Question

Expand each of the following, using suitable identities: 

(i) (x + 2y + 4z) 2 (ii) (2x – y + z) ² (iii) (–2x + 3y + 2z) ² 

(iv) (3a – 7b – c) ² (v) (–2x + 5y – 3z) ² (vi) [ \(\frac{1 }{ 4}\) a - \(\frac{1 }{ 2}\) b + 1]²

Answer 1

Expansion of Squares of Trinomials 

1. Identity Used:

\((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\) \((a - b + c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ca\) (adjust signs as needed)

(i) \( (x + 2y + 4z)^2 \)

\[ x^2 + (2y)^2 + (4z)^2 + 2(x \cdot 2y) + 2(2y \cdot 4z) + 2(4z \cdot x) \]

\[ = x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz \]

(ii) \( (2x - y + z)^2 \)

\[ (2x)^2 + (-y)^2 + z^2 + 2(2x \cdot -y) + 2(-y \cdot z) + 2(z \cdot 2x) \]

\[ = 4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz \]

(iii) \( (-2x + 3y + 2z)^2 \)

\[ (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x \cdot 3y) + 2(3y \cdot 2z) + 2(2z \cdot -2x) \]

\[ = 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8xz \]

(iv) \( (3a - 7b - c)^2 \)

\[ (3a)^2 + (-7b)^2 + (-c)^2 + 2(3a \cdot -7b) + 2(-7b \cdot -c) + 2(-c \cdot 3a) \]

\[ = 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ac \]

(v) \( (-2x + 5y - 3z)^2 \)

\[ (-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x \cdot 5y) + 2(5y \cdot -3z) + 2(-3z \cdot -2x) \]

\[ = 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12xz \]

(vi) \( \left(\frac{1}{4}a - \frac{1}{2}b + 1\right)^2 \)

\[ \left(\frac{1}{4}a\right)^2 + \left(-\frac{1}{2}b\right)^2 + 1^2 + 2\left(\frac{1}{4}a \cdot -\frac{1}{2}b\right) + 2\left(-\frac{1}{2}b \cdot 1\right) + 2\left(1 \cdot \frac{1}{4}a\right) \]

\[ = \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b + \frac{1}{2}a \]

Answer Summary:

• (i) \( x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz \)
• (ii) \( 4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz \)
• (iii) \( 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8xz \)
• (iv) \( 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ac \)
• (v) \( 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12xz \)
• (vi) \( \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b + \frac{1}{2}a \)