Factor using suitable identities: (i) (ii) (iii) (iv) (v)

Understand the Question To factor the given algebraic expressions, we match each expression to standard algebraic identities: Perfect Square Trinomials: Square of a Trinomial: Strategy: Express the square terms as , check that the cross-product terms correctly form (or ), and write the expression in its factored square form. Part (i) Step 1 · Factor Using Express the first and last terms as squares: Check the middle term with and : Using : Answer > (i) Part (ii) Step 1 · Factor Using Express the first and last terms as squares: Check the middle term with and : Using : Answer > (ii) Part (iii) Step 1 · Factor Using Identify the three squared terms: Let , , and . Verify the cross-product terms: Using : Answer > (iii) Part (iv) Step 1 · Factor Using Express the first and last terms as squares: Check the middle term with and : Using : Answer > (iv) Part (v) Step 1 · Factor Using Identify the squared terms: Since the cross-terms containing (i.e. and ) are negative, the term with is negative: . Verify the cross-product terms with , , and : Using : Answer > (v) Common Mistakes Sign Identification in Trinomial Squares: In part (v), look at which variable is common to both negative terms ( and share ) to correctly assign the negative sign to . Missing Middle Term Verification: Always verify that reproduces the middle term of the given expression before applying the identity. Reciprocal Simplification: In part (iv), noticing that clarifies why the middle term is simply the constant without variables.

Exploring Algebraic Identities | Exercise 4.3

Question 2

Factor using suitable identities:

(i) $16y^2 - 24y + 9$

(ii) $\dfrac{9}{4}s^2 + 6st + 4t^2$

(iii) $\dfrac{m^2}{9} + \dfrac{mk}{3} + \dfrac{k^2}{4} + 3nk + 2mn + 9n^2$

(iv) $\dfrac{p^2}{16} - 2 + \dfrac{16}{p^2}$

(v) $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc$

Solution

Understand the Question

To factor the given algebraic expressions, we match each expression to standard algebraic identities:

Perfect Square Trinomials:
- $(a + b)^2 = a^2 + 2ab + b^2$
- $(a - b)^2 = a^2 - 2ab + b^2$
Square of a Trinomial:
- $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

Strategy: Express the square terms as $(a)^2, (b)^2, \dots$ , check that the cross-product terms correctly form $2ab$ (or $2ab, 2bc, 2ca$ ), and write the expression in its factored square form.

(i) $16y^2 - 24y + 9$

Step 1 · Factor Using $(a-b)^2$

Express the first and last terms as squares: $16y^2 = (4y)^2, \quad 9 = 3^2$

Check the middle term with $a = 4y$ and $b = 3$ :

\begin{aligned} 2ab &= 2 \times 4y \times 3 \\[0.6em] &= 24y \end{aligned}

Using $(a - b)^2 = a^2 - 2ab + b^2$ : $16y^2 - 24y + 9 = (4y)^2 - 2(4y)(3) + 3^2 = (4y - 3)^2$

Answer

(i) $(4y - 3)^2$

(ii) $\dfrac{9}{4}s^2 + 6st + 4t^2$

Step 1 · Factor Using $(a+b)^2$

Express the first and last terms as squares: $\dfrac{9}{4}s^2 = \left(\dfrac{3}{2}s\right)^2, \quad 4t^2 = (2t)^2$

Check the middle term with $a = \dfrac{3}{2}s$ and $b = 2t$ :

\begin{aligned} 2ab &= 2 \times \left(\dfrac{3}{2}s\right) \times (2t) \\[0.6em] &= 6st \end{aligned}

Using $(a + b)^2 = a^2 + 2ab + b^2$ : $\dfrac{9}{4}s^2 + 6st + 4t^2 = \left(\dfrac{3}{2}s\right)^2 + 2\left(\dfrac{3}{2}s\right)(2t) + (2t)^2 = \left(\dfrac{3}{2}s + 2t\right)^2$

Answer

(ii) $\left(\dfrac{3}{2}s + 2t\right)^2$

(iii) $\dfrac{m^2}{9} + \dfrac{mk}{3} + \dfrac{k^2}{4} + 3nk + 2mn + 9n^2$

Step 1 · Factor Using $(a+b+c)^2$

Identify the three squared terms: $\dfrac{m^2}{9} = \left(\dfrac{m}{3}\right)^2, \quad \dfrac{k^2}{4} = \left(\dfrac{k}{2}\right)^2, \quad 9n^2 = (3n)^2$

Let $a = \dfrac{m}{3}$ , $b = \dfrac{k}{2}$ , and $c = 3n$ . Verify the cross-product terms:

\begin{aligned} 2ab &= 2\left(\dfrac{m}{3}\right)\left(\dfrac{k}{2}\right) = \dfrac{mk}{3} \\[0.6em] 2bc &= 2\left(\dfrac{k}{2}\right)(3n) = 3nk \\[0.6em] 2ca &= 2(3n)\left(\dfrac{m}{3}\right) = 2mn \end{aligned}

Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ : $\dfrac{m^2}{9} + \dfrac{mk}{3} + \dfrac{k^2}{4} + 3nk + 2mn + 9n^2 = \left(\dfrac{m}{3}\right)^2 + \left(\dfrac{k}{2}\right)^2 + (3n)^2 + 2\left(\dfrac{m}{3}\right)\left(\dfrac{k}{2}\right) + 2\left(\dfrac{k}{2}\right)(3n) + 2(3n)\left(\dfrac{m}{3}\right) = \left(\dfrac{m}{3} + \dfrac{k}{2} + 3n\right)^2$

Answer

(iii) $\left(\dfrac{m}{3} + \dfrac{k}{2} + 3n\right)^2$

(iv) $\dfrac{p^2}{16} - 2 + \dfrac{16}{p^2}$

Step 1 · Factor Using $(a-b)^2$

Express the first and last terms as squares: $\dfrac{p^2}{16} = \left(\dfrac{p}{4}\right)^2, \quad \dfrac{16}{p^2} = \left(\dfrac{4}{p}\right)^2$

Check the middle term with $a = \dfrac{p}{4}$ and $b = \dfrac{4}{p}$ :

\begin{aligned} 2ab &= 2 \times \left(\dfrac{p}{4}\right) \times \left(\dfrac{4}{p}\right) \\[0.6em] &= 2 \end{aligned}

Using $(a - b)^2 = a^2 - 2ab + b^2$ : $\dfrac{p^2}{16} - 2 + \dfrac{16}{p^2} = \left(\dfrac{p}{4}\right)^2 - 2\left(\dfrac{p}{4}\right)\left(\dfrac{4}{p}\right) + \left(\dfrac{4}{p}\right)^2 = \left(\dfrac{p}{4} - \dfrac{4}{p}\right)^2$

Answer

(iv) $\left(\dfrac{p}{4} - \dfrac{4}{p}\right)^2$

(v) $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc$

Step 1 · Factor Using $(x+y+z)^2$

Identify the squared terms: $9a^2 = (3a)^2, \quad 4b^2 = (-2b)^2, \quad c^2 = c^2$

Since the cross-terms containing $b$ (i.e. $-12ab$ and $-4bc$ ) are negative, the term with $b$ is negative: $y = -2b$ .

Verify the cross-product terms with $x = 3a$ , $y = -2b$ , and $z = c$ :

\begin{aligned} 2xy &= 2(3a)(-2b) = -12ab \\[0.6em] 2yz &= 2(-2b)(c) = -4bc \\[0.6em] 2zx &= 2(c)(3a) = 6ac \end{aligned}

Using $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$ : $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc = (3a)^2 + (-2b)^2 + c^2 + 2(3a)(-2b) + 2(-2b)(c) + 2(c)(3a) = (3a - 2b + c)^2$

Answer

(v) $(3a - 2b + c)^2$

Common Mistakes

Sign Identification in Trinomial Squares: In part (v), look at which variable is common to both negative terms ( $-12ab$ and $-4bc$ share $b$ ) to correctly assign the negative sign to $2b$ .
Missing Middle Term Verification: Always verify that $2ab$ reproduces the middle term of the given expression before applying the identity.
Reciprocal Simplification: In part (iv), noticing that $\dfrac{p}{4} \times \dfrac{4}{p} = 1$ clarifies why the middle term is simply the constant $2$ without variables.

Question 2

More questions in Exercise 4.3