Question 1

We use the identity $(a + b)^2 = a^2 + 2ab + b^2$.

Step 1 — Expand $(7x + 4y)^2$

Let's identify $\mathbf{a}$ and $\mathbf{b}$. Here, $\mathbf{a}$ is $\mathbf{7x}$. And $\mathbf{b}$ is $\mathbf{4y}$. We substitute these into the identity.

$(7x + 4y)^2 = (7x)^2 + 2(7x)(4y) + (4y)^2$

$= 49x^2 + 56xy + 16y^2$

$\boxed{49x^2 + 56xy + 16y^2}$

Step 2 — Expand $\left(\frac{7}{5}x + \frac{3}{2}y\right)^2$

Let's identify $\mathbf{a}$ and $\mathbf{b}$. Here, $\mathbf{a}$ is $\mathbf{\frac{7}{5}x}$. And $\mathbf{b}$ is $\mathbf{\frac{3}{2}y}$. We substitute these into the identity.

$\left(\frac{7}{5}x + \frac{3}{2}y\right)^2 = \left(\frac{7}{5}x\right)^2 + 2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right) + \left(\frac{3}{2}y\right)^2$

$= \frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2$

$\boxed{\frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2}$

Step 3 — Expand $(2.5p + 1.5q)^2$

Let's identify $\mathbf{a}$ and $\mathbf{b}$. Here, $\mathbf{a}$ is $\mathbf{2.5p}$. And $\mathbf{b}$ is $\mathbf{1.5q}$. We substitute these into the identity.

$(2.5p + 1.5q)^2 = (2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2$

$= 6.25p^2 + 7.5pq + 2.25q^2$

$\boxed{6.25p^2 + 7.5pq + 2.25q^2}$

<DIAGRAM: Algebraic expansion of a binomial squared, showing terms $a^2$, $2ab$, and $b^2$.>

Step 4 — Expand $\left(\frac{3}{4}s + 8t\right)^2$

Let's identify $\mathbf{a}$ and $\mathbf{b}$. Here, $\mathbf{a}$ is $\mathbf{\frac{3}{4}s}$. And $\mathbf{b}$ is $\mathbf{8t}$. We substitute these into the identity.

$\left(\frac{3}{4}s + 8t\right)^2 = \left(\frac{3}{4}s\right)^2 + 2\left(\frac{3}{4}s\right)(8t) + (8t)^2$

$= \frac{9}{16}s^2 + 12st + 64t^2$

$\boxed{\frac{9}{16}s^2 + 12st + 64t^2}$

<DIAGRAM: Algebraic expansion of a binomial squared, showing terms $a^2$, $2ab$, and $b^2$.>

Step 5 — Expand $\left(x + \frac{1}{2y}\right)^2$

Let's identify $\mathbf{a}$ and $\mathbf{b}$. Here, $\mathbf{a}$ is $\mathbf{x}$. And $\mathbf{b}$ is $\mathbf{\frac{1}{2y}}$. We substitute these into the identity.

$\left(x + \frac{1}{2y}\right)^2 = (x)^2 + 2(x)\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2$

$= x^2 + \frac{x}{y} + \frac{1}{4y^2}$

$\boxed{x^2 + \frac{x}{y} + \frac{1}{4y^2}}$

<DIAGRAM: Algebraic expansion of a binomial squared, showing terms $a^2$, $2ab$, and $b^2$.>

Step 6 — Expand $\left(\frac{1}{x} + \frac{1}{y}\right)^2$

Let's identify $\mathbf{a}$ and $\mathbf{b}$. Here, $\mathbf{a}$ is $\mathbf{\frac{1}{x}}$. And $\mathbf{b}$ is $\mathbf{\frac{1}{y}}$. We substitute these into the identity.

$\left(\frac{1}{x} + \frac{1}{y}\right)^2 = \left(\frac{1}{x}\right)^2 + 2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2$

$= \frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}$

$\boxed{\frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}}$

<DIAGRAM: Algebraic expansion of a binomial squared, showing terms $a^2$, $2ab$, and $b^2$.>

Answer

(i) $49x^2 + 56xy + 16y^2$ (ii) $\frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2$ (iii) $6.25p^2 + 7.5pq + 2.25q^2$ (iv) $\frac{9}{16}s^2 + 12st + 64t^2$ (v) $x^2 + \frac{x}{y} + \frac{1}{4y^2}$ (vi) $\frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}$