verify that 2,3 and 1/2 are the zeroes of the polynomial p (x)= 2x^3-11x^2+17x-6
Answers
Answered by
527
p(x)=2x³-11x²+17x-6
Put x=2,
p(2)=2(2)³-11(2)²+17(2)-6
=2(8)-11(4)+34-6
=16-44+28
=44-44=0
Put x=3,
p(3)=2(3)³-11(3)²+17(3)-6
=2(27)-11(9)+51-6
=54-99+51-6
=99-99=0
Put x=1/2,
p(1/2)=2(1/2)³-11(1/2)²+17(1/2)-6
=2(1/8)-11(1/4)+17/2-6
=2/8-11/4+17/2-6
=1/4-11/4+17/2-6
=(1-11)/4+17/2-6
=-10/4+17/2-6
=(-10+34-24)/4
=(34-34)/4
=0/4
=0
Therefore,2,3 and 1/2 are the zeroes of the given polynomial.
Hence verified
Put x=2,
p(2)=2(2)³-11(2)²+17(2)-6
=2(8)-11(4)+34-6
=16-44+28
=44-44=0
Put x=3,
p(3)=2(3)³-11(3)²+17(3)-6
=2(27)-11(9)+51-6
=54-99+51-6
=99-99=0
Put x=1/2,
p(1/2)=2(1/2)³-11(1/2)²+17(1/2)-6
=2(1/8)-11(1/4)+17/2-6
=2/8-11/4+17/2-6
=1/4-11/4+17/2-6
=(1-11)/4+17/2-6
=-10/4+17/2-6
=(-10+34-24)/4
=(34-34)/4
=0/4
=0
Therefore,2,3 and 1/2 are the zeroes of the given polynomial.
Hence verified
Answered by
50
Answer:
yes
Step-by-step explanation:
put x=2 then you will get zero
then put x=3 then also you will get zero
put x=1/2 then also it will become zero
the divisor which completely divides the dividend is a factor of it hence they are the factors of given quadratic polynomial
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