Certainly! Let's expand each of the given expressions step-by-step using the suitable algebraic identities. Specifically, we'll use the binomial theorem for expanding the square of a trinomial, which states:
[tex]\[(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\][/tex]
### (i) [tex]\((x + 2y + 4z)^2\)[/tex]
Using the identity, we have:
[tex]\[
(x + 2y + 4z)^2 = x^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(x)(4z)
\][/tex]
Calculating each term:
[tex]\[
x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz
\][/tex]
So, the expanded form is:
[tex]\[
x^2 + 4x y + 8x z + 4y^2 + 16y z + 16z^2
\][/tex]
### (ii) [tex]\((2x - y + z)^2\)[/tex]
Using the identity, we have:
[tex]\[
(2x - y + z)^2 = (2x)^2 + (-y)^2 + z^2 + 2(2x)(-y) + 2(-y)(z) + 2(2x)(z)
\][/tex]
Calculating each term:
[tex]\[
4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz
\][/tex]
So, the expanded form is:
[tex]\[
4x^2 - 4x y + 4x z + y^2 - 2y z + z^2
\][/tex]
### (iii) [tex]\((-2x + 3y + 2z)^2\)[/tex]
Using the identity, we have:
[tex]\[
(-2x + 3y + 2z)^2 = (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(-2x)(2z)
\][/tex]
Calculating each term:
[tex]\[
4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8xz
\][/tex]
So, the expanded form is:
[tex]\[
4x^2 - 12x y - 8x z + 9y^2 + 12y z + 4z^2
\][/tex]
### (iv) [tex]\((3a - 7b - c)^2\)[/tex]
Using the identity, we have:
[tex]\[
(3a - 7b - c)^2 = (3a)^2 + (-7b)^2 + (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(3a)(-c)
\][/tex]
Calculating each term:
[tex]\[
9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ac
\][/tex]
So, the expanded form is:
[tex]\[
9a^2 - 42a b - 6a c + 49b^2 + 14b c + c^2
\][/tex]
### (v) [tex]\((-2x + 5y - 3z)^2\)[/tex]
Using the identity, we have:
[tex]\[
(-2x + 5y - 3z)^2 = (-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-2x)(-3z)
\][/tex]
Calculating each term:
[tex]\[
4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12xz
\][/tex]
So, the expanded form is:
[tex]\[
4x^2 - 20x y + 12x z + 25y^2 - 30y z + 9z^2
\][/tex]
### (vi) [tex]\(\left(\frac{1}{4}a - \frac{1}{2}b + 1\right)^2\)[/tex]
Using the identity, we have:
[tex]\[
\left(\frac{1}{4}a - \frac{1}{2}b + 1\right)^2 = \left(\frac{1}{4}a\right)^2 + \left(-\frac{1}{2}b\right)^2 + 1^2 + 2\left(\frac{1}{4}a\right)\left(-\frac{1}{2}b\right) + 2\left(-\frac{1}{2}b\right) (1) + 2\left(\frac{1}{4}a\right) (1)
\][/tex]
Calculating each term:
[tex]\[
\frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b + \frac{1}{2}a
\][/tex]
So, the expanded form is:
[tex]\[
0.0625a^2 - 0.25a b + 0.5a + 0.25b^2 - 1.0b + 1
\][/tex]
