# 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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