Given the polynomial f(x)=x^3+6x^2+6x+5

Question

a. Verify that x=−5 is a solution and explain how you verified this.

b. Find all normal and imaginary solutions and explain how you got your answers.

c. Write the polynomial as the product of all its factors.

d. Verify through multiplication of these factors that the product is f(x)=x^3+6x^2+6x+5 and explain how you verified this.

(when answering pls label the variables (1,2,3.. or a,b,c...) when writing the answers, THANK YOU)

Answer 1

a)

By the Factor Theorem, if f(c) = 0, then c is a factor of f(x).

f(-5) = -5³+6(-5)²+6(-5) + 5 = -125 + 150 - 30 +5 = 0

So, -5 is a solution

b)

Using synthetic division and we already know -5 is a root

1 6 6 5 |-5

-5 -5 -5

--------------

1 1 1 0

This leaves us with x²+x+1 so we can use the quadratic equation

x = (-1 ± √(1-4))/2 = (-1 ± i√3)/2 = -1/2 - i√3/2, -1/2 + i√3/2

c)

(x+(1-i√3)/2)(x+(1+i√3/2)) = x²+x+1

f(x) = (x+5)(x²+x+1) = x³+x²+x+5x²+5x+5 = x³+6x²+6x+5

Answer 2

Synthetic division says:

-5 | 1 6 6 5

-5 -5 -5

_____________________

1 1 1 0

the remainder is zero, so -5 is a solution;

Also f(-5) = (-5)^3 + 6(-5)^2 + 6(-5) + 5

= -125 + 150 + -30 + 5

= 25 + -30 + 5

= -5 + 5

= 0

so the rational root theorem says that -5 is a solution,

which is CLEARLY (-5,0)

the quadratic is x^2 + x + 1 = 0

[-1 +or- sqrt( 1^2 - 4(1)(1))] / 2

[-1 +or- sqrt(1 - 4)]/2

[ -1 +or- sqrt(-3)]/2

[ -1 +or- i sqrt(3)]/2

Then f(x) = (x+5) [( x+1/2) + sqrt(3)/2 * i ][ (x+1) - sqrt(3)/2 * i]

= (x+5) [ (x+1/2)^2 - (3/4)(-1)]

= (x+5) [ x^2 + x + 1/4 + 3/4]

= (x+5)(x^2 + x + 1)

= x^3 + x^2 + x + 5x^2 + 5x + 5

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

x=-5 is the 1st solution

[ -1 +or- i sqrt(3)]/2 are the 2nd and 3rd solutions

and they are CORRECT because they CHECK OUT