Question

homework assignment 3.6: zeros of polynomials
score: 8.25/12 answered: 9/12
question 11
find the zeros and fully factor $f(x)=x^3+2x^2-6x-4$, including factors for irrational zeros. use radicals, not decimal approximations.
the zeros are
the fully factor form is $f(x)=$
question help: video written example

Explanation:

Step1: Test rational zero candidates

Using Rational Root Theorem, possible rational zeros are $\pm1, \pm2, \pm4$. Test $x=2$:
$f(2) = 2^3 + 2(2)^2 -6(2)-4 = 8+8-12-4=0$
So $x=2$ is a zero, and $(x-2)$ is a factor.

Step2: Perform polynomial division

Divide $f(x)=x^3+2x^2-6x-4$ by $(x-2)$:
$$\frac{x^3+2x^2-6x-4}{x-2} = x^2+4x+2$$

Step3: Solve quadratic for zeros

Use quadratic formula on $x^2+4x+2=0$, where $a=1, b=4, c=2$:
$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-4\pm\sqrt{16-8}}{2} = \frac{-4\pm\sqrt{8}}{2} = -2\pm\sqrt{2}$$

Step4: Write fully factored form

Combine the linear factors from all zeros:
$f(x)=(x-2)(x-(-2+\sqrt{2}))(x-(-2-\sqrt{2}))$

Answer:

The zeros are $2$, $-2+\sqrt{2}$, $-2-\sqrt{2}$
The fully factor form is $f(x)=(x-2)(x+2-\sqrt{2})(x+2+\sqrt{2})$