Question

problem 6. find all the zeros given function and one zero of the function. f(x)=x^3 + 2x^2+2x + 1 with knowledge that x=-1 is a zero of the polynomial function.

Explanation:

Step1: Use synthetic division

Since \(x = - 1\) is a zero, we perform synthetic - division on \(f(x)=x^{3}+2x^{2}+2x + 1\) with divisor \(x+1\).
The coefficients of \(f(x)\) are \(1,2,2,1\).

-1 |  1  2  2  1
    |     -1 -1 -1
    |____________
      1  1  1  0

The quotient is \(x^{2}+x + 1\).

Step2: Solve the quadratic equation

We set the quotient \(x^{2}+x + 1 = 0\).
Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 1\), \(b = 1\), \(c = 1\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=1^{2}-4\times1\times1=1 - 4=-3\).
Then \(x=\frac{-1\pm\sqrt{-3}}{2}=\frac{-1\pm i\sqrt{3}}{2}\).

Answer:

The zeros of the function \(f(x)\) are \(x=-1,\frac{-1 + i\sqrt{3}}{2},\frac{-1 - i\sqrt{3}}{2}\)