show that 1/2 is a zero of the polynomial 2x^2+7x-4
Answers
Answered by
35
By remainder theorem:
p(x)=2x²+7x-4
x=1/2
p(1/2)=2(1/2)²+7(1/2)-4
=1/2+7/2-8/2
=(8-8)/2
=0/2
=0
As the remainder is 0, 1/2 is the zero of p(x)
p(x)=2x²+7x-4
x=1/2
p(1/2)=2(1/2)²+7(1/2)-4
=1/2+7/2-8/2
=(8-8)/2
=0/2
=0
As the remainder is 0, 1/2 is the zero of p(x)
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Answered by
18
Let
p(x)=2x^2+7x-4
If x=1/2
p(1/2)=2*(1/2)^2+7*(1/2)-4
=2/4+7/2-4
=1/2+7/2-4
=(1+7-8)/2
=0/2
=0
Therefore
p(x) is equal to zero
1/2 is a zero of p(x)
p(x)=2x^2+7x-4
If x=1/2
p(1/2)=2*(1/2)^2+7*(1/2)-4
=2/4+7/2-4
=1/2+7/2-4
=(1+7-8)/2
=0/2
=0
Therefore
p(x) is equal to zero
1/2 is a zero of p(x)
:)
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