Question #36696

2 Answers
Nov 30, 2017

The desired sum is #-6# and the product, #5.#

Explanation:

Given that #1# of #3# roots of the cubic polynomial

#p(x)=x^3+2x^2-19x-20" is "4.#

Therefore, #(x-4)# must be a factor of #p(x).#

We use this fact to rewrite #p(x),# as

#p(x)=ul(x^3-4x^2)+ul(6x^2-24x)+ul(5x-20),#

#=x^2(x-4)+6x(x-4)+5(x-4),#

#=(x-4)(x^2+6x+5),#

#=(x-4){ul(x^2+5x)+ul(x+5)},#

#=(x-4){x(x+5)+1(x+5)},#

#=(x-4)(x+5)(x+1).#

This shows that the other #2# roots of #p(x)" are "-5 and -1.#

Therefore, the desired sum is #-6# and the product, #5.#

Dec 1, 2017

#"Sum="-6, &, "Product="5.#

Explanation:

Here is another method to solve the Problem.

We are given that #4# is a root of the cubic equation

#p(x)=x^3+2x^2-19x-20=0.#

Let #alpha and beta# be the other #2# roots.

Clearly, #p(x)=(x-4)(x-alpha)(x-beta).#

#:.x^3+2x^2-19x-20=(x-4)(x-alpha)(x-beta).#

Expanding the R.H.S., we get,

#x^3+2x^2-19x-20=x^3-(alpha+beta+4)x^2+(4alpha+alphabeta+4beta)x-4alphabeta.#

Comparing the respective co-effs., we get,

#alpha+beta+4=-2, or, alpha+beta=-6...............<<1>>.#

#4alpha+alphabeta+4beta=-19, i.e.,#

#4(alpha+beta)+alphabeta=-19...................................<<2>>, and,#

#4alphabeta=20, or, alphabeta=5......................................<<3>>.#

Observe that #<<2>># is satisfied by #<<1>> and <<3>>.#

#<<1>> and <<3>># give the desired sum #-6# and the

product #5#.

Enjoy Maths!