Question

expand each of the following using suitable identities X + 2 Y + 4 Z whole square​

Answer 1

(x + 2y +

4z)²

Using identity,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Here, a = x, b = 2y and c = 4z

(x + 2y + 4z)²

 = x² + (2y)² +

(4z)²² + (2×x×2y) + (2×2y×4z) + (2×4z×x)

                   

= x² +

4y² + 16z²+ 4xy + 16yz + 8xz

 

hope it helps u

Answer 2

Step-by-step explanation:

Use suitable identities to find the following products:

(i) (x + 4) (x + 10)

(ii) (x + 8) (x – 10)

(iii) (3x + 4) (3x – 5)

(iv) (y2+ 3/2 ) (y2– 3/2 )

(v) (3 – 2x) (3 + 2x)

Solution:

(i) (x + 4) (x + 10)

Using identity (x + a)(x + b) = x2 + (a + b)x + ab

(x + 4)(x + 10) = x2 + (4 + 10)x + 4×10

= x2 + 14x + 40

(ii) (x + 8) (x – 10)

Using identity (x + a)(x + b) = x2 + (a + b)x + ab

(x + 8) (x – 10) = x2 + (8 – 10)x + 8x(-10)

= x2 – 2x – 80x

(iii) (3x + 4) (3x – 5)

Using identity (x + a)(x + b) = x2 + (a + b)x + ab

(3x + 4) (3x – 5) = (3x)2 + (4 – 5)3x + 4x(-5)

= 9x2 – 3x – 20

(iv) (y2+ 3/2 ) (y2– 3/2 )

Using identity (x + a)(x – a) = x2 – a2

(y2+ 3/2 ) (y2– 3/2 ) = y2 – (3/2)2

= y2 – 9/4

(v) (3 – 2x) (3 + 2x)

Using identity (x + a)(x – a) = x2 – a2

(3 – 2x) (3 + 2x)= 32 – (2x)2 = 92 – 4x2

mmae