NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.4#
$$(x + a)(x + b) = x^2 + (a + b)x + ab$$$$(a + b)(a - b) = a^2 - b^2$$
To factor $x^2 + px + q$: find two numbers whose sum is $p$ and product is $q$, then split the middle term. For $ax^2 + bx + c$, find two numbers whose sum is $b$ and product is $ac$.
1. Fill in the blanks to complete the following identities: #
(i) $s^2 - 11s + 24 = (\_\_\_)(\_\_\_)$
We need two numbers with sum $-11$ and product $24$. They are $-8$ and $-3$. We factor $s^2 - 11s + 24$:
$$= s^2 - 8s - 3s + 24$$$$= s(s - 8) - 3(s - 8)$$
$$= (s - 8)(s - 3)$$
∴ $s^2 - 11s + 24 = $ $(s - 8)(s - 3)$.
(ii) $(\_\_\_)(x + 1) = (3x^2 - 4x - 7)$
For $3x^2 - 4x - 7$, we need two numbers with sum $-4$ and product $3 \times (-7) = -21$. They are $-7$ and $3$. We factor $3x^2 - 4x - 7$:
$$= 3x^2 - 7x + 3x - 7$$$$= x(3x - 7) + 1(3x - 7)$$
$$= (3x - 7)(x + 1)$$
∴ The blank is $(3x - 7)$.
(iii) $10x^2 - 11x - 6 = (2x - \_\_\_)(\_\_\_ + 2)$
We need two numbers with sum $-11$ and product $10 \times (-6) = -60$. They are $-15$ and $4$. We factor $10x^2 - 11x - 6$:
$$= 10x^2 - 15x + 4x - 6$$$$= 5x(2x - 3) + 2(2x - 3)$$
$$= (2x - 3)(5x + 2)$$
∴ The blanks are 3 and $5x$: $10x^2 - 11x - 6 = (2x - 3)(5x + 2)$.
(iv) $6x^2 + 7x + 2 = (\_\_\_)(\_\_\_)$
We need two numbers with sum $7$ and product $6 \times 2 = 12$. They are $3$ and $4$. We factor $6x^2 + 7x + 2$:
$$= 6x^2 + 3x + 4x + 2$$$$= 3x(2x + 1) + 2(2x + 1)$$
$$= (2x + 1)(3x + 2)$$
∴ $6x^2 + 7x + 2 = $ $(2x + 1)(3x + 2)$.
2. Select and use the identity that will help you to find the following products without multiplying directly: #
(i) $(41)^2$
Using $(a + b)^2$ with $41 = 40 + 1$. We find $(41)^2$:
$$= 40^2 + 2(40)(1) + 1^2$$$$= 1600 + 80 + 1$$
$$= 1681$$
(ii) $(27)^2$
Using $(a - b)^2$ with $27 = 30 - 3$. We find $(27)^2$:
$$= 30^2 - 2(30)(3) + 3^2$$$$= 900 - 180 + 9$$
$$= 729$$
(iii) $(23 \times 17)$
23 and 17 are 3 more and 3 less than 20, so we use $(a + b)(a - b) = a^2 - b^2$. We find $23 \times 17$:
$$= (20 + 3)(20 - 3)$$$$= 20^2 - 3^2$$
$$= 400 - 9$$
$$= 391$$
(iv) $(135)^2$
Using Śhrīdharāchārya’s form $a^2 = (a + b)(a - b) + b^2$ with $a = 135$, $b = 5$. We find $(135)^2$:
$$= (135 + 5)(135 - 5) + 5^2$$$$= 140 \times 130 + 25$$
$$= 18200 + 25$$
$$= 18225$$
(v) $(97)^2$
Using $(a - b)^2$ with $97 = 100 - 3$. We find $(97)^2$:
$$= 100^2 - 2(100)(3) + 3^2$$$$= 10000 - 600 + 9$$
$$= 9409$$
(vi) $(18 \times 29)$
Using $(x + a)(x + b) = x^2 + (a + b)x + ab$ with $x = 20$, $a = -2$, $b = 9$. We find $18 \times 29$:
$$= (20 - 2)(20 + 9)$$$$= 20^2 + (-2 + 9)(20) + (-2)(9)$$
$$= 400 + 140 - 18$$
$$= 522$$
(vii) $(34 \times 43)$
Using $(x + a)(x + b)$ with $x = 40$, $a = -6$, $b = 3$. We find $34 \times 43$:
$$= (40 - 6)(40 + 3)$$$$= 40^2 + (-6 + 3)(40) + (-6)(3)$$
$$= 1600 - 120 - 18$$
$$= 1462$$
(viii) $(205)^2$
Using $(a + b)^2$ with $205 = 200 + 5$. We find $(205)^2$:
$$= 200^2 + 2(200)(5) + 5^2$$$$= 40000 + 2000 + 25$$
$$= 42025$$
3. Factor the following: #
(i) $9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc$
The squares are $(3a)^2$, $b^2$ and $(2c)^2$. The terms $-6ab$ and $-4bc$ are negative and $12ac$ is positive, so $b$ takes the minus sign. With $3a$, $-b$, $2c$:
$$2(3a)(-b) = -6ab$$$$2(-b)(2c) = -4bc$$
$$2(2c)(3a) = 12ac$$
So, using $(a + b + c)^2$:
$$9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc = (3a - b + 2c)^2$$(ii) $16s^2 + 25t^2 - 40st$
$16s^2 = (4s)^2$, $25t^2 = (5t)^2$ and $-40st = -2(4s)(5t)$. We factor $16s^2 + 25t^2 - 40st$:
$$= (4s)^2 - 2(4s)(5t) + (5t)^2$$$$= (4s - 5t)^2$$
(iii) $r^2 - r - 42$
We need two numbers with sum $-1$ and product $-42$. They are $-7$ and $6$. We factor $r^2 - r - 42$:
$$= r^2 - 7r + 6r - 42$$$$= r(r - 7) + 6(r - 7)$$
$$= (r - 7)(r + 6)$$
(iv) $49g^2 + 14gh + h^2$
$49g^2 = (7g)^2$ and $14gh = 2(7g)(h)$. We factor $49g^2 + 14gh + h^2$:
$$= (7g)^2 + 2(7g)(h) + h^2$$$$= (7g + h)^2$$
(v) $64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw$
The squares are $(8u)^2$, $(11v)^2$ and $(2w)^2$. The $uv$ and $uw$ terms are negative and the $vw$ term is positive, so $u$ has the opposite sign to $v$ and $w$. With $8u$, $-11v$, $-2w$:
$$2(8u)(-11v) = -176uv$$$$2(-11v)(-2w) = 44vw$$
$$2(-2w)(8u) = -32uw$$
So, using $(a + b + c)^2$:
$$64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw = (8u - 11v - 2w)^2$$