Exploring Algebraic Identities | Exercise 4.2

Question 1

Factor completely:

(i) $9x^2 + 24xy + 16y^2$

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

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

(iv) $64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2$

*(v) $3a^2 + 4ab + \dfrac{4}{3}b^2$

*(vi) $\dfrac{9}{5}s^2 + 6sv + 5v^2$

(Hint: 2 was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?)

Solution

Understand the Question

A trinomial of the form $a^2 + 2ab + b^2$ is a perfect square trinomial and factors directly into $(a + b)^2$ .
To factor each expression completely:
- Identify the first and last terms as squares, $a^2$ and $b^2$ .
- Verify that the middle term matches $2ab$ .
- Apply the identity $a^2 + 2ab + b^2 = (a+b)^2$ .
If the leading coefficient is not a perfect square (as in parts (v) and (vi)), factor out a common numerical factor first before applying the identity.

(i) Factor completely: $9x^2 + 24xy + 16y^2$

Step 1 · Apply the Perfect Square Identity

Identify the square terms and verify the middle term:

9x^2 = (3x)^2, \quad 16y^2 = (4y)^2, \quad 2(3x)(4y) = 24xy

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$

\begin{aligned} 9x^2 + 24xy + 16y^2 &= (3x)^2 + 2(3x)(4y) + (4y)^2 \\[0.6em] &= (3x + 4y)^2 \end{aligned}

Answer

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

(ii) Factor completely: $4s^2 + 20st + 25t^2$

Step 1 · Apply the Perfect Square Identity

Identify the square terms and verify the middle term:

4s^2 = (2s)^2, \quad 25t^2 = (5t)^2, \quad 2(2s)(5t) = 20st

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$

\begin{aligned} 4s^2 + 20st + 25t^2 &= (2s)^2 + 2(2s)(5t) + (5t)^2 \\[0.6em] &= (2s + 5t)^2 \end{aligned}

Answer

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

(iii) Factor completely: $49x^2 + 28xy + 4y^2$

Step 1 · Apply the Perfect Square Identity

Identify the square terms and verify the middle term:

49x^2 = (7x)^2, \quad 4y^2 = (2y)^2, \quad 2(7x)(2y) = 28xy

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$

\begin{aligned} 49x^2 + 28xy + 4y^2 &= (7x)^2 + 2(7x)(2y) + (2y)^2 \\[0.6em] &= (7x + 2y)^2 \end{aligned}

Answer

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

(iv) Factor completely: $64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2$

Step 1 · Apply the Perfect Square Identity

Identify the square terms and verify the middle term:

64p^2 = (8p)^2, \quad \dfrac{4}{9}q^2 = \left(\dfrac{2}{3}q\right)^2, \quad 2(8p)\left(\dfrac{2}{3}q\right) = \dfrac{32}{3}pq

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$

\begin{aligned} 64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2 &= (8p)^2 + 2(8p)\left(\dfrac{2}{3}q\right) + \left(\dfrac{2}{3}q\right)^2 \\[0.6em] &= \left(8p + \dfrac{2}{3}q\right)^2 \end{aligned}

Answer

(iv) $\left(8p + \dfrac{2}{3}q\right)^2$

(v) Factor completely: $3a^2 + 4ab + \dfrac{4}{3}b^2$

Step 1 · Factor Out Common Factor and Apply Identity

Factor out $3$ from all terms:

\begin{aligned} 3a^2 + 4ab + \dfrac{4}{3}b^2 &= 3\left(a^2 + \dfrac{4}{3}ab + \dfrac{4}{9}b^2\right) \\[0.6em] &= 3\left((a)^2 + 2(a)\left(\dfrac{2}{3}b\right) + \left(\dfrac{2}{3}b\right)^2\right) \\[0.6em] &= 3\left(a + \dfrac{2}{3}b\right)^2 \end{aligned}

Answer

(v) $3\left(a + \dfrac{2}{3}b\right)^2$

(vi) Factor completely: $\dfrac{9}{5}s^2 + 6sv + 5v^2$

Step 1 · Factor Out Common Factor and Apply Identity

Factor out $\dfrac{9}{5}$ from all terms:

\begin{aligned} \dfrac{9}{5}s^2 + 6sv + 5v^2 &= \dfrac{9}{5}\left(s^2 + 6 \times \dfrac{5}{9}sv + 5 \times \dfrac{5}{9}v^2\right) \\[0.6em] &= \dfrac{9}{5}\left(s^2 + \dfrac{30}{9}sv + \dfrac{25}{9}v^2\right) \\[0.6em] &= \dfrac{9}{5}\left(s^2 + \dfrac{10}{3}sv + \dfrac{25}{9}v^2\right) \\[0.6em] &= \dfrac{9}{5}\left((s)^2 + 2(s)\left(\dfrac{5}{3}v\right) + \left(\dfrac{5}{3}v\right)^2\right) \\[0.6em] &= \dfrac{9}{5}\left(s + \dfrac{5}{3}v\right)^2 \end{aligned}

Answer

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

Common Mistakes

Skipping Middle Term Check: Always verify $2ab$ before concluding an expression is a perfect square; merely having square terms on the ends does not guarantee it.
Factoring Out Fractions: When factoring out $\dfrac{9}{5}$ , remember that dividing by $\dfrac{9}{5}$ is equivalent to multiplying each remaining term by its reciprocal $\dfrac{5}{9}$ .

Question 1

More questions in Exercise 4.2