Mathematics, 24.07.2019 11:50 andydiaz1227

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

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Answer from: rprest00

Part 1A:

Given

$x^3+x^2+4x+4=0 \\ \\ \Rightarrow x^2(x+1)+4(x+1)=0 \\ \\ \Rightarrow(x^2+4)(x+1)=0 \\ \\ \Rightarrow x= -1\ or\ x=\pm2i$

Part 1B:

Given

x^3+4x^2+x-6=0

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

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 x^3+4x^2+x-6

by x - 1.

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

Thus the other factor is x^2+5x+6=(x+2)(x+3)

Therefore, all the roots of x^3+4x^2+x-6=0

are x = 1, x = -2 and x = -3.

Part 1C:

Given

x^4-4x^3+x^2+12x-12=0

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

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 x^4-4x^3+x^2+12x-12

by x - 2.

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

Thus the other factor is 2x^3-4x^2-6x+12

Thus, we have

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

Therefore, all the roots of x^4-4x^3+x^2+12x-12=0

are x = 2, $x=\pm\sqrt{3}$ .

Part 2A:

Given

y=x^4-6x^2+8

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

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

Part 2B:

Given

y=x^3+6x^2+x+6

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

$x^3+6x^2+x+6=0 \\ \\ \Rightarrow x^2(x+6)+1(x+6)=0 \\ \\ \Rightarrow(x^2+1)(x+6)=0 \\ \\ \Rightarrow x=\pm i\ or\ x=-6$

Part 2C:

Given

g(x)=x^3-4x^2-x+22

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

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 x^3-4x^2-x+22

by x + 2.

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

Thus the other factor is

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

Therefore, all the zeros of g(x)=x^3-4x^2-x+22

are x = -2, $x=3\pm i\sqrt{2}$ .

Part 2D:

Given

y=x^4-x^3-5x^2-x-6

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

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 x^4-x^3-5x^2-x-6

by x + 2.

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

Thus the other factor is

$-2x^3+6x^2-2x+6=0 \\ \\ \Rightarrow x^3-3x^2+x-3=0 \\ \\ \Rightarrow x^2(x-3)+1(x-3)=0 \\ \\ \Rightarrow(x^2+1)(x-3)=0 \\ \\ \Rightarrow x=\pm i\ or\ x=3$

Therefore, all the zeros of y=x^4-x^3-5x^2-x-6