blitzbabeiy

2022-02-01

Factoring $4{x}^{2}-2xy-4x+3y-3$
Why isn't it working?

ataill0k

Expert

Step 1
When $2x=3$, then the expression becomes
$4{x}^{2}-2xy-4x+3y-3=9-3y-6+3y-3=0$
Hence the expression can be factored by $\left(2x-3\right)$, and you can see easily that
$4{x}^{2}-2xy-4x+3y-3=\left(2x-3\right)\left(2x-y+1\right)$

Emilie Booker

Expert

Step 1
Probably not the fastest, but I would like to suggest you use the general factorization method:
We have,
$4{x}^{2}-2xy-4x+3y-3=4{x}^{2}-x\left(2y+4\right)+\left(3y-3\right)$
$⇒\mathrm{\Delta }={\left(y+2\right)}^{2}-4\left(3y-3\right)={\left(y-4\right)}^{2}$
$⇒{x}_{1}=\frac{y+2+\left(y-4\right)}{4}=\frac{y-1}{2}$
$⇒{x}_{2}=\frac{y+2-\left(y-4\right)}{4}=\frac{3}{2}$
$⇒4{x}^{2}-2xy-4x+3y-3=4\left(x-\frac{3}{2}\right)\left(x-\frac{y-1}{2}\right)$
$⇒4{x}^{2}-2xy-4x+3y-3=\left(2x-3\right)\left(2x-y+1\right)$

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