Question

factorise 2y3-4y2-2y+4​

Answer 1

SOLUTION

$= > 2y {}^{3} - 4 {y}^{2} - 2y + 4 \\ = > 2y {}^{2} (y - 2) - 2( y - 2) \\ = > ( {2y}^{2} - 2)(y - 2) \\ = > 2( {y}^{2} - 1)(y - 2) \\ = > 2</u><u>[</u><u>(y) {}^{2} - (1) {}^{2} </u><u>]</u><u>(y - 2) \\ = > 2(y + 1)(y - 1)(y - 2)$

hope it helps ☺️

Answer 2

2y³ - 4y² - 2y + 4

____________ [ GIVEN ]

• We have to find factors of 2y³ - 4y² - 2y + 4.

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→ 2y³ - 4y² - 2y + 4

Take 2 common from all

→ 2(y³ - 2y² - y + 2)

Take y² common from y² - 2y² and -1 common from -y + 2

It can be written like that..

→ 2[y²(y - 2) -1(y - 2)]

y - 2 is common so, write it one time

→ 2[(y² - 1) (y - 2)]

→ 2[(y + 1) (y - 1) (y - 2)]

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2[(y + 1) (y - 1) (y - 2)] are the factors of 2y³ - 4y² - 2y + 4.

_________ [ ANSWER ]

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