Question

X + 2 Y + 4 Z Square factorise​

Answer 1

Answer:

hand side expression.

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Solution:

i)

(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

(ii) (2x – y + z)²

Using identity,

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

Here, a = 2x, b = –y & c

= z

(2x – y + z)²

= (2x)² +

(-y)² + z² + (2×2x×-y) + (2×-y×z) +

(2×z×2x)

= 4x² + y² + z² –

4xy – 2yz + 4xz

(iii) (–2x +

3y + 2z)²

Using identity=

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

Here, a = -2x, b = 3y & c

= 2z

(–2x + 3y + 2z)²

= (-2x)² + (3y)²+

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

= 4x² + 9y² + 4z² –

12xy + 12yz – 8xz

(iv) (3a –

7b – c)²

Using identity,

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

Here, a = 3a, b,= -7b &c = -c

(3a – 7b – c)²

= (3a)² +

(-7b)²+ (-c)² + (2×3a×-7b) +

(2×-7b×-c) + (2×-c×3a)

= 9a² + 49b² + c² –

42ab + 14bc – 6ac

(v) (–2x +

5y – 3z)²

Using identity,

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

Here, a = -2x, b = 5y and c = -3z

(–2x + 5y – 3z)²

= (-2x)² + (5y)² +

(-3z)² + (2×-2x×5y) + (2×5y×-3z) + (2×-3z×-2x)

= 4x² + 25y² + 9z²–

20xy – 30yz + 12xz

(vi) [1/4 a –

1/2 b + 1]²

Using identity,

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

Here, a = a/4 , b = -b/2

& c = 1

[1/4 a – 1/2 b + 1]²

=(a/4)²+(-b/2 )²+1²+

(2×a/4 ×-b/2)+(2×-b/2 ×1) + (2×1×a/4)

= a²/16+ b²/4+ 1 – ab/4– b + a/2

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