Question 1

We will use the algebraic identity for a perfect square trinomial.

Step 1 — Factor $9x^2 + 24xy + 16y^2$

Let's look at the first term. We see that $9x^2$ is the square of $3x$. The last term is $16y^2$. This is the square of $4y$. Now, let's check the middle term. It should be twice the product of $3x$ and $4y$. $2 \times (3x) \times (4y) = 24xy$. This matches our middle term. So, this is a perfect square trinomial. We use the identity $(a+b)^2 = a^2 + 2ab + b^2$.

$9x^2 + 24xy + 16y^2 = (3x)^2 + 2(3x)(4y) + (4y)^2$

$= (3x + 4y)^2$

$\boxed{(3x + 4y)^2}$

Step 2 — Factor $4s^2 + 20st + 25t^2$

Let's examine the first term. We see that $4s^2$ is the square of $2s$. The last term is $25t^2$. This is the square of $5t$. Now, let's check the middle term. It should be twice the product of $2s$ and $5t$. $2 \times (2s) \times (5t) = 20st$. This matches our middle term. So, this is a perfect square trinomial. We use the identity $(a+b)^2 = a^2 + 2ab + b^2$.

$4s^2 + 20st + 25t^2 = (2s)^2 + 2(2s)(5t) + (5t)^2$

$= (2s + 5t)^2$

$\boxed{(2s + 5t)^2}$

Step 3 — Factor $49x^2 + 28xy + 4y^2$

Let's look at the first term. We see that $49x^2$ is the square of $7x$. The last term is $4y^2$. This is the square of $2y$. Now, let's check the middle term. It should be twice the product of $7x$ and $2y$. $2 \times (7x) \times (2y) = 28xy$. This matches our middle term. So, this is a perfect square trinomial. We use the identity $(a+b)^2 = a^2 + 2ab + b^2$.

$49x^2 + 28xy + 4y^2 = (7x)^2 + 2(7x)(2y) + (2y)^2$

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

$\boxed{(7x + 2y)^2}$

Step 4 — Factor $64p^2 + \frac{32}{3}pq + \frac{4}{9}q^2$

Let's examine the first term. We see that $64p^2$ is the square of $8p$. The last term is $\frac{4}{9}q^2$. This is the square of $\frac{2}{3}q$. Now, let's check the middle term. It should be twice the product of $8p$ and $\frac{2}{3}q$. $2 \times (8p) \times \left(\frac{2}{3}q\right) = \frac{32}{3}pq$. This matches our middle term. So, this is a perfect square trinomial. We use the identity $(a+b)^2 = a^2 + 2ab + b^2$.

$64p^2 + \frac{32}{3}pq + \frac{4}{9}q^2 = (8p)^2 + 2(8p)\left(\frac{2}{3}q\right) + \left(\frac{2}{3}q\right)^2$

$= \left(8p + \frac{2}{3}q\right)^2$

$\boxed{\left(8p + \frac{2}{3}q\right)^2}$

Step 5 — Factor $3a^2 + 4ab + \frac{4}{3}b^2$

This expression has fractional coefficients. Let's try to factor out a common term. We can factor out 3 from all terms. This will make the first term $a^2$.

$3a^2 + 4ab + \frac{4}{3}b^2 = 3\left(a^2 + \frac{4}{3}ab + \frac{4}{9}b^2\right)$

Now, let's look at the expression inside the parenthesis. The first term is $a^2$, which is the square of $a$. The last term is $\frac{4}{9}b^2$, which is the square of $\frac{2}{3}b$. The middle term is $\frac{4}{3}ab$. Let's check if it is $2 \times (a) \times \left(\frac{2}{3}b\right)$. $2 \times (a) \times \left(\frac{2}{3}b\right) = \frac{4}{3}ab$. This matches the middle term. So, the expression inside is a perfect square.

$= 3\left((a)^2 + 2(a)\left(\frac{2}{3}b\right) + \left(\frac{2}{3}b\right)^2\right)$

$= 3\left(a + \frac{2}{3}b\right)^2$

$\boxed{3\left(a + \frac{2}{3}b\right)^2}$

Step 6 — Factor $\frac{9}{5}s^2 + 6sv + 5v^2$

This expression also has fractional coefficients. Let's try to factor out a common term. We can factor out $\frac{9}{5}$ from all terms. This will make the first term $s^2$.

$\frac{9}{5}s^2 + 6sv + 5v^2 = \frac{9}{5}\left(s^2 + 6 \times \frac{5}{9}sv + 5 \times \frac{5}{9}v^2\right)$

$= \frac{9}{5}\left(s^2 + \frac{30}{9}sv + \frac{25}{9}v^2\right)$

$= \frac{9}{5}\left(s^2 + \frac{10}{3}sv + \frac{25}{9}v^2\right)$

Now, let's look at the expression inside the parenthesis. The first term is $s^2$, which is the square of $s$. The last term is $\frac{25}{9}v^2$, which is the square of $\frac{5}{3}v$. The middle term is $\frac{10}{3}sv$. Let's check if it is $2 \times (s) \times \left(\frac{5}{3}v\right)$. $2 \times (s) \times \left(\frac{5}{3}v\right) = \frac{10}{3}sv$. This matches the middle term. So, the expression inside is a perfect square.

$= \frac{9}{5}\left((s)^2 + 2(s)\left(\frac{5}{3}v\right) + \left(\frac{5}{3}v\right)^2\right)$

$= \frac{9}{5}\left(s + \frac{5}{3}v\right)^2$

$\boxed{\frac{9}{5}\left(s + \frac{5}{3}v\right)^2}$

Answer

(i) $(3x + 4y)^2$ (ii) $(2s + 5t)^2$ (iii) $(7x + 2y)^2$ (iv) $\left(8p + \frac{2}{3}q\right)^2$ (v) $3\left(a + \frac{2}{3}b\right)^2$ (vi) $\frac{9}{5}\left(s + \frac{5}{3}v\right)^2$