Question
Using factor theorem, factorize the following polynomials:
$y^3 - 2y^2 - 29y - 42$
$y^3 - 2y^2 - 29y - 42$
✓
Answer
Let $f(y) = y^3 - 2y^2 - 29y - 42$ be the given polynomial.
Now, putting $y = -2$, we get
$f(-2) = (-2)^3 - 2(-2)^2 - 29(-2) - 42$
$= -8 - 8 + 58 - 42$
$= -58 + 58 = 0$
Therefore, $(y + 2)$ is a factor of polynomial f(y).
Now,
$f(y) = y^2(y + 2) + 4y(y + 2) - 2(y + 2)$
$= (y + 2)(y^2 - 4y - 21)$
$= (y + 2)(y^2 - 7y + 3y - 21)$
$= (y + 2)(y + 3)(y - 7)$
Hence$ (y + 2), (y + 3)$ and $(y - 7)$ are the factors of polynomial f(y).
Now, putting $y = -2$, we get
$f(-2) = (-2)^3 - 2(-2)^2 - 29(-2) - 42$
$= -8 - 8 + 58 - 42$
$= -58 + 58 = 0$
Therefore, $(y + 2)$ is a factor of polynomial f(y).
Now,
$f(y) = y^2(y + 2) + 4y(y + 2) - 2(y + 2)$
$= (y + 2)(y^2 - 4y - 21)$
$= (y + 2)(y^2 - 7y + 3y - 21)$
$= (y + 2)(y + 3)(y - 7)$
Hence$ (y + 2), (y + 3)$ and $(y - 7)$ are the factors of polynomial f(y).
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