Factorising a quadratic

**Natasha1** · January 8th 2011, 10:47 AM

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.

**snowtea** · January 8th 2011, 10:51 AM

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).

**Natasha1** · January 8th 2011, 10:53 AM

I saw on the net a method called the AC method but that is also confusing

**Archie Meade** · January 8th 2011, 10:55 AM

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 $-1.$

**Also sprach Zarathustra** · January 8th 2011, 10:56 AM

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

**Wilmer** · January 8th 2011, 10:58 AM

Nawwww....I say if not apparent, use the quadratic formula...but that's me!

**dwsmith** · January 8th 2011, 10:59 AM

$(\pm\alpha x\pm\beta)(\pm\lambda x\pm\phi)$

$\beta*\phi=-3\Rightarrow 3*1=3$ 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?

**TheCoffeeMachine** · January 8th 2011, 11:02 AM

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 $a = 2$ .

**TheCoffeeMachine** · January 8th 2011, 11:10 AM

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 $x$ ?

**Natasha1** · January 8th 2011, 11:12 AM

that is freaky! They are the same but it's pure coincidence. I promise

**Archie Meade** · January 8th 2011, 12:30 PM

Originally Posted by Wilmer

Nawwww....I say if not apparent, use the quadratic formula...but that's me!

You remind me of Armand Assante, Wilmer!

"When in doubt, get the ABC Robot out!"

**Wilmer** · January 8th 2011, 04:45 PM

Compliments will get you everywhere, Archie !

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