Question 1 : Factorize ( x 2 − 4 x − 5) ( x 2 − 4 x − 12) − 144


 

Question 1 : Factorize ( x 2 − 4 x − 5) ( x 2 − 4 x − 12) − 144

cross multiplication factorisation, difference of two squares factorisation, greatest common divisor factorisation, how to factorise 4 terms, factorisation method formula

Solution :

We write the given expression;

\[ (x^2-4x-5)(x^2-4x-12)-144 \]

We see that ​\( x^2-4x \)​ is present in first two multiplying terms, so for sake of simplicity we suppose it to be equale to y i-e

suppose ​\( y=x^2-4x \)

==> ​\( (y-5)(y-12)-144 \)

==>​\( =y^2-17y+60-144 \)

==>​\( =y^2-17y+84 \)

Dividing mid term into two parts such that their sum is (-17y) and their product is ​\( 84y^2 \)​. We can think of (-21y) and  (4y) — so;

==> ​\( =y^2 − 21y + 4y − 84 \)

==>​\( = y(y − 21) + 4(y − 21) \)

or

==> ​\( =[y(y − 21) + 4(y − 21)] \)

Taking (y-21) as common;

==>  ​\( =(y-21)(y+4) \)

Since we supposed ​\( y=x^2-4x \)

==> ​\( =(x^2-4x-21)(x^2-4x+4) \)

 

 

 


					
					
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