Question

factorise : 9x^2-p^2+4p-4​

Answer 1

Answer:(3x+P-2)(3X-P+2)

Step-by-step explanation:

$9x^{2}-P^{2}+4P-4$

=$(3x)^{2}-P^{2}+2*2*P-(2)^{2}$

=$(3x)^{2}-(P^{2}-2*2*P+(2)^{2})$

=$(3x)^{2}-(P-2)^{2}$

from, $(a+b)^{2}=a^{2}+2ab+ b^{2}$

=$(3x+(P-2))(3x-(P-2))$

=(3x+P-2)(3X-P+2)

from, $a^{2}-b^{2}=(a+b)(a+b)$

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