Factoring hard trinomial confusion

I have this equation that I need to factor

http://latex.codecogs.com/gif.latex?...x%20+%203

The answer to this I am given is

http://latex.codecogs.com/gif.latex?...3%29%282x-1%29

However no matter how hard I try I cannot seem to get to this solution

I have tried...

Step 1

Multiplying the first term by the last term to get 24

Step 2

Finding out what factor of 24 adds to get -10. The answer I get for this is -6 and -4

Step 3

With this information, I make a new equation

http://latex.codecogs.com/gif.latex?8x%5E2-6x-4x+3

Step 4

I try to do grouping at this point however the answer I get is completely different from

http://latex.codecogs.com/gif.latex?...3%29%282x-1%29

To be specific, I tried to group http://latex.codecogs.com/gif.latex?8x%5E2 with http://latex.codecogs.com/gif.latex?-4x since they share a common factor and did the same logic with grouping http://latex.codecogs.com/gif.latex?-6x with http://latex.codecogs.com/gif.latex?3

However this ultimately leads me with an answer that isn't what I'm given.

Is there something I am overlooking here? If so, please explain. Much obliged

Re: Factoring hard trinomial confusion

$\large ax^2 + bx +c=a(x-r1)(x-r2)$

where

$r1,r2=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

in this case

$r1=\dfrac 3 4, ~r2=\dfrac 1 2, ~a=8$

$8\left(x-\dfrac 3 4\right)\left(x - \dfrac 1 2\right)=$

$8\left(x^2-\dfrac{5x}{4}+\dfrac 3 8\right)=$

$8x^2-10x+3$

and of course you can distribute the factor of 8 between the 2 factors to get rid of fractions as

$8\left(x-\dfrac 3 4\right)\left(x - \dfrac 1 2\right)=$

$4\left(x-\dfrac 3 4\right) \cdot 2\left(x - \dfrac 1 2\right) = $

$(4x-3)(2x-1)$

This isn't generally the fastest way to factor a polynomial but it will always work when all else fails.

Re: Factoring hard trinomial confusion

Hi,

you stopped one step before the finish:

Quote:

Originally Posted by

**DanSmith**

... but of course it would be better if you learn to master the method romsek showed you.