Topic : Factorisation using identities
Problem : Factorise 49p^2 – 36
Solution :
We know that (a + b)^ 2 = a^2 + 2ab + b^2 …………… (I)
(a – b) 2= a^2 – 2ab + b62………….. (II)
(a + b) (a – b) = a^2 – b^2………… (III)
Let us use these identities for factorisation.
There are two terms; both are squares and the second is negative. The expression is of the form (a^2 – b^2). Identity III is applicable here;
49 p^2– 36 = (7p)^ 2 – (6)^2
= (7p – 6) (7p + 6) (required factorisation)
Problem : Factorise 49p^2 – 36
Solution :
We know that (a + b)^ 2 = a^2 + 2ab + b^2 …………… (I)
(a – b) 2= a^2 – 2ab + b62………….. (II)
(a + b) (a – b) = a^2 – b^2………… (III)
Let us use these identities for factorisation.
There are two terms; both are squares and the second is negative. The expression is of the form (a^2 – b^2). Identity III is applicable here;
49 p^2– 36 = (7p)^ 2 – (6)^2
= (7p – 6) (7p + 6) (required factorisation)
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