f(x) = x^3 +*9x^2 + 23x + 15*

*To solve, we will subsitute with the 15 factors which are:*

*1, -1, 3, -3, 5, -5, 15, -15 *

*let us subsitute with x= -5:*

*f(-5) = 5^3 + 9*5^2 + 23*5 + 15 *

* = -125 + 225 - 529 + 15 = 0*

*The (x+5) is one of the factors:*

*==> f(x) = (x+ 5) * g(x) *

*Now we will divide f(x) by (x+5):*

*==> f(x) = (x+ 5) (x^2 + 4x + 3)*

*Now factor (x^2 + 4x + 3)*

*==> f(x) = (x+5)(x+3)(x+1)*

*Now we will determine the zeros:*

*==> x1= -5*

*==> x2= -3*

*==> x3= -1*

To find the zeros of f(x) = x^3 +*9x^2 + 23x + 15.*

*We observe that if we put x= -1, then f(-1) = (-1)+9-23+15 = 0.*

*So x+1 is a facor of the f(x).*

*Therefore x^3+9x^2+23x+15 = (x+1))(x^2 +kx+15) , as this agrees with x^3 and constant terms.*

*Now equating x terms on both sides, 23x = 15x+kx. *

*kx = 23x-15x*

*kx = 8x.*

*k = 8.*

*Therefore x^2+8x+15 is a factor.*

*x^2+8x+15 = (x+5)(x+3).*

*Therefore f(x) = x^3+9x^2+15x+23 = (x+1)(x+3)(x+5).*

*Therefore x+1 = 0 , x+5 = 0 , x+3 = 0 gives the zeros of f(x)*

*Or x = -1, x= -5 , or x= -3 are the zeros of f(x).*