Factorising after product/chain rule Watch

Reena Bansi
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Attachment 880352

**see photo attached**

I have differentiated the equation and have to write it in factorised form without expanding the brackets. Any ideas?
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RDKGames
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(Original post by Reena Bansi)

**see photo attached**

I have differentiated the equation and have to write it in factorised form without expanding the brackets. Any ideas?
From

y' = 24x^3(5x+7)(3x^4 + 1) + 5(3x^4 + 1)^2

you should notice that the two terms which are you are adding share a common factor of 3x^4 + 1.
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dextrous63
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Nope
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Reena Bansi
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(Original post by RDKGames)
From

y' = 24x^3(5x+7)(3x^4 + 1) + 5(3x^4 + 1)^2

you should notice that the two terms which are you are adding share a common factor of 3x^4 + 1.
yes i did see that. i have tried taking it out as a common factor but can't seem to get to the right answer.
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ghostwalker
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(Original post by Reena Bansi)
yes i did see that. i have tried taking it out as a common factor but can't seem to get to the right answer.
What are you getting, and what is considered to be the "right answer"?
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dextrous63
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(Original post by Reena Bansi)
yes i did see that. i have tried taking it out as a common factor but can't seem to get to the right answer.
On that picture of your working out (which you have deleted), you were along the right track. But you needed to have used brackets around the 3x^4+1 before your square brackets, and correct the sloppy stuff inside.

As an example, you wrote abc+ade = a(b+c+d+e) instead of a(bc+de)
Last edited by dextrous63; 1 week ago
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