Exercise 4.2 Practice
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Overview
This page provides comprehensive Ch 4: Algebraic Identities – Exercise 4.2 Practice. Practice factorising perfect square trinomials using $(a + b)^2$ in reverse, and evaluate squares of numbers using the identity $(a - b)^2 = a^2 - 2ab + b^2$. Free step-by-step solutions and explanations.
Factorising Perfect Squares & Using $(a - b)^2$
Q1: Factor Completely
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?)
(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?)
Strategy: Recognise each expression as $(a + b)^2 = a^2 + 2ab + b^2$, where the first and last terms are perfect squares and the middle term is twice their product.
(i) $9x^2 + 24xy + 16y^2$
$= (3x)^2 + 2(3x)(4y) + (4y)^2$ $$= (3x + 4y)^2$$
$= (3x)^2 + 2(3x)(4y) + (4y)^2$ $$= (3x + 4y)^2$$
(ii) $4s^2 + 20st + 25t^2$
$= (2s)^2 + 2(2s)(5t) + (5t)^2$ $$= (2s + 5t)^2$$
$= (2s)^2 + 2(2s)(5t) + (5t)^2$ $$= (2s + 5t)^2$$
(iii) $49x^2 + 28xy + 4y^2$
$= (7x)^2 + 2(7x)(2y) + (2y)^2$ $$= (7x + 2y)^2$$
$= (7x)^2 + 2(7x)(2y) + (2y)^2$ $$= (7x + 2y)^2$$
(iv) $64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2$
$= (8p)^2 + 2(8p)\!\left(\dfrac{2}{3}q\right) + \left(\dfrac{2}{3}q\right)^2$ $$= \left(8p + \frac{2}{3}q\right)^2$$
$= (8p)^2 + 2(8p)\!\left(\dfrac{2}{3}q\right) + \left(\dfrac{2}{3}q\right)^2$ $$= \left(8p + \frac{2}{3}q\right)^2$$
*(v) $3a^2 + 4ab + \dfrac{4}{3}b^2$
Take out the common factor $\dfrac{1}{3}$: $$= \frac{1}{3}\!\left(9a^2 + 12ab + 4b^2\right) = \frac{1}{3}\!\left[(3a)^2 + 2(3a)(2b) + (2b)^2\right]$$ $$= \frac{1}{3}(3a + 2b)^2$$
Take out the common factor $\dfrac{1}{3}$: $$= \frac{1}{3}\!\left(9a^2 + 12ab + 4b^2\right) = \frac{1}{3}\!\left[(3a)^2 + 2(3a)(2b) + (2b)^2\right]$$ $$= \frac{1}{3}(3a + 2b)^2$$
*(vi) $\dfrac{9}{5}s^2 + 6sv + 5v^2$
Take out the common factor $\dfrac{1}{5}$: $$= \frac{1}{5}\!\left(9s^2 + 30sv + 25v^2\right) = \frac{1}{5}\!\left[(3s)^2 + 2(3s)(5v) + (5v)^2\right]$$ $$= \frac{1}{5}(3s + 5v)^2$$
Take out the common factor $\dfrac{1}{5}$: $$= \frac{1}{5}\!\left(9s^2 + 30sv + 25v^2\right) = \frac{1}{5}\!\left[(3s)^2 + 2(3s)(5v) + (5v)^2\right]$$ $$= \frac{1}{5}(3s + 5v)^2$$
(i) $(3x + 4y)^2$
(ii) $(2s + 5t)^2$
(iii) $(7x + 2y)^2$
(iv) $\left(8p + \dfrac{2}{3}q\right)^2$
*(v) $\dfrac{1}{3}(3a + 2b)^2$ *(vi) $\dfrac{1}{5}(3s + 5v)^2$
*(v) $\dfrac{1}{3}(3a + 2b)^2$ *(vi) $\dfrac{1}{5}(3s + 5v)^2$
Q2: Evaluate Using (a − b)²
Find the values of the following using the identity $(a - b)^2 = a^2 - 2ab + b^2$:
(i) $(79)^2$ (ii) $(193)^2$ (iii) $(299)^2$
(i) $(79)^2$ (ii) $(193)^2$ (iii) $(299)^2$
Strategy: Write each number as a difference of a round number and a small number, then apply $(a - b)^2 = a^2 - 2ab + b^2$.
(i) $(79)^2$
Write $79 = 80 - 1$, so $a = 80$, $b = 1$. $$79^2 = (80 - 1)^2 = 80^2 - 2(80)(1) + 1^2$$ $$= 6400 - 160 + 1 = 6241$$
Write $79 = 80 - 1$, so $a = 80$, $b = 1$. $$79^2 = (80 - 1)^2 = 80^2 - 2(80)(1) + 1^2$$ $$= 6400 - 160 + 1 = 6241$$
(ii) $(193)^2$
Write $193 = 200 - 7$, so $a = 200$, $b = 7$. $$193^2 = (200 - 7)^2 = 200^2 - 2(200)(7) + 7^2$$ $$= 40000 - 2800 + 49 = 37249$$
Write $193 = 200 - 7$, so $a = 200$, $b = 7$. $$193^2 = (200 - 7)^2 = 200^2 - 2(200)(7) + 7^2$$ $$= 40000 - 2800 + 49 = 37249$$
(iii) $(299)^2$
Write $299 = 300 - 1$, so $a = 300$, $b = 1$. $$299^2 = (300 - 1)^2 = 300^2 - 2(300)(1) + 1^2$$ $$= 90000 - 600 + 1 = 89401$$
Write $299 = 300 - 1$, so $a = 300$, $b = 1$. $$299^2 = (300 - 1)^2 = 300^2 - 2(300)(1) + 1^2$$ $$= 90000 - 600 + 1 = 89401$$
(i) $79^2 = 6241$ (ii) $193^2 = 37249$ (iii) $299^2 = 89401$