if two zeroes of polynomial p(x)=2x^4-3x^3-3x^2+6x-2 are root 2 and -root 2. find its other zeroes
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here is your answer.
(x-√2)(x+√2)=x^2-2
2x^4-3x^3-3x^2+6x-2 ÷ x^2-2
= 2x^2-3x+1
now,
for zeroes, 2x^2-3x+1=0
2x^2-2x-x+1=0
2x(x-1)-1(x-1)=0
(2x-1)=0,(x-1)=0
x=1/2 & x=1
(x-√2)(x+√2)=x^2-2
2x^4-3x^3-3x^2+6x-2 ÷ x^2-2
= 2x^2-3x+1
now,
for zeroes, 2x^2-3x+1=0
2x^2-2x-x+1=0
2x(x-1)-1(x-1)=0
(2x-1)=0,(x-1)=0
x=1/2 & x=1
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