Question
Factorize:
$9(2 a-b)^2-4(2 a-b)-13$
$9(2 a-b)^2-4(2 a-b)-13$
✓
Answer
$\text { Let } 2 \mathrm{a}-\mathrm{b}=\mathrm{x}$
$=9 \mathrm{x}^2-4 \mathrm{x}-13$
Splitting the middle term,
$=9 x^2-13 x+9 x-13$
$=x(9 x-13)+1(9 x-13)$
$=(9 x-13)(x+1)$
Substituting $\mathrm{x}=2 \mathrm{a}-\mathrm{b}$
$=[9(2 a-b)-13](2 a-b+1)$
$=(18 a-9 b-13)(2 a-b+1)$
$\therefore 9(2 a-b)^2-4(2 a-b)-13$
$=(18 a-9 b-13)(2 a-b+1)$
$=9 \mathrm{x}^2-4 \mathrm{x}-13$
Splitting the middle term,
$=9 x^2-13 x+9 x-13$
$=x(9 x-13)+1(9 x-13)$
$=(9 x-13)(x+1)$
Substituting $\mathrm{x}=2 \mathrm{a}-\mathrm{b}$
$=[9(2 a-b)-13](2 a-b+1)$
$=(18 a-9 b-13)(2 a-b+1)$
$\therefore 9(2 a-b)^2-4(2 a-b)-13$
$=(18 a-9 b-13)(2 a-b+1)$
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