Question #9c3e2

1 Answer
Feb 14, 2017

Answer:

#{-3,-1/2}#

Explanation:

If #1/2# and #3# are zeroes then #y=4x^4-37x^2+9=p_2(x)(x-1/2)(x-3)# where #p_2(x)=a x^2+bx+c# is a generic order #2# polynomial.

Substituting and equating coefficients we have

#4x^4-37x^2+9=(ax^2+bx+c)(x-1/2)(x-3)#

so

#{((3 c)/2=9),(-(3 b)/2 + (7 c)/2=0),( - (3 a)/2 + (7 b)/2 - c=37),((7 a)/2 - b=0),(a=4):}#

now solving for #a,b,c# we obtain

#a = 4, b = 14, c = 6#

so #p_2(x)=4x^2+14x+6# with roots

#x={-3,-1/2}# so finally

#y=4x^4-37x^2+9=4(x^2-3^2)(x^2-(1/2)^2)#