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Maths 3 Marks Question

Factorization Of Algebraic Expressions [NEW] · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 41 questions.

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Question 13 Marks
Simplify:
$\frac{173\times173\times173+127\times127\times127}{173\times173-173\times127+127\times127}$
Answer
$\frac{173\times173\times173+127\times127\times127}{173\times173-173\times127+127\times127}$
$=\frac{173^3+127^3}{173^2-173\times127+127^2}$
$=\frac{(173+127)(173^2-173\times127+127^2)}{173^2-173\times127+127^2}$
$\big[\therefore$ a3 + b3 = (a + b)(a2 − ab + b2)$\big]$
= (173 + 127)
= 300
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Question 23 Marks
Simplify:
$\frac{155\times155\times155-55\times55\times55}{155\times155+155\times55+55\times55}$
Answer
$\frac{155\times155\times155-55\times55\times55}{155\times155+155\times55+55\times55}$
$=\frac{155^3-55^3}{155^2+155\times55+55^2}$
$=\frac{(155-55)(155^2+155\times55+55^2)}{155^2+155\times55+55^2}$
$\big[\therefore$ a3 - b3 = (a - b)(a2 + ab + b2)$\big]$
= (155 - 55)
= 100
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Question 33 Marks
Simplify:
$\frac{1.2\times1.2\times1.2-0.2\times0.2\times0.2}{1.2\times1.2+1.2\times0.2+0.2\times0.2}$
Answer
$\frac{1.2\times1.2\times1.2-0.2\times0.2\times0.2}{1.2\times1.2+1.2\times0.2+0.2\times0.2}$
$=\frac{1.2^3-0.23}{1.2^2+1.2\times0.2+0.2^2}$
$=\frac{(1.2-0.2)((1.2)^2+1.2\times0.2+(0.2)^2)}{1.2^2+1.2\times0.2+0.2^2}$
$\big[\therefore$ a3 - b3 = (a - b)(a2 + ab + b2)$\big]$
= (1.2 - 0.2)
= 1.0
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Question 43 Marks
Multiply:
(9x2 + 25y2 + 15xy + 12x - 20y + 16) by (3x - 5y + 4)
Answer
= (3x - 5y + 4)(9x2 + 25y2 + 15xy + 20y - 12x + 16)
= (3x + (5y) + 4){(3x)2 + (-5y)2 + 42 - 3x(-5y) - (-5y)4 - 4(3x)}
$\big[\because$ (a + b + c)(a2 + b2 + c2 - ab - bc - ca) = a3 + b3 + c3 - 3abc$\big]$
Here, a = 3x, b = -5y, c = 4
= (3x)3 + (-5y)3 + 43 - 3(3x)(-5y)(4)
= 27x3 - 125y3 + 64 + 180xy
$\therefore$ (3x - 5y + 4)(9x2 + 25y2 + 15xy + 20y - 12x + 16)
= 27x3 - 125y3 + 64 + 180xy
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Question 53 Marks
Find the value of x3 + y3 - 12xy + 64, when x + y = -4
Answer
$\because$ x + y = -4
$\therefore$ x + y + 4 = 0 ...(1)
Now, x3 + y3 - 12xy + 64
= x3 + y3 + 64 - 12xy
= (x)3 + y3 + 43 - 3 × x × y × 4
= (x + y + 4)(x2 + y2 + 16 - xy - 4y - 4x)
= 0(x2 + y2 + 16 - xy - 4y - 4x) [from(1)]
= 0
$\therefore$ x3 + y3 - 12xy + 64 = 0 when x + y = -4
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Question 63 Marks
Factorize the following expressions:
x4y4 - xy
Answer
x4y4 - xy
= xy(x3y3 - 1)
= xy((xy)3 - 13)
= xy(xy - 1)((xy)2 + xy × 1 + 12)
$\big[\therefore$x3 - y3 = (x - y)(x2 + xy + y2)$\big]$
= xy(xy - 1)(x2y2 + xy + 1)
$\therefore$ x4y4 - xy = xy(xy - 1)(x2y2 + xy + 1)
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Question 73 Marks
Factorize the following expressions:
x3y3 + 1
Answer
x3y3 + 1
= (xy)3 + 13
= (xy + 1)((xy)2 + xy + 12)
$\big[\therefore$ x3 + y3 = (x + y)(x2 - xy + y2)$\big]$
= (xy + 1)(x2y2 - xy + 1)
$\therefore$ x3y3 + 1 = (xy + 1)(x2y2 - xy + 1)
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Question 83 Marks
Factorize the following expressions:
(x + 2)3 + (x - 2)3
Answer
= (x + 2 + x - 2)((x + 2)2 - (x + 2)(x - 2) + (x - 2)2)
$\big[\therefore$ [a3 + b3 = (a + b)(a2 - ab + b2)$\big]$
= 2x(x2 + 4x + 4 - (x + 2)(x - 2) + x2 - 4x + 4)
= 2x(2x2 + 8 - (x2 - 22))
$\big[\therefore$ (a + b)(a - b) = a2 - b2$\big]$
= 2x(2x2 + 8 - x2 + 4)
= 2x(x2 + 12)
$\therefore$ (x + 2)3 + (x - 2)3 = 2x(x2 + 12)
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Question 93 Marks
Factorize the following expressions:
$\frac{1}{27}\text{x}^3-\text{y}^3+125\text{z}^3+5\text{xyz}$
Answer
$\frac{1}{27}\text{x}^3-\text{y}^3+125\text{z}^3+5\text{xyz}$
$=\Big(\frac{\text{x}}{3}\Big)^3+(-\text{y})^3+(5\text{z})^3-3\times\frac{\text{x}}{3}(-\text{y})(5\text{z})$
$=\Big(\frac{\text{x}}{3}+(-\text{y})+5\text{z}\Big)\Big(\frac{\text{x}}{3}\Big)^2+(-\text{y})^2+(5\text{z})^2-\frac{\text{x}}{3}(-\text{y})-(-\text{y})5\text{z}-5\text{z}\Big(\frac{\text{x}}{3}\Big)\Big)$
$=\Big(\frac{\text{x}}{3}-\text{y}+5\text{z}\Big)\Big(\frac{\text{x}^2}{9}+\text{y}^2+25\text{z}^2+\frac{\text{xy}}{3}+5\text{yz}-\frac{5}{3}\text{zx}\Big)$
$\therefore\frac{1}{27}\text{x}^3-\text{y}^3+125\text{z}^3+5\text{xyz}$
$=\Big(\frac{\text{x}}{3}-\text{y}+5\text{z}\Big)\Big(\frac{\text{x}^2}{9}+\text{y}^2+25\text{z}^2+\frac{\text{xy}}{3}+5\text{yz}-\frac{5}{3}\text{zx}\Big)$
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Question 103 Marks
Factorize the following expressions:
$\Big(\text{a}^3-\frac{1}{\text{a}^3}\Big)-2\text{a}+\frac{2}{\text{a}}$
Answer
$=\Big(\text{a}^3-\frac{1}{\text{a}^3}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}^3-\Big(\frac{1}{\text{a}^3}\Big)\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\text{a}\times\frac{1}{\text{a}}+\Big(\frac{1}{\text{a}}\Big)^2\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$\big[\therefore$ a3 - b3 = (a - b)(a2 + ab + b2)$\big]$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}-2\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
$\therefore\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+2\text{a}$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
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Question 113 Marks
Factorize the following expressions:

8x3 - 125y3 + 180xy + 216

Answer
8x3 - 125y3 + 180xy + 216
or, 8x3 - 125y3+ 216 + 180xy
= (2x)3 + (-5y)3 + 63 - 3 × (2x)(-5y)(6)
= (2x + (-5y) + 6)((2x)2 + (-5y)2 +62 - 2x(-5y) - (-5y)6 - 6(2x))
= (2x - 5y + 6)(4x2 + 25y2 + 36 + 10xy + 30y -12x)
$\therefore$ 8x3 - 125y3 + 180xy + 216
= (2x - 5y + 6)(4x2 + 25y2 + 36 + 10xy + 30y -12x)
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Question 123 Marks
Factorize the following expressions:
a3 + b3 + a + b
Answer
a3 + b3 + a + b
= (a3 + b3) + 1(a + b)
= (a + b)(a2 - ab + b2) + 1(a + b)
$\big[\therefore$ a3 + b3 = (a + b)(a2 - ab + b2)$\big]$
= (a + b)(a2 - ab + b2 + 1)
$\therefore$ a3 + b3 + a + b = (a + b)(a2 - ab + b2 + 1)
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Question 133 Marks
Factorize the following expressions:
a3 + 3a2b + 3ab2 + b3 - 8
Answer
= (a + b)3 - 8 
$\big[\therefore$ a3 + 3a2b + 3ab2 + b3 = (a + b)3$\big]$
= (a + b)3 - 23
= (a + b - 2)((a + b)2 + (a + b) × 2 + 22)
= (a + b - 2)(a² + 2ab + b² + 2a + 2b + 4)
$\therefore$ a3 + 3a2b + 3ab2 + b3 - 8 = (a + b - 2)(a² + 2ab + b² + 2a + 2b + 4)
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Question 143 Marks
Factorize the following expressions:
8x2y3 - x5
Answer
8x2y3 - x5
= x2((2y)3 - x3)
= x2(2y - x)((2y)2 + 2y × x + x2)
$\big[\therefore$ x3 - y3 = (x - y)(x2 + xy + y2)$\big]$
= x2(2y - x)(4y2 + 2xy + x2)
$\therefore$ 8x2y3 - x5 = x2(2y - x)(4y2 + 2xy + x2)
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Question 153 Marks
Factorize the following expressions:
8a3 - b3 - 4ax + 2bx
Answer
= (2a)3 - b3 - 2x(2a - b)
= (2a - b)((2a)2 + 2a × b + b2) - 2x(2a - b) 
$\big[\therefore$ a3 - b3 = (a - b)(a2 + ab + b2)$\big]$
= (2a - b)(4a2 + 2ab + b2 - 2x)
$\therefore$ 8a3 - b3 - 4ax + 2bx = (2a - b)(4a2 + 2ab + b2 - 2x)
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Question 163 Marks
Factorize the following expressions:
54x6y + 2x3y4
Answer
54x6y + 2x3y4
= 2x3y(27x3 + y3)
= 2x3y((3x)3 + y3)
= 2x3y(3x + y)((3x)2 - 3x × y + y2)
$\therefore$ [a3 + b3 = (a + b)(a2 - ab + b2)]
= 2x3y(3x + y)(9x2 - 3xy + y2)
$\therefore$ 54x6y + 2x3y4 = 2x3y(3x + y)(9x2 - 3xy + y2)
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Question 173 Marks
Factorize the following expressions:
32a3 + 108b3
Answer
32a3 + 108b3
= 4(8a3 + 27b3)
= 4((2a)3 + (3b)3)
= 4[(2a + 3b)((2a)2 - 2a × 3b + (3b)2
$\therefore$ [a3 + b3 = (a + b)(a2 - ab + b2)]
= 4(2a + 3b)(4a2 - 6ab + 9b2)
$\therefore$ 32a3 + 108b3 = 4(2a + 3b)(4a2 - 6ab + 9b2)
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Question 183 Marks
Factorize the following expressions:
125 + 8x3 - 27y3 + 90xy
Answer
125 + 8x3 - 27y3 + 90xy
= 53 + (2x)3 + (-3y)3 - 3 × 5 × 2x × (-3y)
= (5 + 2x + (-3y))(52 + (2x)2 + (-3y)2 - 5(2x) - 2x(-3y) - (-3y)5)
= (5 + 2x + -3y)(25 + 4x2 + 9y2 - 10x + 6xy + 15y)
$\therefore$ 125 + 8x3 - 27y3 + 90xy
= (5 + 2x + -3y)(25 + 4x2 + 9y2 - 10x + 6xy + 15y)
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Question 193 Marks
Factorize the following expressions:
1029 - 3x3
Answer
1029 - 3x3
= 3(343 - x3)
= 3((7)3 - x3)
= 3(7 - x)(72 + 7x + x2) 
$\big[\therefore$ a3 - b3 = (a - b)(a2 + ab + b2)$\big]$
= 3(7 - x)(49 + 7x + x2)
$\therefore$ 1029 - 3x3 = 3(7 - x)(49 + 7x + x2)
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Question 203 Marks
Factorize:
xy9 - yx9
Answer
The given expression to be factorized is
xy9 - yx9
This can be wriiten in the form
xy9 - yx9 = x.y.y8 - y.x.x8
Take common xy from the two terms of the above expression
xy9 - yx9 = xy(y8 - x8)
= xy(y8 - x8)
= {xy(y4)2 - (x4)2)}
= xy(y4 + x4)(y4 - x4)
xy9 - yx9 = xy(y4 + x4){(y2)2 - (x2)2}
= xy(y4 + x4)(y2 + x2)(y2 - x2)
= xy(y4 + x4)(y2 + x2){(y)2 - (x)2}
= xy(y4 + x4)(y2 + x2)(y + x)(y - x)
We cannot further factorize the expression.
So, the required factorization of xy9 - yx9 is xy(y4 + x4)(y2 + x2)(y + x)(y - x)
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Question 213 Marks
Factorize:
x(x3 - y3) + 3xy(x - y)
Answer
x(x3 - y3) + 3xy(x - y)
Elaborating x3 - y3 using the identity
x3 - y3 = (x - y)(x2 + xy + y2)
= x(x - y)(x2 + xy + y2) + 3xy(x - y)
Taking common x(x - y) in both the terms
= x(x - y)(x2 + xy + y2 + 3y)
$\therefore$ x(x3 - y3) + 3xy(x - y)
= x(x - y)(x2 + xy + y2 + 3y)
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Question 223 Marks
Factorize:
x4 + x2 + 25.
Answer
The given expression to be factorized is x4 + x2 + 25
This can be written in the form
x4 + x2 + 25 = (x2)2 + 2.x2.5 + (5)2 - 9x2
= {(x2)2 + 2.x2.5 + (5)2} - (3x)2
= (x2 + 5)2 - (3x)2
= (x2 + 5 + 3x)(x2 + 5 - 3x)
We cannot further factorize the expression.
So, the required factorization is x4 + x2 + 25 = (x2 + 5 + 3x)(x2 + 5 - 3x).
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Question 233 Marks
Factorize:
x3 + x - 3x2 - 3
Answer
x3 + x - 3x2 - 3
Taking x common in x3 + x
= x(x2 + 1) - 3x2 - 3
Taking -3 common in -3x2 - 3
= x(x2 + 1) - 3(x2 + 1)
Now, we take (x2 + 1) common
= (x2 + 1)(x - 3)
$\therefore$ x3 + x - 3y2 - 3
= (x2 + 1)(x - 3)
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Question 243 Marks
Factorize:
x3 - 2x2y + 3xy2 - 6y3
Answer
x3 - 2x2y + 3xy2 - 6y3
Taking x2 common in (x3 - 2x2y) and +3y2 common in (3xy2 - 6y3)
= x2(x - 2y) + 3y2(x - 2y)
Taking (x - 2y) common in the terms
= (x - 2y)(x2 + 3y2)
$\therefore$ x3 - 2x2y + 3xy2 - 6y3 = (x - 2y)(x2 + 3y2)
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Question 253 Marks
Factorize:
x2 + y - xy - x
Answer
x2 + y - xy - x
On rearranging
x2 - xy - x + y
Taking x common in the (x2 - xy) and -1 in (-x + y)
= x(x - y) - 1(x - y)
Taking (x - y) common in the terms
= (x - y)(x - 1)
$\therefore$ x2 + y - xy - x = (x - y)(x - 1)
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Question 263 Marks
Factorize:
x2 - y2 - 4xz + 4z2
Answer
x2 - y2 - 4xz + 4z2
On rearranging the terms
= x2 - 4xz + 4z2 - y2
= (x)2 - 2 × x × 2z + (2z)2 - y2
Using the identity x2 - 2xy + y2 = (x - y)2
= (x - 2z)2 - y2
Using the identity p2 - q2 = (p + q)(p - q)
= (x - 2z + y)(x - 2z - y)
$\therefore$ x2 - y2 - 4xz + 4z2 = (x - 2z + y)(x - 2z - y)
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Question 273 Marks
Factorize:
x2 - 1 - 2a - a2
Answer
The given expression to be factorized is x2 - 1 - 2a - a2
Take common -1 from the last three terms and then we have
x2 - 1 - 2a - a2
= x2 - (1 + 2a + a2)
= x2 - {(1)2 + 2.1.a + (a)2}
= x2 - (1 + a)2
= (x)2 - (1 + a)2
= {x + (1 + a)}{x - (1 + a)}
= (x + 1 + a)(x - 1 - a)
We cannot further factorize the expression.
So, the required factorization is x2 - 1 - 2a - a2 = (x + 1 + a)(x - 1 - a).
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Question 283 Marks
Factorize:
(x + 2)(x2 + 25) - 10x2 - 20x
Answer
(x + 2)(x2 + 25) - 10x2 - 20x
(x + 2)(x2 + 25) - 10x (x + 2)
Taking (x + 2) common in both the terms
= (x + 2)(x2 + 25 - 10x)
= (x + 2)(x2 - 10x + 25)
Splitting the middle term of (x2 - 10x + 25)
= (x + 2)(x2 - 5x - 5x + 25)
= (x + 2){x(x - 5)-5 (x - 5)}
= (x + 2)(x - 5)(x - 5)
$\therefore$ (x + 2)(x2 + 25) - 10x2 - 20x = (x + 2)(x - 5)(x - 5)
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Question 293 Marks
Factorize:
$5\sqrt{5}\text{x}^2+20\text{x}+3\sqrt{5}$
Answer
$5\sqrt{5}\text{x}^2+20\text{x}+3\sqrt{5}$
Splitting the middle term,
$=5\sqrt{5}\text{x}^2+15\text{x}+5\text{x}+3\sqrt{5}$
$\big[\therefore20=15+5 \ \text{and} \ 15\times5=5\sqrt{5}\times3\sqrt{5}\big]$
$=5\text{x}\big(\sqrt{5}\text{x}+3\big)+\sqrt{5}\big(\sqrt{5}\text{x}+3\big)$
$=\big(\sqrt{5}\text{x}+3\big)\big(5\text{x}+\sqrt{5}\big)$
$\therefore5\sqrt{5}\text{x}^2+20\text{x}+3\sqrt{5}$
$=\big(\sqrt{5}\text{x}+3\big)\big(5\text{x}+\sqrt{5}\big)$
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Question 303 Marks
Factorize:
$2\text{x}^2-\frac{5}{6}\text{x}+\frac{1}{12}$
Answer
$2\text{x}^2-\frac{5}{6}\text{x}+\frac{1}{12}$
Splitting the middle term,
$=2\text{x}^2-\text{x}^2-\text{x}^3+\frac{1}{12}$
$\Big[\therefore-\frac{5}{6}=-\frac{1}{2}-\frac{1}{3} \ \text{also} \ -\frac{1}{2}\times-\frac{1}{3}=2\times\frac{1}{12}\Big]$
$=\text{x}\Big(2\text{x}-\frac{1}{2}\Big)-\frac{1}{6}\Big(2\text{x}-\frac{1}{2}\Big)$
$=\Big(2\text{x}-\frac{1}{2}\Big)\Big(\text{x}-\frac{1}{6}\Big)$
$\therefore2\text{x}^2-56\text{x}+\frac{1}{12}=\Big(2\text{x}-\frac{1}{2}\Big)\Big(\text{x}-\frac{1}{6}\Big)$
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Question 313 Marks
Factorize:
$2\text{x}^2+3\sqrt{5}\text{x}+5$
Answer
$2\text{x}^2+3\sqrt{5}\text{x}+5$
Splitting the middle term,
$=2\text{x}^2+2\sqrt{5}\text{x}+\sqrt{5}\text{x}+5$
$=2\text{x}\big(\text{x}+\sqrt{5}\big)+\sqrt{5}\big(\text{x}+\sqrt{5}\big)$
$=\big(\text{x}+\sqrt{5}\big)\big(2\text{x}+\sqrt{5}\big)$
$\therefore2\text{x}^2+3\sqrt{5}\text{x}+5$
$=\big(\text{x}+\sqrt{5}\big)\big(2\text{x}+\sqrt{5}\big)$
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Question 323 Marks
Factorize:
$2\text{a}^2+2\sqrt{6}\text{ab}+3\text{b}^2$
Answer
$2\text{a}^2+2\sqrt{6}\text{ab}+3\text{b}^2$
$=\big(\sqrt{2}\text{a}\big)^2+2\times\sqrt{2}\text{a}\times\sqrt{3}\text{b}+\big(\sqrt{3}\text{b}\big)^2$
Using the identity (p + q)2 = p2 + q2 + 2pq
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)^2$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)$
$\therefore2\text{a}^2+2\sqrt{6}\text{ab}+3\text{b}^2$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)$
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Question 333 Marks
Factorize:
$21\text{x}^2-2\text{x}+\frac{1}{21}$
Answer
$21\text{x}^2-2\text{x}+\frac{1}{21}$
$=\big(\sqrt{21\text{x}}\big)^2-2\sqrt{21}\text{x}\times\frac{1}{\sqrt{21}}+\Big(\frac{1}{\sqrt{21}}\Big)^2$
Using the identity (x - y)2 = x2 + y2 - 2xy
$\Big(\sqrt{21}\text{x}-\frac{1}{\sqrt{21}}\Big)^2$
$\therefore21\text{x}^2-2\text{x}+\frac{1}{21}=\Big(\sqrt{21}\text{x}-\frac{1}{\sqrt{21}}\Big)^2$
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Question 343 Marks
Factorize:
a3 - 3a2b + 3ab2 - b3 + 8
Answer
a3 - 3a2b + 3ab2 - b3 + 8
= (a - b)3 + 8) $\big[\because$ a3 - b3 + 3a2b + 3ab2 = (a - b)3$\big]$
= (a - b)3 + 23
= (a - b + 2)((a - b)2 - (a - b)2 + 22) $\big[\because$ a3 + b3 = (a + b)(a2 - ab + b2)$\big]$
= (a - b + 2)(a2 + b2 - 2ab - 2a + 2b + 4)
$\therefore$ a3 - 3a2b + 3ab2 - b3 + 8
= (a - b + 2)(a2 + b2 - 2ab - 2a + 2b + 4)
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Question 353 Marks
Factorize:
a2x2 + (ax2 + 1)x + a
Answer
a2x2 + (ax2 + 1)x + a
We multiply x(ax2 + 1) = ax3 + x
= a2x2 + ax3 + x + a
Taking common ax2 in (a2x2 + ax3) and 1 in (x + a)
= ax2(a + x) + 1(x + a)
= ax2(a + x) + 1(a + x)
Taking (a + x) common in both the terms
= (a + x)(ax2 + 1)
$\therefore$ a2x2 + (ax2 + 1)x + a
= (a + x)(ax2 + 1)
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Question 363 Marks
Factorize:
a2 + 4b2 - 4ab - 4c2
Answer
The given expression to be factorized is:
a2 + 4b2 - 4ab - 4c2
This can be arrange in the form
a2 + 4b2 - 4ab - 4c2
= (a2 - 4ab + 4b2) - 4c2
= {(a)2 - 2.a.2b + (2b)2} - 4c2
= (a - 2b)2 - 4c2
Substitute x = (a - 2b)
a2 + 4b2 - 4ab - 4c2 = x2 - 4c2
= x2 - (2c)2
= (x + 2c)(x - 2c)
Put x = (a - 2b)
a2 + 4b2 - 4ab - 4c2 = {(a - 2b) + 2c}{(a - 2b) - 2c}
= (a - 2b + 2c)(a - 2b - 2c)
we cannot further factorize the expresion.
So, the required factorization of a2 + 4b2 - 4ab - 4c2 is (a - 2b + 2c)(a - 2b - 2c)
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Question 373 Marks
Factorize:
(a - b + c)2 + (b - c + a)2 + 2(a - b + c)(b - c + a)
Answer
(a - b + c)2 + (b - c + a)2 + 2(a - b + c)(b - c + a)
Let (a - b + c) = x and (b - c + a) = y
= x2 + y2 + 2xy
Using the identity (a + b)2 = a2 + b2 + 2ab
= (x + y)2
Now, substituting x and y
(a - b + c + b - c + a)2
Cancelling -b, +b & + c, -c
= (2a)2
= 4a2
$\therefore$ (a - b + c)2 + (b - c + a)2 + 2(a - b + c)(b - c + a) = 4a2
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Question 383 Marks
Factorize:
9(2a - b)2 - 4(2a - b) - 13
Answer
Let 2a - b = x
= 9x2 - 4x - 13
Splitting the middle term,
= 9x2 - 13x + 9x - 13
= x(9x - 13) + 1(9x - 13)
= (9x - 13)(x + 1)
Substituting x = 2a - b
= [9(2a - b) - 13](2a - b + 1)
= (18a - 9b - 13)(2a - b + 1)
$\therefore$ 9(2a - b)2 - 4(2a - b) - 13
= (18a - 9b - 13)(2a - b + 1)
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Question 393 Marks
Factorize:
7(x - 2y)2 - 25(x - 2y) + 12
Answer
Let x - 2y = P
= 7P2 - 25P + 12
Splitting the middle term,
= 7P2 - 21P - 4P + 12
= 7P(P - 3) - 4(P - 3)
= (P - 3)(7P - 4)
Substituting P = x - 2y
= (x - 2y - 3)(7(x - 2y) - 4)
= (x - 2y - 3)(7x - 14y - 4)
$\therefore$ 7(x - 2y)2 - 25(x - 2y) + 12
= (x - 2y - 3)(7x - 14y - 4)
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Question 403 Marks
Factorize:
6ab - b2 + 12ac - 2bc
Answer
6ab - b2 + 12ac - 2bc
Taking b common in (6ab - b2) and 2c in (12ac - 2bc)
= b(6a - b) + 2c (6a - b)
Taking (6a - b) common in the terms
= (6a - b)(b + 2c)
$\therefore$ 6ab - b2 + 12ac - 2bc = (6a - b)(b + 2c)
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Question 413 Marks
Factorize:
4(x - y)2 - 12(x - y)(x + y) + 9(x + y)2
Answer
4(x - y)2 - 12(x - y)(x + y) + 9(x + y)2
Let(x - y) = x,(x + y) = y
= 4x2 - 12xy + 9y2
Splitting the middle term -12 = -6 - 6 also 4 × 9 = -6 × -6
= 4x2 - 6xy - 6xy + 9y2
= 2x(2x - 3y) - 3y(2x - 3y)
= (2x - 3y)(2x - 3y)
= (2x - 3y)2
Substituting x = x - y & y = x + y
= [2(x - y) - 3(x + y)]2 = [2x - 2y - 3x - 3y]2
= (2x - 3x - 2y - 3y)²
= [-x - 5y]2
= [(-1)(x + 5y)]2
= (x + 5y)2 [(-1)2 = 1]
$\therefore$ 4(x - y)2 - 12(x - y)(x + y) + 9(x + y)2 = (x + 5y)2
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