# Thread: Factoring hard trinomial confusion

1. ## Factoring hard trinomial confusion

I have this equation that I need to factor
$8x^2 - 10x + 3$

The answer to this I am given is
$(4x-3)(2x-1)$

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
$8x^2-6x-4x+3$

Step 4
I try to do grouping at this point however the answer I get is completely different from
$(4x-3)(2x-1)$

To be specific, I tried to group $8x^2$ with $-4x$ since they share a common factor and did the same logic with grouping $-6x$ with $3$

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

2. ## 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.

3. ## Re: Factoring hard trinomial confusion

Hi,

you stopped one step before the finish:

Originally Posted by DanSmith
I have this equation that I need to factor
$8x^2 - 10x + 3$

Step 3
With this information, I make a new equation
$8x^2-6x-4x+3$

...
$\displaystyle 8x^2-4x-6x+3 = 4x(2x-1) - 3(2x-1) = (2x-1)(4x-3)$

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