Math, asked by ranialisha128, 1 year ago

factorise;
2a^7-128a

Answers

Answered by MaheswariS
10

\underline{\textbf{Given:}}

\mathsf{2\,a^7-128\,a}

\underline{\textbf{To find:}}

\mathsf{Factors\;of\;2\,a^7-128\,a}

\underline{\textbf{Solution:}}

\mathsf{Consider,}

\mathsf{2\,a^7-128\,a}

\textsf{This can be written as,}

\mathsf{=2a(a^6-64)}

\mathsf{=2a\{(a^3)^2-8^2\}}

\mathsf{Using\;the\;identity,}\;\;\boxed{\bf\,a^2-b^2=(a-b)(a+b)}

\mathsf{=2a\,(a^3-8)\,(a^3+8)}

\mathsf{=2a\,(a^3-2^3)\,(a^3+2^3)}

\textsf{Using the following identities,}

\boxed{\begin{minipage}{5cm}$\\\mathsf{a^3-b^3=(a-b)(a^2+ab+b^2)}\\\\\mathsf{a^3+b^3=(a+b)(a^2-ab+b^2)}\\$\end{minipage}}

\mathsf{=2a\,(a-2)\,(a^2+2a+4)\,(a+2)\,(a^2-2a+4)}

\implies\boxed{\boxed{\mathsf{2\,a^7-128\,a=2a\,(a-2)\,(a^2+2a+4)\,(a+2)\,(a^2-2a+4)}}}

Answered by 8c36tanzilahmad
1

Given: 2a^7 - 128a

Answer: 2a (a + 2)(a - 2)(a2 - 2a + 4)(a2 + 2a + 4)

Step-by-step explanation:

2a (a^6 - 64)

2a [(a^3)^2 - (8)^2]

using formula a^2 - b^2 = (a+b)(a-b)

2a (a^3 + 8)(a^3 - 8)

2a [(a)^3 + (2)^3)][(a)^3 - (2)^3]

using formula a^3 + b^3 = (a + b)(a^2 - ab + b^2)

and a^3 - b^3 = (a - b)(a^2 + ab + b^2)

2a (a + 2)(a^2 - 2a + 4)(a - 2)(a^2 + 2a + 4)

2a (a + 2)(a - 2)(a^2 - 2a + 4)(a^2 + 2a + 4)

Similar questions