Question

Factorize 9x2+y2+z2_6xy+2yz_6xzthen find its value when x=1 ,y=3,z=_1

Answer 1

9x² + y² + z² - 6xy + 2yz - 6xz

Whenever there are three perfect squares and six terms in a polynomial, then just recall the identity a² + b² + c² + 2ab + 2bc + 2ca = ( a + b + c)²

Here given polynomial can be written as

( 3x )² + (y)² + (z)² - 2(3x)(y) + 2(y)(z) - 2 (3x)(z)

Compare it with a² + b² + c² + 2ab + 2bc + 2ca , you'll observe that

terms representing 2ab and 2ca in the given polynomial are negative , and these terms have 'a' common ∴ we can say that 'a' is negative

∴ ( -3x)² + (y)² + (z)² + 2( -3x )(y) + 2(y)(z) + 2 ( -3x )(z)

Which can be written as

( -3x + y + z )²  [∵ a² + b² + c² + 2ab + 2bc + 2ca = (a + b +  c)²]

Now as given in question, putting x = 1, y = 3 and z = -1

= [ -3(1) + 3 + ( -1)]²

= ( -3 + 3 -1 )²

= 1

∴ Value of given polynomial for x = 1, y = 3 and z = - 1 is 1