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Maths 4 Marks Questions

Factorization Of Algebraic Expressions [NEW] · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 8 questions.

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Question 14 Marks
Factorize the following expressions:
x3 + 6x2 + 12x + 16
Answer
= x3 + 6x2 + 12x + 8 + 8
= x3 + 3 × x2 × 2 + 3 × x × 22 + 23 + 8
= (x + 2)3 + 8
$\big[\therefore$ a3 + 3a2b + 3ab2 + b3 = (a + b)3$\big]$
= (x + 2)3 + 23
= (x + 2 + 2)((x + 2)2 - 2(x + 2) + 22)
$\big[\therefore$ a3 + b3 = (a + b)(a2 - ab + b2)$\big]$
= (x + 2 + 2)(x2 + 4 + 4x - 2x - 4 + 4)
$\big[\therefore$ (a + b)2 = a2 + b2 + 2ab$\big]$
= (x + 4)(x2 + 4 + 2x)
$\therefore$ x3 + 6x2 + 12x + 16 = (x + 4)(x2 + 4 + 2x)
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Question 24 Marks
Factorize the following expressions:
$\Big[\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big]^3+\Big[\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big]^3+\Big[-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big]^3$
Answer
$\Big[\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big]^3+\Big[\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big]^3+\Big[-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big]^3$
Let $\Big(\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big)=\text{a},\Big(\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big)=\text{b},\Big(-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big)=\text{c}$
$\text{a}+\text{b}+\text{c}=\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}+\frac{\text{x}}{3}-\frac{\text{2y}}{3}+\text{z}-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=\Big(\frac{\text{x}}{2}+\frac{\text{x}}{3}-\frac{5\text{x}}{6}\Big)+\Big(\text{y}-\frac{\text{2y}}{3}-\frac{\text{y}}{3}\Big)+\Big(\frac{\text{z}}{3}+\text{z}-\frac{4\text{z}}{3}\Big)$
$\text{a}+\text{b}+\text{c}=\frac{3\text{x}}{6}+\frac{2\text{x}}{6}-\frac{5\text{x}}{6}+\frac{3\text{y}}{3}-\frac{2\text{y}}{3}-\frac{\text{y}}{3}+\frac{\text{z}}{3}+\frac{3\text{z}}{3}-\frac{4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=\frac{5\text{x}-5\text{x}}{6}+\frac{3\text{y}-3\text{y}}{3}+\frac{4\text{z}-4\text{z}}{3}$
$\text{a}+\text{b}+\text{c}=0$
$\because \ \text{a}+\text{b}+\text{c}=0$
$\therefore \ \text{a}^3+\text{b}^3+\text{c}^3=3\text{abc}$
$\therefore\Big[\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big]^3+\Big[\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big]^3+\Big[-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big]^3$
$=3\Big(\frac{\text{x}}{2}+\text{y}+\frac{\text{z}}{3}\Big)\Big(\frac{\text{x}}{3}-\frac{2\text{y}}{3}+\text{z}\Big)\Big(-\frac{5\text{x}}{6}-\frac{\text{y}}{3}-\frac{4\text{z}}{3}\Big)$
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Question 34 Marks
Factorize the following expressions:
(a + b)3 - 8(a - b)3
Answer
= (a + b)3 - [2(a - b)]3
= (a + b)3 - [2a - 2b]3
= (a + b - (2a - 2b))((a + b)2 + (a + b)(2a - 2b) + (2a - 2b)2)
$\therefore$ [a3 - b3 = (a - b)(a2 + ab + b2)]
= (a + b - 2a + 2b)(a2 + b2 + 2ab + (a + b)(2a - 2b) + (2a - 2b)2)
= (a + b - 2a + 2b)(a2 + b2 + 2ab + 2a2 - 2ab + 2ab - 2b2 + (2a - 2b)2)
= (3b - a)(3a2 + 2ab - b2 + (2a - 2b)2)
= (3b - a)(3a2 + 2ab - b2 + 4a2 + 4b2 - 8ab)
= (3b - a)(3a2 + 4a2 - b2 + 4b2 - 8ab + 2ab)
= (3b - a)(7a2 +3b2 - 6ab)
$\therefore$ (a + b)3 - 8(a - b)3 = (3b - a)(7a2 + 3b2 - 6ab)
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Question 44 Marks
Factorize:
x(x - 2)(x - 4) + 4x - 8
Answer
x(x - 2)(x - 4) + 4x - 8
= x(x - 2)(x - 4) + 4(x - 2)
Taking (x - 2) common in both the terms
=(x - 2){x(x - 4) + 4}
=(x - 2){x2 - 4x + 4}
Now splitting the middle term of x2 - 4x + 4
= (x - 2){x2 - 2x - 2x + 4}
= (x - 2){x( x - 2) -2(x - 2)}
= (x - 2){(x - 2)(x - 2)}
= (x - 2)(x - 2)(x - 2)
= (x - 2)3
$\therefore$ x(x - 2)(x - 4) + 4x - 8 = (x - 2)3
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Question 54 Marks
Factorize:
$\text{x}^2+\frac{12}{35}\text{x}+\frac{1}{35}$
Answer
$\text{x}^2+\frac{12}{35}\text{x}+\frac{1}{35}$
Splitting the middle term,
$=\text{x}^2+\frac{5}{35}\text{x}+\frac{7}{35}\text{x}+\frac{1}{35}$
$\Big[\therefore\frac{12}{35}=\frac{5}{35}+\frac{7}{35} \ \text{and} \ \frac{5}{35}\times\frac{7}{35}=\frac{1}{35}\Big]$
$=\text{x}^2+\frac{\text{x}}{7}+\frac{\text{x}}{5}+\frac{1}{35}$
$=\text{x}\Big(\text{x}+\frac{1}{7}\Big)+\frac{1}{5}\Big(\text{x}+\frac{1}{7}\Big)$
$=\Big(\text{x}+\frac{1}{7}\Big)\Big(\text{x}+\frac{1}{5}\Big)$
$\therefore\text{x}^2+\frac{12}{35}\text{x}+\frac{1}{35}=\Big(\text{x}+\frac{1}{7}\Big)\Big(\text{x}+\frac{1}{5}\Big)$
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Question 64 Marks
Factorize:
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-4\Big(\text{x}+\frac{1}{\text{x}}\Big)+6$
Answer
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-4\Big(\text{x}+\frac{1}{\text{x}}\Big)+6$
$=\text{x}^2+\frac{1}{\text{x}^2}-4\text{x}-\frac{4}{\text{x}}+4+2$
$=\text{x}^2+\frac{1}{\text{x}^2}+4+2-4\text{x}-4\text{x}$
$=\big(\text{x}^2\big)+\Big(\frac{1}{\text{x}}\Big)^2+(-2)^2+2\times\text{x}\times\frac{1}{\text{x}}+2\times\frac{1}{\text{x}}\times(-2)+2(-2)\text{x}$
Using identity
x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2
We get,
$=\Big[\text{x}+\frac{1}{\text{x}}+(-2)\Big]^2$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]^2$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]\Big[\text{x}+\frac{1}{\text{x}}-2\Big]$
$\therefore\Big[\text{x}^2+\frac{1}{\text{x}^2}\Big]-4\Big[\text{x}+\frac{1}{\text{x}}\Big]+6$
$=\Big[\text{x}+\frac{1}{\text{x}}-2\Big]\Big[\text{x}+\frac{1}{\text{x}}-2\Big]$
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Question 74 Marks
Factorize:
a2 + 2ab + b2 - c2
Answer
a2 + 2ab + b2 - c2
Using the identity (p + q)2 = p2 + q2 + 2pq
= (a + b)2 - c2
Using the identity p2 - q2 = (p + q)(p - q)
= (a + b + c)(a + b - c)
$\therefore$ a2 + 2ab + b2 - c2 = (a + b + c)(a + b - c)
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Question 84 Marks
Factorize:
a(a + b)3 - 3a2b(a + b)
Answer
a(a + b)3 - 3a2b(a + b)
Taking (a + b) common in the two terms
= (a + b){a(a + b)2 - 3a2b}
Now, using (a + b)2 = a2 + b2 + 2ab
= (a + b){a(a2 + b2 + 2ab) - 3a2b}
= (a + b){a3 + ab2 + 2a2b - 3a2b}
= (a + b){a3 + ab2 - a2b}
= (a + b)p{a2 + b2 - ab}
= p(a + b)(a2 + b2 - ab)
$\therefore$ a(a + b)3 - 3a2b(a + b)
= a(a + b)(a2 + b2 - ab)
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