Question

Factorise:2y³+y²-2y-1 using factor theorem or splitting middle term concept​

Answer 1

✤Qᴜᴇꜱᴛɪᴏɴ :-

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Factorise:2y³+y²-2y-1 using factor theorem or splitting middle term concept

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✤ᴡʜᴀᴛ ᴛᴏ ᴅᴏ :-

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We have to factorize. We will factorize using splitting middle term concept and using Identities

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✤ɪᴅᴇɴᴛɪᴛɪᴇꜱ ᴛᴏ ʙᴇ ᴜꜱᴇᴅ :-

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$\bf \pink{ {a}^{2} - {b}^{2} } = \purple{(a + b)(a - b)}$

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✤ꜱᴏʟᴜᴛɪᴏɴ :-

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$\bf \to {2y}^{3} + {y}^{2} - 2y - 1$

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$\bf \to {y}^{2} (2y + 1) - 1(2y + 1)$

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$\bf \to (2y + 1)( {y}^{2} - 1)$

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$\bf \to (2y + 1)( {y}^{2} - {1}^{2} )$

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Using Identity

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$\bf \pink{ {a}^{2} - {b}^{2} } = \purple{(a + b)(a - b)}$

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$\bf \to (2y + 1)(y + 1)(y - 1)$

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✏ᴀᴅᴅɪᴛɪᴏɴᴀʟʟʏ :-

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Step1: Find two numbers that multiply to give ac (in other words a times c), and add to give b.

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Step2: Rewrite the middle with those numbers

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Step3: Factor the first two and last two terms separately:

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Step4: If we've done this correctly, our two new terms should have a clearly visible common factor.

Answer 2

=2y³+y²-2y-1

=y²(2y+1)-1(2y+1)

=(2y+1)(y²-1)

=(2y+1)(y-1)(y+1)