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