Question 2

We will use suitable algebraic identities to factor each expression.

Step 1 — Factoring $16y^2 - 24y + 9$

Let's look at the expression. It has three terms. The first term is a perfect square: $16y^2 = (\mathbf{4y})^2$. The last term is a perfect square: $9 = (\mathbf{3})^2$. This looks like the identity $(a-b)^2 = a^2 - 2ab + b^2$. Let's check the middle term. Here, $a = \mathbf{4y}$ and $b = \mathbf{3}$. The middle term should be $2ab$. $2ab = 2 \times (\mathbf{4y}) \times (\mathbf{3})$ $= \mathbf{24y}$ This matches the given middle term, $\mathbf{-24y}$. So, we use the identity $(a-b)^2$. $16y^2 - 24y + 9 = (4y)^2 - 2(4y)(3) + (3)^2$

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

Step 2 — Factoring $\frac{9}{4}s^2 + 6st + 4t^2$

Let's examine this expression. It also has three terms. The first term is a perfect square: $\frac{9}{4}s^2 = (\mathbf{\frac{3}{2}s})^2$. The last term is a perfect square: $4t^2 = (\mathbf{2t})^2$. This suggests the identity $(a+b)^2 = a^2 + 2ab + b^2$. Let's identify $a$ and $b$. Here, $a = \mathbf{\frac{3}{2}s}$ and $b = \mathbf{2t}$. We check the middle term $2ab$. $2ab = 2 \times (\mathbf{\frac{3}{2}s}) \times (\mathbf{2t})$ $= \mathbf{6st}$ This matches the middle term in the expression. So, we apply the identity $(a+b)^2$. $\frac{9}{4}s^2 + 6st + 4t^2 = (\frac{3}{2}s)^2 + 2(\frac{3}{2}s)(2t) + (2t)^2$

$\boxed{(\frac{3}{2}s + 2t)^2}$

Step 3 — Factoring $\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2$

This expression has six terms. This form reminds us of $(a+b+c)^2$. The identity is $a^2+b^2+c^2+2ab+2bc+2ca$. Let's find the squared terms. We have $\frac{m^2}{9} = (\mathbf{\frac{m}{3}})^2$. We have $\frac{k^2}{4} = (\mathbf{\frac{k}{2}})^2$. We have $9n^2 = (\mathbf{3n})^2$. So, let $a = \mathbf{\frac{m}{3}}$, $b = \mathbf{\frac{k}{2}}$, and $c = \mathbf{3n}$. Let's verify the cross-product terms. $2ab = 2(\frac{m}{3})(\frac{k}{2}) = \mathbf{\frac{mk}{3}}$ $2bc = 2(\frac{k}{2})(3n) = \mathbf{3nk}$ $2ca = 2(3n)(\frac{m}{3}) = \mathbf{2mn}$ All terms match the given expression. So, we can factor it using $(a+b+c)^2$. $\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2 = (\frac{m}{3})^2 + (\frac{k}{2})^2 + (3n)^2 + 2(\frac{m}{3})(\frac{k}{2}) + 2(\frac{k}{2})(3n) + 2(3n)(\frac{m}{3})$

$\boxed{(\frac{m}{3} + \frac{k}{2} + 3n)^2}$

Step 4 — Factoring $\frac{p^2}{16} - 2 + \frac{16}{p^2}$

Let's look at this expression. It has three terms. The first term is a perfect square: $\frac{p^2}{16} = (\mathbf{\frac{p}{4}})^2$. The last term is a perfect square: $\frac{16}{p^2} = (\mathbf{\frac{4}{p}})^2$. This looks like the identity $(a-b)^2 = a^2 - 2ab + b^2$. Let's identify $a$ and $b$. Here, $a = \mathbf{\frac{p}{4}}$ and $b = \mathbf{\frac{4}{p}}$. We check the middle term $2ab$. $2ab = 2 \times (\mathbf{\frac{p}{4}}) \times (\mathbf{\frac{4}{p}})$ $= \mathbf{2}$ This matches the middle term, which is $\mathbf{-2}$. So, we use the identity $(a-b)^2$. $\frac{p^2}{16} - 2 + \frac{16}{p^2} = (\frac{p}{4})^2 - 2(\frac{p}{4})(\frac{4}{p}) + (\frac{4}{p})^2$

$\boxed{(\frac{p}{4} - \frac{4}{p})^2}$

Step 5 — Factoring $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc$

This expression has six terms. It suggests the identity $(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$. Let's find the squared terms. We have $9a^2 = (\mathbf{3a})^2$. We have $4b^2 = (\mathbf{2b})^2$. We have $c^2 = (\mathbf{c})^2$. Now, let's consider the signs of the cross-product terms. We have $\mathbf{-12ab}$, $\mathbf{+6ac}$, $\mathbf{-4bc}$. The terms with $b$ are negative. This means the term involving $b$ must be negative. Let $x = \mathbf{3a}$, $y = \mathbf{-2b}$, and $z = \mathbf{c}$. Let's verify the cross-product terms with these choices. $2xy = 2(\mathbf{3a})(\mathbf{-2b}) = \mathbf{-12ab}$ $2yz = 2(\mathbf{-2b})(\mathbf{c}) = \mathbf{-4bc}$ $2zx = 2(\mathbf{c})(\mathbf{3a}) = \mathbf{6ac}$ All terms match the given expression. So, we can factor it using $(x+y+z)^2$. $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc = (3a)^2 + (-2b)^2 + (c)^2 + 2(3a)(-2b) + 2(-2b)(c) + 2(c)(3a)$

$\boxed{(3a - 2b + c)^2}$

Answer

(i) $(4y - 3)^2$ (ii) $(\frac{3}{2}s + 2t)^2$ (iii) $(\frac{m}{3} + \frac{k}{2} + 3n)^2$ (iv) $(\frac{p}{4} - \frac{4}{p})^2$ (v) $(3a - 2b + c)^2$