Applying properties of operations on complex numbers to solve problems

One solution to the equation  $x^2+px+q=0$   is x = 1 + 4i. What would the factored form (x - m)(x - n)=0 of the quadratic be?

•  $(X-(1+4i))(X-(1-4i))\ =0$   
  

•  $(X+(1+4i))(X+(1-4i))=0$  

•  $(X-(1+4i))(X-(1+4i))=0$  

•  $(X-(1-4i))(X-(1-4i))=0$  

Answer explanation

Factored form is (x - m)(x - n) = 0, where m and n are solutions (aka x-values) that make the equation equal zero.
If one solution is x = 1 +4i, then you plug that in for m. 
The other solution will be the conjugate x = 1 -4i and you plug that in for n.
 $(X-(1-4i))(X-(1-4i))=0$  

For each function, determine the real zeros and state the multiplicity of any repeated zeros.

• {0 mult. 3, 5, 2}

• {−2 mult. 3, 3, 1}

• {0 mult. 3, 3, 2}

• {2 mult. 3, 3, 3}

One zero of 2x³-9x²-123x-220 is -4.
What is another zero?

• 18
• 4
• 11
• -7

What is the total number of possible negative roots for the following equation?

y = 4x⁶ - 12x⁵ - x⁴ + 2x³ - 6x² - 5x + 10

• 4

• 3

• 2

• 5

• 1

• 2

• 3

• 4

Zero of the zero polynomial is

• Zero
• One

• Infinity
• Not define

 $g\left(x\right)\ =x^4-\ x^3-11x^2+31x-20$  
How many real roots does the function have?

Solve for a

• a= -3 or a=8

• a=3 or a = -8

• a=- 3 or a= -8

• a= -3 pr a=8