Factorize:

Question:

Factorize:

$24 x^{2}-41 x+12$

 

Solution:

We have:

$24 x^{2}-41 x+12$

We have to split $(-41)$ into two numbers such that their sum is $(-41)$ and their product is 288 , i.e., $24 \times 12$.

Clearly, $(-32)+(-9)=-41$ and $(-32) \times(-9)=288$

$\therefore 24 x^{2}-41 x+12=24 x^{2}-32 x-9 x+12$

$=8 x(3 x-4)-3(3 x-4)$

$=(3 x-4)(8 x-3)$

 

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