# Find the real and complex zeros of the following function. f(x)=x^3 + x^2 + 11x + 51

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You need to remember that the zeroes of polynomial may be found among the divisor of the coefficient 51 over the divisors of leading coefficients such that:

`(D_51)/(D_1) = {+-1;+-3;+-17;+-51}`

ence, substituting +-1 for x, you will not get a zero.

Considering x = -3, yields:

`f(-3)=-27+9-33+51 = 0 =gt x = -3` is a zero of polynomial

Henceyou may write the factored form of the polynomial such that:

`f(x) = (x+3)q(x)`

Notice that q(x) is a polynomial of second order and reminder is zero.

You need to find the polynomial q(x) such that you need to equate f(x) = (x+3)q(x):

`x^3 + x^2 + 11x + 51 = (x + 3)(ax^2 + bx + c)`

`x^3 + x^2 + 11x + 51 = ax^3 + bx^2 + cx + 3ax^2 + 3bx + ``3c`

`` `x^3 + x^2 + 11x + 51 =ax^3 + x^2(b + 3a) + x(c + 3b) + 3c`

Equating the coefficients of like powers yields:

a = 1

b + 3a = 1 => b + 3 = 1 => b = 1-3=> b = -2

`c + 3b = 11 =gt c = 11+ 6 =gt c = 17`

Hence, the polynomial `q(x) = x^2 - 2x + 17` . If f(x)=0, then (x+3)q(x) = 0 => q(x) = 0.

Since q(x)=0 if f(x)=0, then the zeroes of q(x) cancel f(x).

You may find the zeroes of q(x) using the quadratic formula:

`x_(1,2) = (2+-sqrt(4 - 68))/2=gt x_(1,2) = (2+-sqrt(-64))/2`

`x_(1,2) = (2+-8i)/2` (use complex number theory such that `sqrt(-1)=i` )

`x_1 = 1 + 4i ; x_2 = 1 - 4i`

**Hence, evaluating the zeroes of polynomial `f(x)=x^3 + x^2 + 11x + 51 ` yields the complex roots`x_1 = 1 + 4i ; x_2 = 1 - 4i` and the real root `x_3 = -3` .**