Exploring Algebraic Identities | Exercise 4.4

Question 2

Select and use the identity that will help you find the following products without multiplying directly:

(i) $(41)^2$

(ii) $(27)^2$

(iii) $(23 \times 17)$

(iv) $(135)^2$

(v) $(97)^2$

(vi) $(18 \times 29)$

(vii) $(34 \times 43)$

(viii) $(205)^2$

Solution

Understand the Question

To evaluate products and squares of numbers without direct multiplication, express each number as a sum or difference relative to a convenient base (like multiples of $10$ or $100$ ), then apply the appropriate algebraic identity:

Square of a sum: $(a + b)^2 = a^2 + 2ab + b^2$
Square of a difference: $(a - b)^2 = a^2 - 2ab + b^2$
Product of sum and difference: $(a + b)(a - b) = a^2 - b^2$
Product of two binomials: $(x + a)(x + b) = x^2 + (a + b)x + ab$

(i) $(41)^2$

Step 1 · Evaluate $(41)^2$

Write $41 = 40 + 1$ . Diagram 1

Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 40$ and $b = 1$

\begin{aligned} (41)^2 &= (40 + 1)^2 \\ &= (40)^2 + 2 \times 40 \times 1 + (1)^2 \\ &= 1600 + 80 + 1 \\ &= 1681 \end{aligned}

Answer

(i) $1681$

(ii) $(27)^2$

Step 1 · Evaluate $(27)^2$

Write $27 = 30 - 3$ .

Using $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 30$ and $b = 3$

\begin{aligned} (27)^2 &= (30 - 3)^2 \\ &= (30)^2 - 2 \times 30 \times 3 + (3)^2 \\ &= 900 - 180 + 9 \\ &= 729 \end{aligned}

Answer

(ii) $729$

(iii) $(23 \times 17)$

Step 1 · Evaluate $23 \times 17$

Write $23 = 20 + 3$ and $17 = 20 - 3$ .

Using $(a + b)(a - b) = a^2 - b^2$ with $a = 20$ and $b = 3$

\begin{aligned} 23 \times 17 &= (20 + 3)(20 - 3) \\ &= (20)^2 - (3)^2 \\ &= 400 - 9 \\ &= 391 \end{aligned}

Answer

(iii) $391$

(iv) $(135)^2$

Step 1 · Evaluate $(135)^2$

Write $135 = 100 + 35$ .

Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 100$ and $b = 35$

\begin{aligned} (135)^2 &= (100 + 35)^2 \\ &= (100)^2 + 2 \times 100 \times 35 + (35)^2 \\ &= 10000 + 7000 + 1225 \\ &= 18225 \end{aligned}

Answer

(iv) $18225$

(v) $(97)^2$

Step 1 · Evaluate $(97)^2$

Write $97 = 100 - 3$ .

Using $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 100$ and $b = 3$

\begin{aligned} (97)^2 &= (100 - 3)^2 \\ &= (100)^2 - 2 \times 100 \times 3 + (3)^2 \\ &= 10000 - 600 + 9 \\ &= 9409 \end{aligned}

Answer

(v) $9409$

(vi) $(18 \times 29)$

Step 1 · Evaluate $18 \times 29$

Write $18 = 20 - 2$ and $29 = 20 + 9$ .

Using $(x + a)(x + b) = x^2 + (a + b)x + ab$ with $x = 20$ , $a = -2$ , and $b = 9$

\begin{aligned} 18 \times 29 &= (20 - 2)(20 + 9) \\ &= (20)^2 + (-2 + 9) \times 20 + (-2) \times 9 \\ &= 400 + (7) \times 20 - 18 \\ &= 400 + 140 - 18 \\ &= 522 \end{aligned}

Answer

(vi) $522$

(vii) $(34 \times 43)$

Step 1 · Evaluate $34 \times 43$

Write $34 = 38 - 4$ and $43 = 38 + 5$ .

Using $(x + a)(x + b) = x^2 + (a + b)x + ab$ with $x = 38$ , $a = -4$ , and $b = 5$

\begin{aligned} 34 \times 43 &= (38 - 4)(38 + 5) \\ &= (38)^2 + (-4 + 5) \times 38 + (-4) \times 5 \\ &= 1444 + (1) \times 38 - 20 \\ &= 1444 + 38 - 20 \\ &= 1462 \end{aligned}

Answer

(vii) $1462$

(viii) $(205)^2$

Step 1 · Evaluate $(205)^2$

Write $205 = 200 + 5$ .

Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 200$ and $b = 5$

\begin{aligned} (205)^2 &= (200 + 5)^2 \\ &= (200)^2 + 2 \times 200 \times 5 + (5)^2 \\ &= 40000 + 2000 + 25 \\ &= 42025 \end{aligned}

Answer

(viii) $42025$

Common Mistakes

Sign in $(a - b)^2$ : Confusing the signs and writing $(a - b)^2 = a^2 - 2ab - b^2$ . The last term is always positive ( $+b^2$ ).
Signs in $(x + a)(x + b)$ : Forgetting to include negative signs when multiplying $ab$ or adding $(a + b)$ if terms involve subtractions.
Inefficient Base Choice: Choosing bases that are not easy multiples of $10$ or $100$ , which defeats the purpose of simplifying mental multiplication.

Question 2

More questions in Exercise 4.4