Question

show that 1, 2, and 3 are
zeros of p(x)= x^3- 6x² + 11x-6​

Answer 1

Step-by-step explanation:

p(x)=x³-6x²+11x-6

p(1)=(1)³-6(1)²+11(1)-6

p(1)=1-6+11-6

p(1)=-5+5

p(1)=0

p(2)=(2)³-6(2)²+11(2)-6

p(2)=8-24+22-6

p(2)=-16+16

p(2)=0

p(3)=(3)³-6(3)²+11(3)-6

p(3)=27-54+33-6

p(3)=-27+27

p(3)=0

here when we put x=1,2,3 the ans is zero therefore they are the roots of the given eqn i.e they are the zeros of given equation

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