Question

factorise : (a) 9x² - 30x + 25
(b) x² - x - 56​

Answer 1

$\large\sf\underline{Given\::}$

• $\sf\:9x^{2} - 30x +25$

• $\sf\:x^{2} - x - 56$

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$\large\sf\underline{To\::}$

• Factorise both the expressions.

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$\large\sf\underline{How\:do\:we\:factorise\:?}$

Factorising a quadratic equation means to break the given expression into its product form.

To solve $\sf\:ax^{2}+bx+c$ in its product form we need to :

• Multiply c and a if the expression is complicated otherwise we can proceed simply by searching for such term whose sum or difference gives us the middle term.

• The number which we get as product by multiplying c and a , need to be broken in such a way that the sum or difference of those two numbers gives us the middle number.

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$\large\sf\underline{Solution\::}$

1]

$\sf\:9x^{2} - 30x +25$

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• Multiplying 9 and 25 we get 225 . So we need to find such a number whose sum or difference is 30 and whose product gives us 225 .

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The two numbers are 15 and 15 and their sum gives the middle term that is 30 whereas their product 15 × 15 = 225 .

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$\sf\implies\:9x^{2} - (15+15)x +25$

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• Opening the brackets

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$\sf\implies\:9x^{2} - 15x - 15x +25$

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• Taking 3x as common from first two terms and 5 from other two terms.

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$\sf\implies\:3x(3x - 5) - 5(3x -5)$

$\small{\underline{\boxed{\mathrm\pink{\implies\:(3x-5)(3x-5)}}}}$ ★

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2]

$\sf\:x^{2} - x - 56$

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• Multiplying 56 and 1 we get 56 . So we need to find such a number whose sum or difference is 1 and whose product gives us 56 .

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The two numbers are 8 and 7 and their difference gives the middle term that is 1 whereas their product 8 × 7 = 56 .

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$\sf\implies\:x^{2} - (8-7)x -56$

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• Opening the brackets

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$\sf\implies\:x^{2} - 8x + 7x -56$

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• Taking x as common from first two terms and 7 from other two terms.

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$\sf\implies\:x(x - 8) + 7(x -8)$

$\small{\underline{\boxed{\mathrm\pink{\implies\:(x-8)(x+7)}}}}$ ★

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!! Hope it helps !!