Question

y³ − 2y² − 29y − 42

Answer 1

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).