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2y 3 + y 2 - 2y - 1 Let p(y) = 2y 3 + y 2 - 2y - 1 By trial, we find that p(1) = 2(1) 3 + (1) 2 - 2(1) - 1 = 2 + 1 - 2 - 1 = 0 ∴ By Factor Theorem, (y - 1) is a factor of p(y). Now, 2y 3 + y 2 - 2y - 1 = 2y 2 (y - 1) + 3y(y - 1) + 1(y - 1) = (y - 1)(2y 2 + 3y + 1) = (y - 1)(2y 2 + 2y + y+ 1) = (y - 1){2y(y + 1) + 1(y + 1)} = (y - 1)(y + 1) (2y + 1).

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Polynomials

Short Answer Type

121. Factorise
2y³ + y² - 2y - 1

2y³ + y² - 2y - 1
Let p(y) = 2y³ + y² - 2y - 1
By trial, we find that
p(1) = 2(1)³ + (1)² - 2(1) - 1
= 2 + 1 - 2 - 1 = 0
∴ By Factor Theorem, (y - 1) is a factor of p(y).
Now,
2y³ + y² - 2y - 1
= 2y²(y - 1) + 3y(y - 1) + 1(y - 1)
= (y - 1)(2y² + 3y + 1)
= (y - 1)(2y² + 2y + y+ 1)
= (y - 1){2y(y + 1) + 1(y + 1)}
= (y - 1)(y + 1) (2y + 1).