Question
Using factor theorem, factorize the following polynomials:
2y3 + y2 - 2y - 1
2y3 + y2 - 2y - 1
✓
Answer
Let p(y) = 2y3 + y2 - 2y - 1
By hit and trial method
p(1) = 2(1)3 + (1)2 - 2(1) - 1
= 2 + 1 - 2 - 1 = 0
So, y -1 is a factor of this polynomial.
Now,
By long division method,
$\therefore$ 2y3 + y2 - 2y - 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)
By hit and trial method
p(1) = 2(1)3 + (1)2 - 2(1) - 1
= 2 + 1 - 2 - 1 = 0
So, y -1 is a factor of this polynomial.
Now,
By long division method,
$\therefore$ 2y3 + y2 - 2y - 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)
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