Question

verify that 3,-2, 1 are the zeroes
of the Polynomial P(x)=x2 - 2x2- 5x+6
& verify the relation between its
zeros and coefficient ​

Answer 1

Answer:

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Step-by-step explanation:

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) = x^3 - 2x^2 - 5x + 6

As

given zeroes are = 3, -2, 1

Put x = 3

p(3) = (3)^3 - 2(3)^2 - 5(3) + 6

p(3) = 27 - 2*9 - 15 + 6

p(3) = 27 - 18 - 9

p(3) = 27 - 27

p(3) = 0

Hence verified that 3 is a zero of given polynomial

Now, put x = -2

p(-2) = (-2)^3 - 2(-2)^2 - 5(-2) + 6

p(-2) = -8 - 2(4) - 5(-2) + 6

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

p(-2) = -16 + 16

p(-2) = 0

Hence verified that -2 is a zero of given polynomial

Now, put x = 1

p(1) = (1)^3 - 2(1)^2 - 5(1) + 6

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

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

p(1) = -1 + 1

p(1) = 0

Hence, verified that 1 is a zero of given polynomial

Therefore, it's verified that 3, -2, 1 are zeroes of x^3 - 2x^2 - 5x + 6

Relationship between zeroes and coefficients:-

Sum of zeroes = 3 + (-2) + 1 = 2 => -(-2)/1 = (-coefficient of x^2) / coefficient of x^3

Product of zeroes = 3*(-2)*1 = -6 => -6/1 = (-constant term)/ coefficient of x^3

Sum of product of pair of zeroes = (3)(-2) + (-2)(1) + (1)(3) = (-5) => (-5)/1 = coefficient of x / coefficient of x^3

Hope it will help you

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