Factorise : (x 2 - 3x)(x 2 - 3x - 1) - 20

Given, (x 2 - 3x)(x 2 - 3x - 1) - 20 Substituting x 2 - 3x = a, we get : ⇒ a(a - 1) - 20 = a 2 - a - 20 = a 2 - 5a + 4a - 20 = a(a - 5) + 4(a - 5) = (a - 5)(a + 4) = (x 2 - 3x - 5)(x 2 - 3x + 4). Hence, (x 2 - 3x)(x 2 - 3x - 1) - 20 = (x 2 - 3x - 5)(x 2 - 3x + 4).

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Mathematics

Factorise :
(x² - 3x)(x² - 3x - 1) - 20

Factorisation

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Answer

Given,

(x² - 3x)(x² - 3x - 1) - 20

Substituting x² - 3x = a, we get :

⇒ a(a - 1) - 20

= a² - a - 20

= a² - 5a + 4a - 20

= a(a - 5) + 4(a - 5)

= (a - 5)(a + 4)

= (x² - 3x - 5)(x² - 3x + 4).

Hence, (x² - 3x)(x² - 3x - 1) - 20 = (x² - 3x - 5)(x² - 3x + 4).

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Mathematics

Factorise :
(x² - 3x)(x² - 3x - 1) - 20

Factorisation

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Mathematics

Factorise :(x2 - 3x)(x2 - 3x - 1) - 20

Factorisation

Answer

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Factorise :
(x² - 3x)(x² - 3x - 1) - 20

Factorise :
x(6x - 5y) - 4(6x - 5y)²

Factorise :
$\dfrac{1}{35} + \dfrac{12}{35}a + a^2$

For each trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
(i) x² - 3x - 54
(ii) 2x² - 7x - 15
(iii) 2x² + 2x - 75
(iv) 3x² + 4x - 10
(v) x(2x - 1) - 1

Factorise :
(i) $4\sqrt{3}x^2 + 5x - 2\sqrt{3}$
(ii) $7\sqrt{2}x^2 - 10x - 4\sqrt{2}$