Question

please help me , thank you ​

Answer 1

Hello BinibiningDel!

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

• Factorise the polynomials.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\tt \: 1) \: \: \: {x}^{2} - 49 \\ \tt \: = (x) ^{2} - ( 7) ^{2} \\ = \boxed{ \bf \: (x + 7)(x - 7) }$

Method used :-

• Use the algebraic identity ⇨ a² - b² = (a + b) (a - b ).
• Here, a = x & b = 7.

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$\tt 2) \: \: \: {a}^{2} + 10a+ 25 \\ = \boxed{\bf(a + 5)^{2} }$

Method used :-

• Use the algebraic identity ⇨ a² + 2ab + b² = (a + b)².
• Here, a = a & b = 5.

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$\tt \:3) \: \: \: {x}^{2} + x - 30 \\ = \tt ({x}^{2} - 5x) + (6x - 30) \\ = \tt \: x(x - 5) + 6(x - 5) \\ = \boxed{ \bf(x + 6)(x - 5)}$

Method used :-

• Use the splitting the middle term method to factorise this.

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$\tt \: 4) \: \: \: {4x}^{2} - {y}^{2} \\ = \tt \: (2x) ^{2} - (y) ^{2} \\ = \boxed{ \bf(2x - y)(2x + y)}$

Method used :-

• Use the algebraic identity ⇨ a² - b² = (a + b) (a - b ).
• Here, a = 2x & b = y.

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$\tt \: 5) \: \: \: 8 {m}^{2} n ^{3} - 6mn ^{2} \\ = \tt \: 2\left(4m^{2}n^{3}-3mn\right) \\ = \boxed{ \bf \: 2mn\left(4mn^{2}-3\right) }$

Method used :-

• Take the common factor out & factorise it.

__________________

$\tt \: 6) \: \: \: \tt \: 2 d ^ { 2 } + 5 d - 12 \\ = \tt \: \left(2d^{2}-3d\right)+\left(8d-12\right) \\ \tt \: = d\left(2d-3\right)+4\left(2d-3\right) \\ = \boxed{ \bf \: \left(2d-3\right)\left(d+4\right) }$

Method used :-

• Use the splitting the middle term method to factorise this.

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$\tt \:7) \: \: \: y ^ { 3 } + 27 \\ = \tt (y) ^{3} + ( {3})^{3} \\ = \boxed{ \bf\left(y+3\right)\left(y^{2}-3y+9\right) }$

Method used :-

• Use the algebraic identity ⇨ a³ + b³ = (a + b) (a² - ab + b²).
• Here, a = y & b = 3.

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$\tt \: 8) \: \: \: 20 b ^ { 3 } c ^ { 5 } + 35 b ^ { 4 } c ^ { 2 } \\ = \tt \: 5\left(4b^{3}c^{5}+7b^{4}c^{2}\right) \\ = \boxed{ \bf \: 5b^{3}c^{2}\left(4c^{3}+7b\right) }$

Method used :-

• Take the common factor out & factorise it.

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Hope it'll help you!

ℓu¢αzz ッ