find a degree 3 polynomial

Question

Find a degree 3 polynomial with coefficient of x³ equal to 1 and zeros 5, -4i and 4i 

Answer 1

Hi Dalia,

 

(x-5)(x-(-4i))(x-(4i))

=(x-5)(x+4i)(x-4i)

=(x-5)(x²-4i²)

=(x-5)(x²-16i²)        

now using √-1=i,so i²=-1

=(x-5)(x²+16)

=x³-5x²+16x-80

 

 

Ans: A degree 3 polynomial P(x)=x³-5x²+16x-80

 

Meena from Strongsville, OH     

Answer 2

Hi Dalia,

 

By the factor theorem,the required polynomial have factors (x-5),(x+4i) and (x-4i).

 

Therefore the solution is P(x)=(x-5)(x+4i)(x-4i)

                                         

                                         =(x-5)((x)²-(4i)²)

 

                                         =(x-5)(x²+16)           (i²=-1)

 

                                         =x³+16x-5x²-80

                                           

                                         =x³-5x²+16x-80

Answer 3

Hi Dalia;

My mistake...see below...Thank you Shelly and Meena...

f(x)=(x²....)(x....)

We know that the first parenthetical equation must have the results of -4i and 4i because...

x²+2=0.....x²=-16

f(x)=(x²+16)(x...)

The second parenthetical equation must be (x-5) to render the result of...

0=x-5...x=5

f(x)=(x²+16)(x-5)