The roots of the polynomial equation #2x^3-8x^2+3x+5=0# are #alpha#, #beta# and #gamma#. What is the polynomial equation with roots #alpha^2#, #beta^2# and #gamma^2#?
3 Answers
Explanation:
Note that in general:
#(x-alpha)(x-beta)(x-gamma)#
#=x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma#
So if we know the elementary symmetric polynomials:
#alpha+beta+gamma#
#alphabeta+betagamma+gammaalpha#
#alphabetagamma#
then we know the coefficients of a monic cubic polynomial with these zeros.
If the roots are
#2x^3-8x^2+3x+5#
#= 2(x-alpha)(x-beta)(x-gamma)#
#= 2x^3-2(alpha+beta+gamma)x^2+2(alphabeta+betagamma+gammaalpha)x-2alphabetagamma#
Equating coefficients, we find:
#{ (alpha+beta+gamma = 4), (alphabeta+betagamma+gammaalpha = 3/2), (alphabetagamma = -5/2) :}#
So:
#alpha^2+beta^2+gamma^2#
#= (alpha+beta+gamma)^2 - 2(alphabeta+betagamma+gammaalpha)#
#= 4^2-2*3/2 = 16-3 = 13#
#alpha^2beta^2 + beta^2gamma^2 + gamma^2alpha^2#
#=(alphabeta+betagamma+gammaalpha)^2 - 2alphabetagamma(alpha+beta+gamma)#
#=(3/2)^2 - 2(-5/2)(4) = 9/4+20 = 89/4#
#alpha^2beta^2gamma^2 = (alphabetagamma)^2 = (-5/2)^2 = 25/4#
So we can write a cubic equation:
#x^3-13x^2+89/4x-25/4 = 0#
or multiply by
#4x^3-52x^2+89x-25 = 0#
It is:
Explanation:
If the roots of the polynomial are
Consider now:
So:
But on the other hand:
So, substituting
Explanation:
when the new roots are symmetrical, as in this case, we can use the following 'approach, which sometimes is quicker.
roots
substitute into the cubic
group and then square to eliminate fractional powers