One of the factors of (x² – 4x)(x² – 4x – 1) – 20 is: 1. x - 1 2. x - 2 3. x - 4 4. x + 5

Given, (x2 - 4x)(x2 - 4x - 1) - 20 Let us consider y = (x2 - 4x) ⇒ (y)(y - 1) - 20 ⇒ y2 - y - 20 ⇒ y2 - 5y + 4y - 20 ⇒ y(y - 5) + 4(y - 5) ⇒ (y + 4)(y - 5) ⇒ (x2 - 4x + 4)(x2 - 4x - 5) ⇒ [(x)2 - 2(2)(x) + (2)2](x2 - 5x + x - 5) ⇒ (x - 2)2[x(x - 5) + 1(x - 5)] ⇒ (x - 2)(x - 2)(x + 1)(x - 5) Hence, option 2 is correct option.

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Mathematics

One of the factors of (x² - 4x)(x² - 4x - 1) - 20 is :
x - 1
x - 2
x - 4
x + 5

Factorisation

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Answer

Given,

(x² - 4x)(x² - 4x - 1) - 20

Let us consider y = (x² - 4x)

⇒ (y)(y - 1) - 20

⇒ y² - y - 20

⇒ y² - 5y + 4y - 20

⇒ y(y - 5) + 4(y - 5)

⇒ (y + 4)(y - 5)

⇒ (x² - 4x + 4)(x² - 4x - 5)

⇒ [(x)² - 2(2)(x) + (2)²](x² - 5x + x - 5)

⇒ (x - 2)²[x(x - 5) + 1(x - 5)]

⇒ (x - 2)(x - 2)(x + 1)(x - 5)

Hence, option 2 is correct option.

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Mathematics

One of the factors of (x² - 4x)(x² - 4x - 1) - 20 is :
x - 1
x - 2
x - 4
x + 5

Factorisation

Answer

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One of the factors of (x² - 4x)(x² - 4x - 1) - 20 is :
x - 1
x - 2
x - 4
x + 5

Assertion(A): One of the factors of (5x + 1)² + (25x² - 1) is 2x.
Reason(R): (a + b)² = (a + b)(a + b) and a² - b² = (a + b)(a - b)
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Assertion(A): $a^3 - 2\sqrt{2}b^3$ can be factorized as $(a - 2\sqrt{2})(a^2 + \sqrt{2}ab + 2b^2)$
Reason(R): x³ - y³ = (x - y)(x² + xy + y²).
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

x⁴ + 7x² + 16 can be factorized as:
(x² + x + 4)(x² - x + 4)
(x² + x - 4)(x² - x + 4)
(x² - x - 4)(x² + x + 4)
none of these

If $x = \dfrac{a + b + c}{3}$ , then (x - a)³ + (x - b)³ + (x - c)³ can be factorized as:
(x - a)(x - b)(x - c)
$\dfrac{(x - a)(x - b)(x - c)}{3}$
3(x - a)(x - b)(x - c)
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