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Exercise 4(a) · Q1

Q.Form the monic polynomial equation of degree 33 whose roots are 2,3,62, 3, 6.

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Step 1. The required equation is (x−2)(x−3)(x−6)=0(x-2)(x-3)(x-6)=0, since a root of the equation must make one of these factors zero.

Step 2. Compute the elementary symmetric functions of the roots:

S1=2+3+6=11,S_1=2+3+6=11,

S2=2⋅3+3⋅6+6⋅2=6+18+12=36,S_2=2\cdot3+3\cdot6+6\cdot2=6+18+12=36,

S3=2⋅3⋅6=36.S_3=2\cdot3\cdot6=36.

Step 3. The monic cubic with these roots is x3−S1x2+S2x−S3=0x^3-S_1x^2+S_2x-S_3=0, i.e.

x3−11x2+36x−36=0.x^3-11x^2+36x-36=0.

✓Final answer

The required monic cubic equation is x3−11x2+36x−36=0x^3-11x^2+36x-36=0.

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