Question

Find HCF of: (x–y)^2+4xy​

Answer 1

The H.C.F. of numerical coefficients = The H.C.F. of 4 and 6.

Since, 4 = 2 × 2 = 22 and 6 = 2 × 3 = 21 × 31

Therefore, the H.C.F. of 4 and 6 is 2

The H.C.F. of literal coefficients = The H.C.F. of x2y3 and xy2z = xy2

Since, in x2y3 and xy2z, x and y are common.

The lowest power of x is x.

The lowest power of y is y2.

Therefore, the H.C.F. of x2y3 and xy2z is xy2.

Thus, the H.C.F. of 4x2y3 and 6xy2z

= The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 2 × (xy2)

= 2xy2.

Answer 2

Step-by-step explanation:

ANSWER:-

Factorizing 9x

2

−4x

2

and 6x

2

+4xy,

9x

2

−4x

2

=(3x+2y)(3x−2y)

6x

2

+4xy=2x(3x+2y)

H.C.F. =3x+2y

L.C.M. =2x(3x+2y)(3x−2y)