Exercise 4.4 Practice
00:00
Overview
This page provides comprehensive Ch 4: Algebraic Identities – Exercise 4.4 Practice. Solve problems on filling in binomial factor blanks, finding products without direct multiplication, and factoring trinomial and quadratic expressions with step-by-step solutions.
Trinomial Factorisation & Identity Products
Q1: Fill in the Blanks
Fill in the blanks to complete the following identities:
(i) $s^2 - 11s + 24 = (\underline{\quad\quad}) (\underline{\quad\quad})$
(ii) $(\underline{\quad\quad}) (x + 1) = (3x^2 - 4x - 7)$
(iii) 10x^2 - 11x - 6 = (2x - \underline{\quad\quad}) (\underline{\quad\quad} + 2)$
(iv) 6x^2 + 7x + 2 = (\underline{\quad\quad}) (\underline{\quad\quad})$
(i) $s^2 - 11s + 24 = (\underline{\quad\quad}) (\underline{\quad\quad})$
(ii) $(\underline{\quad\quad}) (x + 1) = (3x^2 - 4x - 7)$
(iii) 10x^2 - 11x - 6 = (2x - \underline{\quad\quad}) (\underline{\quad\quad} + 2)$
(iv) 6x^2 + 7x + 2 = (\underline{\quad\quad}) (\underline{\quad\quad})$
(i) $s^2 - 11s + 24$:
Find two numbers that multiply to $24$ and add to $-11$. These are $-8$ and $-3$.
$$s^2 - 11s + 24 = \mathbf{(s - 8) (s - 3)}$$
Find two numbers that multiply to $24$ and add to $-11$. These are $-8$ and $-3$.
$$s^2 - 11s + 24 = \mathbf{(s - 8) (s - 3)}$$
(ii) $(\underline{\quad\quad}) (x + 1) = (3x^2 - 4x - 7)$:
Split the middle term: $3x^2 - 7x + 3x - 7 = x(3x - 7) + 1(3x - 7) = (3x - 7)(x + 1)$.
Therefore, the blank is:
$$\mathbf{(3x - 7)} (x + 1) = (3x^2 - 4x - 7)$$
Split the middle term: $3x^2 - 7x + 3x - 7 = x(3x - 7) + 1(3x - 7) = (3x - 7)(x + 1)$.
Therefore, the blank is:
$$\mathbf{(3x - 7)} (x + 1) = (3x^2 - 4x - 7)$$
(iii) $10x^2 - 11x - 6 = (2x - \underline{\quad\quad}) (\underline{\quad\quad} + 2)$:
Factorise $10x^2 - 11x - 6$. We need two numbers multiplying to $-60$ and adding to $-11$, which are $-15$ and $4$.
$$10x^2 - 15x + 4x - 6 = 5x(2x - 3) + 2(2x - 3) = (2x - 3)(5x + 2)$$
Comparing with $(2x - \text{blank}_1)(\text{blank}_2 + 2)$, we get:
$$\text{blank}_1 = 3, \quad \text{blank}_2 = 5x$$
Factorise $10x^2 - 11x - 6$. We need two numbers multiplying to $-60$ and adding to $-11$, which are $-15$ and $4$.
$$10x^2 - 15x + 4x - 6 = 5x(2x - 3) + 2(2x - 3) = (2x - 3)(5x + 2)$$
Comparing with $(2x - \text{blank}_1)(\text{blank}_2 + 2)$, we get:
$$\text{blank}_1 = 3, \quad \text{blank}_2 = 5x$$
(iv) $6x^2 + 7x + 2 = (\underline{\quad\quad}) (\underline{\quad\quad})$:
Factorise $6x^2 + 7x + 2$. Split the middle term: $6x^2 + 4x + 3x + 2 = 2x(3x + 2) + 1(3x + 2) = (2x + 1)(3x + 2)$.
Therefore:
$$6x^2 + 7x + 2 = \mathbf{(2x + 1) (3x + 2)}$$
Factorise $6x^2 + 7x + 2$. Split the middle term: $6x^2 + 4x + 3x + 2 = 2x(3x + 2) + 1(3x + 2) = (2x + 1)(3x + 2)$.
Therefore:
$$6x^2 + 7x + 2 = \mathbf{(2x + 1) (3x + 2)}$$
(i) $(s - 8) (s - 3)$ (ii) $(3x - 7)$ (iii) $(2x - 3)(5x + 2)$ (iv) $(2x + 1)(3x + 2)$
Q2: Evaluate Products Mentally
Select and use the identity that will help you to 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) $(205)^2$
(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) $(205)^2$
(i) $(41)^2$:
$$= (40 + 1)^2 = 40^2 + 2(40)(1) + 1^2 = 1600 + 80 + 1 = \mathbf{1681}$$
$$= (40 + 1)^2 = 40^2 + 2(40)(1) + 1^2 = 1600 + 80 + 1 = \mathbf{1681}$$
(ii) $(27)^2$:
$$= (30 - 3)^2 = 30^2 - 2(30)(3) + 3^2 = 900 - 180 + 9 = \mathbf{729}$$
$$= (30 - 3)^2 = 30^2 - 2(30)(3) + 3^2 = 900 - 180 + 9 = \mathbf{729}$$
(iii) $(23 \times 17)$:
$$= (20 + 3)(20 - 3) = 20^2 - 3^2 = 400 - 9 = \mathbf{391}$$
$$= (20 + 3)(20 - 3) = 20^2 - 3^2 = 400 - 9 = \mathbf{391}$$
(iv) $(135)^2$:
Using Śhrīdharāchārya's method (since it ends in 5, offset is 5):
$$= (135 + 5)(135 - 5) + 5^2 = (140 \times 130) + 25 = 18200 + 25 = \mathbf{18225}$$
Using Śhrīdharāchārya's method (since it ends in 5, offset is 5):
$$= (135 + 5)(135 - 5) + 5^2 = (140 \times 130) + 25 = 18200 + 25 = \mathbf{18225}$$
(v) $(97)^2$:
$$= (100 - 3)^2 = 100^2 - 2(100)(3) + 3^2 = 10000 - 600 + 9 = \mathbf{9409}$$
$$= (100 - 3)^2 = 100^2 - 2(100)(3) + 3^2 = 10000 - 600 + 9 = \mathbf{9409}$$
(vi) $(18 \times 29)$:
$$= (20 - 2)(30 - 1) = 20(30) - 20(1) - 2(30) + 2 = 600 - 20 - 60 + 2 = \mathbf{522}$$
$$= (20 - 2)(30 - 1) = 20(30) - 20(1) - 2(30) + 2 = 600 - 20 - 60 + 2 = \mathbf{522}$$
(vii) $(34 \times 43)$:
$$= (40 - 6)(40 + 3) = 40^2 + 40(3 - 6) - 18 = 1600 - 120 - 18 = \mathbf{1462}$$
$$= (40 - 6)(40 + 3) = 40^2 + 40(3 - 6) - 18 = 1600 - 120 - 18 = \mathbf{1462}$$
(viii) $(205)^2$:
$$= (200 + 5)^2 = 200^2 + 2(200)(5) + 5^2 = 40000 + 2000 + 25 = \mathbf{42025}$$
$$= (200 + 5)^2 = 200^2 + 2(200)(5) + 5^2 = 40000 + 2000 + 25 = \mathbf{42025}$$
(i) 1681 (ii) 729 (iii) 391 (iv) 18225 (v) 9409 (vi) 522 (vii) 1462 (viii) 42025
Q3: Factorise Algebraic Expressions
Factor the following:
(i) $9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc$
(ii) $16s^2 + 25t^2 - 40st$
(iii) $r^2 - r - 42$
(iv) $49g^2 + 14gh + h^2$
(v) $64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw$
(i) $9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc$
(ii) $16s^2 + 25t^2 - 40st$
(iii) $r^2 - r - 42$
(iv) $49g^2 + 14gh + h^2$
(v) $64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw$
(i) $9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc$:
Write as: $(3a)^2 + (-b)^2 + (2c)^2 + 2(3a)(-b) + 2(3a)(2c) + 2(-b)(2c)$.
This matches the expansion $(x+y+z)^2$:
$$= \mathbf{(3a - b + 2c)^2}$$
Write as: $(3a)^2 + (-b)^2 + (2c)^2 + 2(3a)(-b) + 2(3a)(2c) + 2(-b)(2c)$.
This matches the expansion $(x+y+z)^2$:
$$= \mathbf{(3a - b + 2c)^2}$$
(ii) $16s^2 + 25t^2 - 40st$:
Write as: $(4s)^2 - 2(4s)(5t) + (5t)^2$.
This matches the expansion $(a - b)^2$:
$$= \mathbf{(4s - 5t)^2}$$
Write as: $(4s)^2 - 2(4s)(5t) + (5t)^2$.
This matches the expansion $(a - b)^2$:
$$= \mathbf{(4s - 5t)^2}$$
(iii) $r^2 - r - 42$:
Find two numbers that multiply to $-42$ and add to $-1$. These are $-7$ and $6$.
$$= r^2 - 7r + 6r - 42 = r(r - 7) + 6(r - 7) = \mathbf{(r - 7)(r + 6)}$$
Find two numbers that multiply to $-42$ and add to $-1$. These are $-7$ and $6$.
$$= r^2 - 7r + 6r - 42 = r(r - 7) + 6(r - 7) = \mathbf{(r - 7)(r + 6)}$$
(iv) $49g^2 + 14gh + h^2$:
Write as: $(7g)^2 + 2(7g)(h) + h^2$.
This matches the expansion $(a + b)^2$:
$$= \mathbf{(7g + h)^2}$$
Write as: $(7g)^2 + 2(7g)(h) + h^2$.
This matches the expansion $(a + b)^2$:
$$= \mathbf{(7g + h)^2}$$
(v) $64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw$:
Recognise squares: $(8u)^2 + (-11v)^2 + (-2w)^2$.
Check cross terms:
• $2(8u)(-11v) = -176uv$
• $2(8u)(-2w) = -32uw$
• $2(-11v)(-2w) = 44vw$
This fits perfectly:
$$= \mathbf{(8u - 11v - 2w)^2}$$
Recognise squares: $(8u)^2 + (-11v)^2 + (-2w)^2$.
Check cross terms:
• $2(8u)(-11v) = -176uv$
• $2(8u)(-2w) = -32uw$
• $2(-11v)(-2w) = 44vw$
This fits perfectly:
$$= \mathbf{(8u - 11v - 2w)^2}$$
(i) $(3a - b + 2c)^2$ (ii) $(4s - 5t)^2$ (iii) $(r - 7)(r + 6)$ (iv) $(7g + h)^2$ (v) $(8u - 11v - 2w)^2$