Answer :
Given the function:
[tex]f(x)=3x^4-18x^3-21x^2+144x-108[/tex]You can find the x-intercepts by following these steps:
1. Make the function equal to zero:
[tex]3x^4-18x^3-21x^2+144x-108=0[/tex]2. Factor the equation:
- Notice that all the terms have 3 in common. Therefore, you can factor it out:
[tex]3(x^4-6x^3-7x^2+48x-36)=0[/tex]- Rewrite these terms as the Difference between two terms:
[tex]\begin{gathered} -6x^3=-x^3-5x^3 \\ \\ -7x^2=5x^2-12x^2 \\ \\ 48x=12x+36x \end{gathered}[/tex]Then:
[tex]3(x^4-x^3-5x^3+5x^2-12x^2+12x+36x-36)=0[/tex]- Make groups of two terms using parentheses:
[tex]3\lbrack(x^4-x^3)-(5x^3-5x^2)-(12x^2-12x)+(36x-36)\rbrack=0[/tex]- Factor the Greatest Common Factor of each group out (the greatest factor the terms have in common):
[tex]3\lbrack x^3(x-1)-5x^2(x-1)-12x(x-1)+36(x-1)\rbrack=0[/tex]- Factor this common factor out:
[tex]x-1[/tex]Then:
[tex]3(x-1)(x^3-5x^2-12x+36)=0[/tex]- Rewrite these terms as follows:
[tex]\begin{gathered} -5x^2=-2x-3x^2 \\ \\ -12x=6x-18x \end{gathered}[/tex]Then:
[tex]3(x-1)(x^3-2x^2-3x^2+6x-18x+36)=0[/tex]- Make groups of terms:
[tex]3(x-1)\lbrack(x^3-2x^2)-(3x^2-6x)-(18x-36)\rbrack=0[/tex]- Factor the Greatest Common Factor of each group out:
[tex]3(x-1)\lbrack x^2(x-2)-3x(x-2)-18(x-2)\rbrack=0[/tex]- Factor this factor out:
[tex]x-2[/tex]Then:
[tex]3(x-1)(x-2)(x^2-3x-18)=0[/tex]- Factor the Quadratic Expression by following two terms whose Sum is -3 and whose Product is -18. These are 3 and -6. Then:
[tex]3(x-1)(x-2)(x+3)(x-6)=0[/tex]3. Divide both sides of the equation by 3:
[tex]\frac{3(x-1)(x-2)(x+3)(x-6)}{3}=\frac{0}{3}[/tex][tex](x-1)(x-2)(x+3)(x-6)=0[/tex]4. Solving for "x", you get the following values:
[tex]\begin{gathered} x-1=0\Rightarrow x=1 \\ \\ x-2=0\Rightarrow x=2 \\ \\ x+3=0\Rightarrow x=-3 \\ \\ x-6=0\Rightarrow x=6 \end{gathered}[/tex]Hence, knowing that you have to reject the negative values, the answer is:
[tex]\begin{gathered} x_1=1 \\ x_2=2 \\ x_3=6 \end{gathered}[/tex]