Respuesta :
Given:
[tex]\mleft(x^3-5x^2\mright)+\mleft(2x-10\mright)[/tex]You can find all its zeros, as follows:
1. Make the expression equal to zero:
[tex](x^3-5x^2)+(2x-10)=0[/tex]2. Distribute the positive sign. Remember the Sign Rules for Multiplication:
[tex]\begin{gathered} +\cdot+=+ \\ -\cdot-=- \\ +\cdot-=- \\ -\cdot+=- \end{gathered}[/tex]Then:
[tex]x^3-5x^2+2x-10=0[/tex]3. Factor the expression:
- Make two groups using parentheses:
[tex](x^3-5x^2)+(2x-10)=0[/tex]- Identify the Greatest Common Factor of each group. For the first group:
[tex]GCF=x^2[/tex]And for the second group:
[tex]GCF=2[/tex]- Factor them out:
[tex]x^2(x^{}-5^{})+2(x-5)=0[/tex]4. Notice that this expression is common in both terms:
[tex]x-5[/tex]Then, you can factor it out:
[tex](x-5)(x^2+2)=0[/tex]5. Now you can set up these two equations:
[tex]\begin{gathered} x-5=0\text{ (Equation 1)} \\ \\ x^2+2=0\text{ (Equation 2)} \end{gathered}[/tex]6. Solve for "x" from each equation:
- For Equation 1:
[tex]x_1=5_{}[/tex]- For Equation 2:
[tex]\begin{gathered} x^2=-2 \\ x_{}=\pm\sqrt[]{-2} \end{gathered}[/tex]By definition:
[tex]\sqrt[]{-1}=i[/tex]Then, you get:
[tex]x=\pm\sqrt[]{-2}\Rightarrow\begin{cases}x_2=i\sqrt[]{2} \\ \\ x_3=-i\sqrt[]{2}\end{cases}_{}[/tex]Hence, the answer is:
[tex]\begin{gathered} x_1=5_{} \\ \\ x_2=i\sqrt[]{2} \\ \\ x_3=-i\sqrt[]{2} \end{gathered}[/tex]