Question

factorise 16x²-20 x-6​

Answer 1

To factorise :

• 16x² – 20x – 6

Solution :

16x² – 20x – 6

Sum = –20 , Product = –96 (–24 , 4)

→ (16x² – 24x) + (4x – 6)

→ [4x (4x – 6) ] + [1 (4x – 6) ]

→ (4x + 1) (4x – 6)

→ (4x + 1) and (4x – 6)

∴ Required answer :

Factors are :

• 4x + 1 and 4x – 6

Verification :

(4x + 1)(4x – 6)

→ (4x × 4x) + (4x × –6) + 4x – 6

→ 16x² – 24x + 4x – 6

→ 16x² – 20x – 6

• Hence verified !

Answer 2

Given: 16x² - 20x - 6

Need to find: Factorization of 16x² - 20x - 6

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Here, In this question, We are asked to find the factors for factorising 16x² - 20x - 6. In order to find the factors, We need to Factorize the expression.

❍ Note - Factor the first two and last two terms separately.

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$\implies \sf {16x}^{2} - 20x - 6 \\ \\ \\ \implies \sf 2( {8x}^{2} - 10x - 3) \\ \\ \\ \implies \sf 2( {8x}^{2} + 2x - 12x - 3) \\ \\ \\\implies \sf2(2x(4x + 1) - 3(4x + 1)) \\ \\ \\ \implies\sf\boxed{\underline{\frak{ \purple{2(2x - 3)(4x + 1)}}}}$

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$\therefore\:{\underline{\sf{Hence, By \: factorising \:{\bf {16x}^{2} - 20x - 6} \: we \: get \: {\bf 2(2x - 3)(4x + 1)} \: \sf{respectively}.}}}$

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$\boxed{\begin{array}{cc}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{array}}$