Question

homework assignment 5.5 zeros of polynomials
score: 4.33/12 answered: 5/12
question 6
find all zeros of f(x)=x^3 - 7x^2 + 13x - 7. enter the zeros separated by commas. enter exact value, not decimal approximations.
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Explanation:

Step1: Try rational - root theorem

The rational - root theorem states that if a polynomial \(a_nx^n+\cdots+a_1x + a_0\) has a rational root \(p/q\), then \(p\) is a factor of \(a_0\) and \(q\) is a factor of \(a_n\). For the polynomial \(f(x)=x^{3}-7x^{2}+13x - 7\), \(a_n = 1\) and \(a_0=-7\). The possible rational roots are \(\pm1,\pm7\).

Step2: Test possible roots

Evaluate \(f(x)\) at the possible rational roots:
When \(x = 1\), \(f(1)=1^{3}-7\times1^{2}+13\times1 - 7=1 - 7+13 - 7=0\). So \(x - 1\) is a factor of \(f(x)\).

Step3: Perform polynomial long - division

Divide \(x^{3}-7x^{2}+13x - 7\) by \(x - 1\).
\((x^{3}-7x^{2}+13x - 7)\div(x - 1)=x^{2}-6x + 7\).

Step4: Solve the quadratic equation

We need to solve \(x^{2}-6x + 7=0\). Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\). Here \(a = 1\), \(b=-6\), \(c = 7\).
\[

$$\begin{align*} x&=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times7}}{2\times1}\\ &=\frac{6\pm\sqrt{36 - 28}}{2}\\ &=\frac{6\pm\sqrt{8}}{2}\\ &=\frac{6\pm2\sqrt{2}}{2}\\ &=3\pm\sqrt{2} \end{align*}$$

\]

Answer:

\(1,3 + \sqrt{2},3-\sqrt{2}\)