Question

Using factor theorem, factorize each of the following polynomials

$y^3-2y^2$ - 29y -42

Answer 1

$y^3-2y^2$ - 29y -42

Let f(y) = $y^3-2y^2$ - 29y - 42

Factors of constant term -42 are

$\pm 1,\pm 2,\pm 3,\pm 6,\pm 7,\pm 14,\pm 21,\pm 42$

Let y = -1, then

f(-1) $=(-1{)}^3-2(-1{)}^2$ - 29(-1) - 42

= - 1 - 2 + 29 - 42 = 29 - 45 = - 36 $\ne$ 0

$\therefore y\ne$ 1 is not its factor

Let y =-2, then

f(-2) $=(-2{)}^3-2(-2{)}^2$ - 29(-2)-42

= - 8 - 8 + 58 - 42 = 58 - 58 = 0

$\therefore$ y + 2 is its factor

Now dividing f(y) by y + 2, we get

f(y) = (y+2) $(y^2$-4y-21)

=$(y+2)[y^2-7y+3y-21]$

$=(y+2)\left[y\right(y-7)+3(y-7\left)\right]$

=(y+2) (y-7)(y+3)

Hence, $y^3-2y^2-29y-42$

= (y+2) (y+3) (y-7)