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Maths 2 Marks Questions

Factorization Of Algebraic Expressions [NEW] · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 48 questions.

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Question 12 Marks
What are the possible expression for the cuboid having volume 3x2 - 12x.
Answer
Volume = 3x2 - 12x
= 3x(x - 4)
= 3 × x(x - 4)
Also volume = Length × Breadth × Height
$\therefore$ Possible expression for dimensions of cuboid are = 3, x, (x - 4)
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Question 22 Marks
Multiply:
(x2 + y2 + z2 - xy + xz + yz) by (x + y -z)
Answer
(x2 + y2 + z2 - xy + xz + yz) by (x + y -z)
= (x + y - z)(x2 + y2 + z2 - xy + xz + yz)
= (x + y + (-z))(x2 + y2 + (-z)2 - xy - y(-z) - (-z)x)
= x3 + y3 + (-z)3 - 3xyz(-z) $\big[\because$ (a + b + c)(a2 + b2 + c2 - ab - bc - ca) = a3 + b3 + c3 - 3abc$\big]$
= x3 + y3 - z3 + 3xyz
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Question 32 Marks
Multiply:
(x2 + 4y2 + z2 + 2xy + xz - 2yz) by (x - 2y - z)
Answer
= (x - 2y - z)(x2 + 4y2 + z2 + 2xy + xz - 2yz)
= (x + (-2y) + (-z))(x2 + (-2y)2 + (-z)2 - x(-2y) - (-2y)(-z) - (-z)x)
= x3 + (-2y)3 + (-z)3 - 3 × x(-2y)(-z) $\big[\because$ (a + b + c)(a2 + b2 + c2 - ab - bc - ca) = a3 + b3 + c3 - 3abc$\big]$
= x3 - 8y3 - z3 + 3 × x × 2yz
= x3 - 8y3 - z3 - 6xyz
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Question 42 Marks
Multiply:
(x2 + 4y2 + 2xy - 3x + 6y + 9) by (x - 2y + 3)
Answer
= (x - 2y + 3)(x2 + 4y2 + 9 + 2xy + 6y - 3x)
= (x + (-2y) + 3) (x2 + (-2y)2 + 32 - x(-2y) - (-2y)3 - 3x)
= x3 + (-2y)3 + 33 - 3 × x (-2y)3$\big[\because$ (a + b + c)(a2 + b2 + c2 - ab - bc - ca) = a3 + b3 + c3 - 3abc$\big]$
= x3 - 8y3 + 27 + 18xy
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Question 52 Marks
If a2 + b2 + c2 = 250 and ab + bc + ca = 3, find a + b + c.
Answer
Recall the formula
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Given that
a2 + b2 + c2 = 250, ab + bc + ca = 3
Then we have
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(a + b + c)2 = 250 + 2.(3)
(a + b + c)2 = 256
(a + b + c) = ±16
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Question 62 Marks
If a2 + b2 + c2 = 20 and a + b + c = 0, find ab + bc + ca.
Answer
Recall the formula
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Given that
a2 + b2 + c2 = 20
(a + b + c) = 0
Then we have
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(0)2 = 20 + 2(ab + bc + ca)
20 + 2(ab + bc + ca) = 0
2(ab + bc + ca) = -20
(ab + bc + ca) = -10
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Question 72 Marks
If a + b + c = 9 and ab + bc + ca = 40, find a2 + b2 +c2.
Answer
Recall the formula

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

Given that

(a + b + c) = 9, ab + bc + ca = 40,

Then we have

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

(9)2 = a2 + b2 + c2 + 2.(40)

a2 + b2 + c2 + 80 = 81

a2 + b2 + c2 = 81 - 80

a2 + b2 + c2 = 1

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Question 82 Marks
If a + b + c = 0, then write the value of a3 + b3 + c3.
Answer
Recall the formula
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
When, (a + b + c) = 0, we have
a3 + b3 + c3 - 3abc = 0.(a2 + b2 + c2 - ab - bc - ca)
= 0
a3 + b3 + c3 - 3abc = 0
⇒ a3 + b3 + c3 = 3abc
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Question 92 Marks
Give the possible expression for the length & breadth of the rectangle having 35y2 - 13y - 12 as its area.
Answer
Area is given as 35y2 - 13y - 12
Splitting the middle term,
Area = 35y2 + 218y - 15y - 12
= 7y(5y + 4) - 3(5y + 4)
= (5y + 4)(7y - 3)
We also know that area of rectangle = length × breadth
$\therefore$ Possible length = (5y + 4) and breadth = (7y - 3)
Or possible length = (7y - 3) and breadth = (5y + 4)
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Question 102 Marks
Factorize the following expressions:
y3 + 125
Answer
y3 + 125
= y3 + 53
$\therefore$ [a3 + b3 = (a + b)(a2 - ab + b2)]
= (y + 5)(y2 - 5y + 52)
= (y + 5)(y2 - 5y + 25)
$\therefore$ y3 + 125
= (y + 5)(y2 - 5y + 25)
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Question 112 Marks
Factorize the following expressions:
x6 + y6
Answer
= (x2)3 + (y2)3
= (x2 + y2)((x2)2 - x2y2 + (y2)2)
= (x2 + y2)(x4 - x2y2 + y4) 
$\big[\therefore$ a3 + b3 = (a + b)(a2 - ab + b2)$\big]$
$\therefore$ x6 + y6 = (x2 + y2)(x4 - x2y2 + y4)
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Question 122 Marks
Factorize the following expressions:
x3 - 8y3 + 27z3 + 18xyz
Answer
x3 - 8y3 + 27z3 + 18xyz
= x3 + (-2y)3 + (3z)3 - 3 × x ×(-2y)(3z)
= (x + (-2y) + 3z)(x2 + (-2y)2 +(3z)2 - x(-2y) - (-2y)(3z) - 3z(x))
$\big[\because$ a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)$\big]$
= (x - 2y + 3z)(x2 + 4y2 + 9z2 + 2xy + 6yz - 3zx)
$\therefore$ x3 - 8y3 + 27z3 + 18xyz = (x - 2y + 3z)(x2 + 4y2 + 9z2 + 2xy + 6yz - 3zx)
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Question 132 Marks
Factorize the following expressions:
$\frac{\text{x}^3}{216}=8\text{y}^3$
Answer
$\frac{\text{x}^3}{216}-8\text{y}^3$
$=\frac{\text{x}^3}{6}-(2\text{y})^3$
$=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\Big(\frac{\text{x}}{6}\Big)^2+\frac{\text{x}}{6}\times2\text{y}+(2\text{y})^2\Big)$
$\therefore\big[\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2)\big]$
$=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}^2}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
$\therefore\frac{\text{x} ^3}{216}-8\text{y} ^3=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
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Question 142 Marks
Factorize the following expressions:
$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
Answer
$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
$=\big(\sqrt{3}\text{a}\big)^3+(-\text{b})^3+\big(-\sqrt{5}\text{c}\big)^3-3\times\big(\sqrt{3}\text{a}\big)(-\text{b})\big(-\sqrt{5}\text{c}\big)$
$=\big(\sqrt{3}\text{a}+(-\text{b})+\big(-\sqrt{5}\text{c}\big)\big)\Big(\big(\sqrt{3}\text{a}\big)^2+(-\text{b})^2+\big(-\sqrt{5}\text{c}\big)^2\\-\sqrt{3}\text{a}(-\text{b})-(-\text{b})\big(-\sqrt{5}\text{c}\big)-\big(-\sqrt{5}\text{c}\big)\sqrt{3}\text{a}\Big)$
$=\big(\sqrt{3}\text{a}+(-\text{b})-\sqrt{5}\text{c}\big)\big(3\text{a}^2+\text{b}^2+5\text{c}^2+\sqrt{3}\text{ab}-\sqrt{5}\text{bc}+\sqrt{15}\text{ac})$
$\therefore3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
$=\big(\sqrt{3}\text{a}+(-\text{b})-\sqrt{5}\text{c}\big)\big(3\text{a}^2+\text{b}^2+5\text{c}^2+\sqrt{3}\text{ab}-\sqrt{5}\text{bc}+\sqrt{15}\text{ac})$
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Question 152 Marks
Factorize the following expressions:
$2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$
Answer
$2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$

$=\big(\sqrt{2}\text{a}\big)^3+\big(\sqrt{3}\text{b}\big)^3+\text{c}^3-3\times\sqrt{2}\text{a}\times\sqrt{3}\text{b}\times\text{c}$

$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\Big(\big(\sqrt{2}\text{a}\big)^2+\big(\sqrt{3}\text{b}\big)^2+\\\text{c}^2-(\sqrt{2}\text{a}\big)(\sqrt{3}\text{b}\big)-(\sqrt{3}\text{b}\big)\text{c}-(\sqrt{2}\text{a}\big)\text{c}\Big)$

$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\big(2\text{a}^2+3\text{b}^2+\text{c}^2-\sqrt{6}\text{ab}-\sqrt{3}\text{bc}-\sqrt{2}\text{ac}\big)$

$\therefore2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$

$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\big(2\text{a}^2+3\text{b}^2+\text{c}^2-\sqrt{6}\text{ab}-\sqrt{3}\text{bc}-\sqrt{2}\text{ac}\big)$

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Question 162 Marks
Factorize the following expressions:
$2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
Answer
$2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
$=\big(\sqrt{2}\text{a}\big)^3+\big(2\sqrt{2}\text{b}\big)^3+(\text{c})^3-3\times\sqrt{2}\text{a}\times2\sqrt{2}\text{b}\times\text{c}$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b})+\text{c}\big)\Big(\big(\sqrt{2}\text{a}\big)^2+\big(2\sqrt{2}\text{b})^2+\text{c}^2\\\big(\sqrt{2}\text{a}\big)\big(2\sqrt{2}\text{b}\big)-\big(2\sqrt{2}\text{b}\big)\text{c}-\big(\sqrt{2}\text{a}\big)\text{c}\Big)$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\big(2\text{a}^2+8\text{b}^2+\text{c}^2-4\text{ab}-2\sqrt{2}\text{bc}-\sqrt{2}\text{ac}\big)$
$\therefore2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\big(2\text{a}^2+8\text{b}^2+\text{c}^2-4\text{ab}-2\sqrt{2}\text{bc}-\sqrt{2}\text{ac}\big)$
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Question 172 Marks
Factorize the following expressions:
p3 + 27
Answer
p3 + 27
= p3 + 33
$\therefore$ [a3 + b3 = (a + b)(a2 - ab + b2)]
= (p + 3)(p² - 3p - 9)
$\therefore$ p3 + 27 = (p + 3)(p² - 3p - 9)
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Question 182 Marks
Factorize the following expressions:
a3 + 8b3 + 64c3 - 24abc
Answer
a3 + 8b3 + 64c3 - 24abc
= (a)3 + (2b)3 + (4c)3 - 3 × 2b × 4c
= (a + 2b + 4c)(a2 + (2b)2 + (4c)2 - a × 2b - 2b × 4c - 4c × a)
$\big[\because$ a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)$\big]$
= (a + 2b + 4c)(a2 + 4b2 + 16c2 - 2ab - 8bc - 4ac)
$\therefore$ a3 + 8b3 + 64c3 - 24abc = (a + 2b + 4c)(a2 + 4b2 + 16c2 - 2ab - 8bc - 4ac)
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Question 192 Marks
Factorize the following expressions:
a12 + b12
Answer
= (a4)3 + (b4)3
= (a4 + b4)((a4)2 - a4 × b4 + (b4)2)
$\big[\therefore$a3 + b3 = (a + b)(a2 - ab + b2)$\big]$
= (a4 + b4)(a8 - a4b4 + b8)
$\therefore$ a12 + b12 = (a4 + b4)(a8 - a4b4 + b8)
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Question 202 Marks
Factorize the following expressions:
(a - 3b)3 + (3b - c)3 + (c - a)3
Answer
 (a - 3b)3 + (3b - c)3 + (c - a)3

Let (a - 3b) = x, (3b - c) = y, (c - a) = z

x + y + z = a - 3b + 3b - c + c - a = 0

$\because$ x + y + z = 0

$\therefore$ x3 + y3 + z3 = 3xyz

$\therefore$ (a - 3b)3 + (3b - c)3 + (c - a)3

= 3(a - 3b)(3b - c)(c - a) 

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Question 212 Marks
Factorize the following expressions:
(a - 2b)3 - 512b3
Answer
(a - 2b)3 - 512b3
= (a - 2b)3 - (8b)3
= (a - 2b - 8b)((a - 2b)2 + (a - 2b)8b + (8b)2)
$\therefore$ [a3 - b3 = (a - b)(a2 + ab + b2)]
= (a - 10b)(a2 + 4b2 - 4ab + 8ab - 16b2 + 64b2)
= (a - 10b)(a2 + 52b2 + 4ab)
$\therefore$ (a - 2b)3 − 512b3 = (a - 10b)(a2 + 52b2 + 4ab)
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Question 222 Marks
Factorize the following expressions:
8x3y3 + 27a3
Answer
8x3y3 + 27a3
= (2xy)3 + (3a)3
= (2xy + 3a)((2xy)2 - 2xy × 3a + (3a)2)
$\therefore$ [a3 + b3 = (a + b)(a2 - ab + b2)]
= (2xy + 3a)(4x2y2 - 6xya + 9a2)
$\therefore$ 8x3y3 + 27a3
= (2xy + 3a)(4x2y2 - 6xya + 9a2)
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Question 232 Marks
Factorize the following expressions:
8x3 + 27y3 - 216z3 + 108xyz
Answer
8x3 + 27y3 - 216z3 + 108xyz
= (2x)3 + (3y)3 + (-6z)3 - 3(2x)(3y)(-6z)
= (2x + 3y - 6z)((2x)2 + (3y)2 + (-6z)2 - 2x × 3y - 3y(-6z)-(-6z)2x)
= (2x + 3y - 6z)(4x2 + 9y2 + 36z2 - 6xy + 18yz + 12zx)
$\therefore$ 8x3 + 27y3 - 216z3 + 108xyz
= (2x + 3y - 6z)(4x2 + 9y2 + 36z2 - 6xy + 18yz + 12zx)
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Question 242 Marks
Factorize the following expressions:
64a3 - b3
Answer
64a3 - b3
= (4a)3 - b3
= (4a - b)((4a)2 + 4a × b + b2)
$\therefore$ [a3 - b3 = (a - b)(a2 + ab + b2)]
= (4a − b)(16a2 + 4ab + b2)
$\therefore$ 64a3 - b3 = (4a - b)(16a2 + 4ab + b2)
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Question 252 Marks
Factorize the following expressions:
(3x - 2y)3 + (2y - 4z)3 + (4z - 3x)3
Answer
(3x - 2y)3 + (2y - 4z)3 + (4z - 3x)3
Let (3x - 2y) = a, (2y - 4z) = b, (4z - 3x) = c
$\therefore$ a + b + c = 3x - 2y + 2y - 4z + 4z -3x = 0
$\because$ a + b + c = 0 
$\therefore$ a3 + b3 + c3 + 3abc
$\therefore$ (3x - 2y)3 + (2y - 4z)3 + (4z - 3x)3
= 3(3x - 2y)(2y - 4z)(4z - 3x)
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Question 262 Marks
Factorize the following expressions:
(2x - 3y)3 + (4z - 2x)3 + (3y - 4z)3
Answer
(2x - 3y)3 + (4z - 2x)3 + (3y - 4z)3
Let 2x - 3y = a, 4z - 2x = b, 3y - 4z = c
$\therefore$ a + b + c = 2x - 3y + 4z - 2x + 3y - 4z = 0
$\because$ a + b + c = 0 
$\therefore$ a3 + b3 + c3 = 3abc
$\therefore$ (2x - 3y)3 + (4z - 2x)3 + (3y - 4z)3
= 3(2x - 3y)(4z - 2x)(3y - 4z)
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Question 272 Marks
Factorize the following expressions:
27x3 - y3 - z3 - 9xyz
Answer
We know that
x3 + y3 + z3 - 3xyz = ( x + y + z)(x2 + y2 + z2 - xy - yz -zx)
$\therefore$ 27x3 - y3 - z3 - 9xyz
= (3x)3 + (-y)3 + (-z)3 - 3(3x)(-y)(-z)
= [3x +(-y) + (-z)][(3x)2 + (-y)2 + (-z)2 - (3x)(-y)(-z)-(-z)(3x)]
= (3x - y - z)(9x2 + y2 + z2 + 3xy - yz + 3zx)
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Question 282 Marks
Factorize the following expressions:
1 - 27a3
Answer
1 - 27a3
= (1)3 - (3a)3
= (1 - 3a)(12 + 1 × 3a + (3a)2)
$\therefore$ [a3 - b3 = (a - b)(a2 + ab + b2)]
= (1 - 3a)(12 + 3a + 9a2)
$\therefore$ 1 - 27a3 = (1 - 3a)(12 + 3a + 9a2)
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Question 292 Marks
Factorize the following expressions:
10x4y - 10xy4
Answer
10x4y - 10xy4
= 10xy(x3 - y3)
= 10xy(x - y)(x2 + xy + y2)
$\therefore$ [x3 - y3 = (x - y)(x2 + xy + y2)]
$\therefore$ 10x4y - 10xy4 = 10xy(x - y)(x2 + xy + y2)
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Question 302 Marks
Factorize:
x4 + x2y2 + y4
Answer
x4 + x2y2 + y4
Adding x2y2 and subtracting x2y2 to the given equation
= x4 + x2y2 + y4 + x2y2 - x2y2
= x4 + 2x2y2 + y4 - x2y2
= (x2)2 + 2 × x2 × y2 + (y2)2 - (xy)2
Using the identity (p + q)2 = p2 + q2 + 2pq
= (x2 + y2)2 - (xy)2
Using the identity p2 - q2 = (p + q)(p - q)
= (x2 + y2 + xy)(x2 + y2 - xy)
$\therefore$ x4 + x2y2 + y4 = (x2 + y2 + xy)(x2 + y2 - xy)
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Question 312 Marks
Factorize:
x3 + 8y3 + 6x2y + 12xy2
Answer
x3 + 8y3 + 6x2y + 12xy2
= (x)3 + (2y)3 + 3 × x2 × 2y + 3 × x × (2y)2
= (x + 2y)3 $\big[\because$ a3 + b3 + 3a2b + 3ab2 = (a + b)3$\big]$
= (x + 2y)(x + 2y)(x + 2y)
$\therefore$ x3 + 8y3 + 6x2y + 12xy2
= (x + 2y)(x + 2y)(x + 2y)
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Question 322 Marks
Factorize:
x3 - 12x(x - 4) - 64
Answer
x3 - 12x(x - 4) - 64

= x3 - 12x2 + 48x - 64

= (x)3 - 3 × x2 × 4 + 3 × 42 × x - 43 $\big[\because$ a3 - 3a2b + 3ab2 - b3= (a - b)3$\big]$

= (x - 4)(x - 4)(x - 4)

$\therefore$ x3 - 12x(x - 4) - 64

= (x - 4)(x - 4)(x - 4)

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Question 332 Marks
Factorize:
$\text{x}^2-\sqrt{3}\text{x}-6$
Answer
$\text{x}^2-\sqrt{3}\text{x}-6$
Splitting the middle term,
$=\text{x}^2-2\sqrt{3}\text{x}+\sqrt{3}\text{x}-6$
$\big[\therefore-\sqrt{3}=-2\sqrt{3}+\sqrt{3} \ \text{also} \ -2\sqrt{3}\times\sqrt{3}=-6\big]$
$=\text{x}\big(\text{x}-2\sqrt{3}\big)+\sqrt{3}\big(\text{x}-2\sqrt{3}\big)$
$=\big(\text{x}-2\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)$
$\therefore\text{x}^2-\sqrt{3}\text{x}-6$
$=\big(\text{x}-2\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)$
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Question 342 Marks
Factorize:
$\text{x}^2+6\sqrt{2}\text{x}+10$
Answer
$\text{x}^2+6\sqrt{2}\text{x}+10$
Splitting the middle term,
$=\text{x}^2+5\sqrt{2}\text{x}+\sqrt{2}\text{x}+10$
$\big[\therefore6\sqrt{2}=5\sqrt{2}+\sqrt{2} \ \text{and} \ 5\sqrt{2}\times\sqrt{2}=10\big]$
$=\text{x}\big(\text{x}+5\sqrt{2}\big)+\sqrt{2}\big(\text{x}+5\sqrt{2}\big)$
$=\big(\text{x}+5\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)$
$\therefore\text{x}^2+6\sqrt{2}\text{x}+10$
$=\big(\text{x}+5\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)$
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Question 352 Marks
Factorize:
$\text{x}^2+5\sqrt{5}\text{x}+30$
Answer
$\text{x}^2+5\sqrt{5}\text{x}+30$
Splitting the middle term,
$=\text{x}^2+2\sqrt{5}\text{x}+3\sqrt{5}\text{x}+30$
$\big[\therefore5\sqrt{5}=2\sqrt{5}+3\sqrt{5} \ \text{also} \ 2\sqrt{5}\times3\sqrt{5}=30\big]$
$=\text{x}\big(\text{x}+2\sqrt{5}\big)+3\sqrt{5}\big(\text{x}+2\sqrt{5}\big)$
$=\big(\text{x}+2\sqrt{5}\big)\big(\text{x}+3\sqrt{5}\big)$
$\therefore\text{x}^2+5\sqrt{5}\text{x}+30$
$=\big(\text{x}+2\sqrt{5}\big)\big(\text{x}+3\sqrt{5}\big)$
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Question 362 Marks
Factorize:
$\text{x}^2+2\sqrt{3}\text{x}-24$
Answer
$\text{x}^2+2\sqrt{3}\text{x}-24$
Splitting the middle term,
$=\text{x}^2+4\sqrt{3}\text{x}-2\sqrt{3}\text{x}-24$
$\big[\therefore2\sqrt{3}=4\sqrt{3}-2\sqrt{3} \ \text{also} \ 4\sqrt{3}\big(-2\sqrt{3}\big)=-24\big]$
$=\text{x}\big(\text{x}+4\sqrt{3}\big)-2\sqrt{3}\big(\text{x}+4\sqrt{3}\big)$
$=\big(\text{x}+4\sqrt{3}\big)\big(\text{x}-2\sqrt{3}\big)$
$\therefore\text{x}^2+2\sqrt{3}\text{x}-24$
$=\big(\text{x}+4\sqrt{3}\big)\big(\text{x}-2\sqrt{3}\big)$
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Question 372 Marks
Factorize:
$\text{x}^2-2\sqrt{2}\text{x}-30$
Answer
$\text{x}^2-2\sqrt{2}\text{x}-30$
Splitting the middle term,
$=\text{x}^2=5\sqrt{2}\text{x}+3\sqrt{2}\text{x}-30$
$\big[\therefore-2\sqrt{2}=-5\sqrt{2}+3\sqrt{2} \ \text{also} \ -5\sqrt{2}\times3\sqrt{2}=-30\big]$
$=\text{x}\big(\text{x}-5\sqrt{2}\big)+3\sqrt{2}\big(\text{x}-5\sqrt{2}\big)$
$=\big(\text{x}-5\sqrt{2}\big)\big(\text{x}+3\sqrt{2}\big)$
$\therefore\text{x}^2-2\sqrt{2}\text{x}-30$
$=\big(\text{x}-5\sqrt{2}\big)\big(\text{x}+3\sqrt{2}\big)$
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Question 382 Marks
Factorize:
$\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$
Answer
 $\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$

$=\Big(\frac{2}{3}\text{x}\Big)^3+(1)^3+3\times\Big(\frac{2}{3}\text{x}\Big)^2\times1+3(1)^2\times\Big(\frac{2}{3}\text{x}\Big)$

$=\Big(\frac{2}{3}\text{x}+1\Big)^3$$\big[\because$ a3 + b3 + 3a2b + 3ab2 = (a + b)3$\big]$

$=\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)$

$\therefore\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$

$=\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)$ 

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Question 392 Marks
Factorize:
a3x3 - 3a2bx2 + 3ab2x - b3
Answer
a3x3 - 3a2bx2 + 3ab2x - b3
= (ax)3 - 3(ax)2 × b + 3(ax)b2 - b3
= (ax - b)3 $\big[\because$ a3 - 3a2b + 3ab2 - b3= (a - b)3$\big]$
= (ax - b)(ax - b)(ax - b)
$\therefore$ a3x3 - 3a2bx2 + 3ab2x - b3
= (ax - b)(ax - b)(ax - b)
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Question 402 Marks
Factorize:
a2 - b2 + 2bc - c2
Answer
a2 - b2 + 2bc - c2
a2 - (b2 - 2bc + c2)
Using the identity (a - b)2 = a2 + b2 - 2ab
= a2 - (b - c)2
Using the identity a2 - b2 = (a + b)(a - b)
= (a + b - c)(a - (b - c))
= (a + b - c)(a - b + c)
$\therefore$ a2 - b2 + 2bc - c2
= (a + b - c)(a - b + c)
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Question 412 Marks
Factorize:
a2 + b2 + 2(ab + bc + ca)
Answer
= a2 + b2 + 2ab + 2bc + 2ca
Using the identity (p + q)2 = p2 + q2 + 2pq
We get,
= (a + b)2 + 2bc + 2ca
= (a + b)2 + 2c(b + a)
Or (a + b)2 + 2c(a + b)
Taking (a + b) common
= (a + b)(a + b + 2c)
$\therefore$ a2 + b2 + 2(ab + bc + ca) = (a + b)(a + b + 2c)
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Question 422 Marks
Factorize:
8x3 + y3 + 12x2y + 6xy2
Answer
8x3 + y3 + 12x2y + 6xy2
= (2x)3 + y3 + 3 × (2x)2 × y + 3(2×) × y2
= (2x + y)3 $\big[\because$ a3 + b3 + 3a2b + 3ab2 = (a + b)3$\big]$
= (2x + y)(2x + y)(2x + y)
$\therefore$ 8x3 + y3 + 12x2y + 6xy2
= (2x + y)(2x + y)(2x + y)
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Question 432 Marks
Factorize:
8x3 + 27y3 + 36x2y + 54xy2
Answer
8x3 + 27y3 + 36x2y + 54xy2
= (2x)3 + (3y)3 + 3 × (2x)2 × 3y + 3 × (2x)(3y)2
= (2x + 3y)3 $\big[\because$ a3 + b3 + 3a2b + 3ab2 = (a + b)3$\big]$
= (2x + 3y)(2x + 3y)(2x + 3y)
$\therefore$ 8x3 + 27y3 + 36x2y + 54xy2
= (2x + 3y)(2x + 3y)(2x + 3y)
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Question 442 Marks
Factorize:
8a3 - 27b3 - 36a2b + 54ab2
Answer
8a3 - 27b3 - 36a2b + 54ab2
= (2a)3 - (3b)3 - 3 × (2a)2 × 3b + 3 × (2a)(3b)2
= (2a - 3b)3 $\big[\because$ a3 - b3 - 3a2b + 3ab2 = (a - b)3$\big]$
= (2a - 3b)(2a - 3b)(2a - 3b)
$\therefore$ 8a3 - 27b3 - 36a2b + 54ab2
= (2a - 3b)(2a - 3b)(2a - 3b)
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Question 452 Marks
Factorize:
8a3 + 27b3 + 36a2b + 54ab2
Answer
8a3 + 27b3 + 36a2b + 54ab2
= (2a)3 + (3b)3 + 3 × (2a)2 × 3b + 3 × (2a)(3b)2
= (2a + 3b)3 $\big[\because$ a3 + b3 + 3a2b + 3ab2 = (a + b)3$\big]$
= (2a + 3b)(2a + 3b)(2a + 3b)
$\therefore$ 8a3 + 27b3 + 36a2b + 54ab2
= (2a + 3b)(2a + 3b)(2a + 3b)
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Question 462 Marks
Factorize:
64a3 + 125b3 + 240a2b + 300ab2
Answer
64a3 + 125b3 + 240a2b + 300ab2
= (4a)3 + (5b)3 + 3 × (4a)2 × 5b + 3(4a)(5b)2
= (4a + 5b)3 $\big[\because$ a3 + b3 + 3a2b + 3ab2 = (a + b)3$\big]$
= (4a + 5b)(4a + 5b)(4a + 5b)
$\therefore$ 64a3 + 125b3 + 240a2b + 300ab2
= (4a + 5b)(4a + 5b)(4a + 5b)
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Question 472 Marks
Factorize:
2(x + y)2 - 9(x + y) - 5
Answer
Let x + y = z
= 2z2 - 9z - 5
Splitting the middle term,
= 2z2 - 10z + z - 5
= 2z(z - 5) + 1(z - 5)
= (z - 5)(2z + 1)
Substituting z = x + y
= (x + y - 5)(2(x + y) + 1)
= (x + y - 5)(2x + 2y + 1)
$\therefore$ 2(x + y)2 - 9(x + y) - 5 = (x + y - 5)(2x + 2y + 1)
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Question 482 Marks
Factorize:
125x3 - 27y3 - 225x2y + 135xy2
Answer
125x3 - 27y3 - 225x2y + 135xy2
= (5x)3 - (3y)3 - 3 × (5x)2 × 3y + 3 × (5x)(3y)2
= (5x - 3y)3 $\big[\because$ a3 - b3 - 3a2b + 3ab2 = (a - b)3$\big]$
= (5x - 3y)(5x - 3y)(5x - 3y)
$\therefore$ 125x3 - 27y3 - 225a2y + 135xy2
= (5x - 3y)(5x - 3y)(5x - 3y)
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2 Marks Questions - Maths STD 9 Questions - Vidyadip