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प्रश्न
y3 − 2y2 − 29y − 42
उत्तर
Let f(y) = y3 − 2y2 − 29y − 42 be the given polynomial.
Now, putting y = -2we get
`f(-2) = (-2)^ -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) -21(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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