1.without using a calculator, find all the roots of the equation.a. x^3+x^2+4x+4=0b. x^3+4x^2+x-6=0c. x^4-4x^3+x^2+12x-12=0 2.find all the zeros of the function a. y=x^4-6x^2+8b. y=x^3+6x^2+x+6c. g(x)=x^3-4x^2-x+22d. y=x^4-x^3-5x^2-x-6

Part 1A:

Given

[tex]x^3+x^2+4x+4=0 \\ \\ \Rightarrow x^2(x+1)+4(x+1)=0 \\ \\ \Rightarrow(x^2+4)(x+1)=0 \\ \\ \Rightarrow x= -1\ or\ x=\pm2i[/tex]



Part 1B:

Given

[tex]x^3+4x^2+x-6=0[/tex]

Using rational roots theorem, we can see that the possible roots of the the given equation are: [tex]\pm1,\ \pm2,\ \pm3,\ \pm6[/tex]

By substituting the possible roots, we can see that x = 1 is a root, thus x - 1 is a factor.
We can get the other factors by using sythetic division to divide [tex]x^3+4x^2+x-6[/tex] by x - 1.

 1  | 1     4     1     -6
     |
     |        1     5     6
     |_____________
       1     5     6     0

 Thus the other factor is [tex]x^2+5x+6=(x+2)(x+3)[/tex]

Therefore, all the roots of [tex]x^3+4x^2+x-6=0[/tex] are x = 1, x = -2 and x = -3.



Part 1C:

Given

[tex]x^4-4x^3+x^2+12x-12=0[/tex]

Using rational roots theorem, we can see that the possible roots of the the given equation are: [tex]\pm1,\ \pm2,\ \pm3,\ \pm4,\ \pm6,\ \pm12[/tex]

By substituting the possible roots, we can see that x = 2 is a root, thus x - 2 is a factor.
We can get the other factors by using sythetic division to divide [tex]x^4-4x^3+x^2+12x-12[/tex] by x - 2.

2  | 1     -4     1     12  -12
    |
    |         2    -4     -6    12
    |____________________
      1     -2    -3     6      0

Thus the other factor is [tex]2x^3-4x^2-6x+12[/tex]

Thus, we have

[tex]2x^3-4x^2-6x+12=0 \\ \\ \Rightarrow x^3-2x^2-3x+6=0 \\ \\ \Rightarrow x^2(x-2)-3(x-2)=0 \\ \\ \Rightarrow(x^2-3)(x-2)=0 \\ \\ \Rightarrow x=\pm\sqrt{3}\ or\ x=2[/tex]

Therefore, all the roots of [tex]x^4-4x^3+x^2+12x-12=0[/tex] are x = 2, [tex]x=\pm\sqrt{3}[/tex].



Part 2A:

Given

[tex]y=x^4-6x^2+8[/tex]

The zero of the function is when y = 0, thus we have

[tex]x^4-6x^2+8=0 \\ \\ \Rightarrow x^4-2x^2-4x^2+8=0 \\ \\ \Rightarrow x^2(x^2-2)-4(x^2-2)=0 \\ \\ \Rightarrow(x^2-4)(x^2-2)=0 \\ \\ \Rightarrow x=\pm2\ or\ x=\pm\sqrt{2}[/tex]



Part 2B:

Given

[tex]y=x^3+6x^2+x+6[/tex]

The zero of the funtion is when y = 0, thus we have

[tex]x^3+6x^2+x+6=0 \\ \\ \Rightarrow x^2(x+6)+1(x+6)=0 \\ \\ \Rightarrow(x^2+1)(x+6)=0 \\ \\ \Rightarrow x=\pm i\ or\ x=-6[/tex]



Part 2C:

Given

[tex]g(x)=x^3-4x^2-x+22[/tex]

Using rational zeros theorem, we can see that the possible zeros of the the given equation are: [tex]\pm1,\ \pm2,\ \pm11,\ \pm22[/tex]

By substituting the possible zeros, we can see that x = -2 is a zero, thus x + 2 is a factor.
We can get the other factors by using sythetic division to divide [tex]x^3-4x^2-x+22[/tex] by x + 2.

-2  | 1     -4     -1     22
     |
     |        -2     12   -22
     |____________________
       1     -6     11     0

Thus the other factor is

[tex]-2x^2+12x-22=0\\ \\ \Rightarrow x^2-6x+11=0 \\ \\ \Rightarrow x= \frac{-(-6)\pm\sqrt{(-6)^2-4(11)}}{2} \\ \\ = \frac{6\pm\sqrt{-8}}{2} =\frac{6\pm2\sqrt{2}i}{2}=3\pm i\sqrt{2}[/tex]

Therefore, all the zeros of [tex]g(x)=x^3-4x^2-x+22[/tex] are x = -2, [tex]x=3\pm i\sqrt{2}[/tex].



Part 2D:

Given

[tex]y=x^4-x^3-5x^2-x-6[/tex]

Using rational zeros theorem, we can see that the possible zeros of the the given equation are: [tex]\pm1,\ \pm2,\ \pm3,\ \pm6[/tex]

By substituting the possible zeros, we can see that x = -2 is a zero, thus x + 2 is a factor.
We can get the other factors by using sythetic division to divide [tex]x^4-x^3-5x^2-x-6[/tex] by x + 2.

-2  | 1     -1     -5     -1     -6
     |
     |        -2      6     -2      6
     |____________________
       1     -3      1     -3      0

Thus the other factor is

[tex]-2x^3+6x^2-2x+6=0 \\ \\ \Rightarrow x^3-3x^2+x-3=0 \\ \\ \Rightarrow x^2(x-3)+1(x-3)=0 \\ \\ \Rightarrow(x^2+1)(x-3)=0 \\ \\ \Rightarrow x=\pm i\ or\ x=3[/tex]

Therefore, all the zeros of [tex]y=x^4-x^3-5x^2-x-6[/tex] are x = -2, x = 3, [tex]x=\pm i[/tex].