Question

verify that 3, - 2, 1 are the zeros of the cubic polynomial p x is equals to x cube minus 2 x square - 5 x + 6 and verify the relation between its zeros and coefficients​

Answer 1

Given p(x)=x³-2x²-5x+6

p(3)=27-18-15+6=0

p(-2)=-8-8+10+6=0

p(1)=1-2-5+6=0

Hence the given numbers are the zeroes of the given cubic polynomial.

Let the coefficients be q,r,s,t and roots a,b,c

Relation between roots

sum of the roots=a+b+c=-r/q=-(-2/1)=2=3+(-2)+1

product of the roots=abc=-t/q=-6=3×-2×1

sum of the products of pair of roots=ab+bc+ca=s/q=-5=3(-2)+(-2)1+1(3)

Answer 2

Step-by-step explanation:

Given: P(X)=(x³-2x²-5x+6)

Therefore, P(3) =(3³-2×3²×5×3+6)

=(27-18-15+6)=0

P(-2)=[(-2³)-2×(-2)²-5×(-2)+6]

=(-8-8+10+6)=0

P(1)=(1³-2×1²-5×1+6)

=(1-2-5+6)=0

Therefore , 3,-2 and 1 are the zeroes of P(x).

let ,

a = 3

b= -2

c= 1

then,

(a+b+c) =(3-2+1)=2 = - (coefficient of x²)

---------------------------

(coefficient of x²)

(ab+bc +ac) = (-6-2+3)= -5 = - coefficient of x)

----- -----------------------

1. coefficient of x²)

abc= {3×(-2)×1}= -6 = -constant term

---- -------------------------

1 coefficient of x³

Thanks