Question

Factorise using appropriate identities: (a2−6a+9)−25

Answer 1

Answer:

( a + 2 ) ( a - 8 )

Step-by-step explanation:

a^2 - 6a + 9

= a^2 - 3a - 3a + 9

= a ( a - 3 ) - 3 ( a - 3 )

= ( a - 3 ) ( a - 3 )

= ( a - 3 )^2

25 = 5^2

( a^2 - 6a + 9 ) - 25

= ( a - 3 )^2 - 5^2

a^2 - b^2 = ( a + b ) ( a - b )  { Identity / formula }

Here,

a = a - 3

b = 5

( a^2 - 6a + 9 ) - 25

= ( a - 3 )^2 - 5^2

= ( a - 3 + 5 ) ( a - 3 - 5 )

= ( a + 2 ) ( a - 8 )

Hence, factorized.

Answer 2

Answer:

=》 a =》 8 and (-2 ).

Step-by-step explanation:

Given :-

a^2 - 6a + 9 - 25 .

May be, a^2 - 6a - 16 = 0

Solution, 》a^2 - (8-2) a - 16 =0

》 a^2 -8a + 2a - 16 =0

》 a( a-8 ) + 2( a-8 ) = 0

》( a-8 ) ( a+2 ) = 0

then, a -8 =0 a +2 =0

》 a = 8 》 a = -2 .

Two value of "a " are -》8 and (-2 ).