda=1$

Now you need to decide if 3 is beta and 1 is phi, is alpha 1 or 2 and is lambda 1 or 2?
• Jan 8th 2011, 12:02 PM
TheCoffeeMachine
Quote:

Originally Posted by Also sprach Zarathustra
First guess that one root is x=1(it is an easy guess) then do long division: {2x^2 + x -3}/{x-1} you will get x+3/2

Indeed. Or instead of long division write:

$2x^2+x-3 = (x-1)(ax+b)$ -- it's clear by inspection that $b = 3$ and $a = 2$.
• Jan 8th 2011, 12:10 PM
TheCoffeeMachine
Quote:

Originally Posted by Natasha1
2x^2 + x - 3

I know the answer is (2x + 3) (x - 1) from my textbook but I don't understand the method used.

Isn't this just the problem from this thread with just the letter $w$ replaced with $x$?
• Jan 8th 2011, 12:12 PM
Natasha1
that is freaky! They are the same but it's pure coincidence. I promise
• Jan 8th 2011, 01:30 PM