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
George C.
Sep 25, 2017

#4x^3-52x^2+89x-25 = 0#

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 #alpha, beta, gamma# then:

#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 #4# to get integer coefficients:

#4x^3-52x^2+89x-25 = 0#

Andrea S.
Sep 25, 2017

It is: #4x^3-52x^2+89x-25#

Explanation:

If the roots of the polynomial are #a,b,c# then:

#P(x) = 2x^3-8x^2+3x+5 = 2(x-a)(x-b)(x-c)#

Consider now:

#P(-x) = -2x^3-8x^2-3x+5 = 2(-x-a)(-x-b)(-x-c) = -2(x+a)(x+b)(x+c)#

So:

#P(x)P(-x) = (2x^3-8x^2+3x+5)(-2x^3-8x^2-3x+5)#

#P(x)P(-x) =-4x^6+52x^4-89x^2+25#

But on the other hand:

#P(x)P(-x) = -4(x-a)(x-b)(x-c)(x+a)(x+b)(x+c) = -4(x^2-a^2)(x^2-b^2)(x^2-c^2)#

So, substituting #x# for #x^2# and changing sign we have:

#4x^3-52x^2+89x-25 = 4(x-a^2)(x-b^2)(x-c^2)#

sjc
Sep 25, 2017

#4x^3-52x^2+89x-25=0#

Explanation:

when the new roots are symmetrical, as in this case, we can use the following 'approach, which sometimes is quicker.

#2x^3-8x^2+3x+5=0#

roots

#alpha^2,beta^2,gamma^2#

#"let" y=x^2#

#=>x=y^(1/2)#

substitute into the cubic

#2y^(3/2)-8y^(2/2)+3y^(1/2)+5=0#

group and then square to eliminate fractional powers

#(2y^(3/2)+3y^(1/2))^2=(8y-5)^2#

#4y^3+12y^2+9y=64y^2-80y+25#

#=>4y^3-52y^2+89y-25=0#

#y# is a dummy variable so;

#4x^3-52x^2+89x-25=0#

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