If A&B are the zeroes of the polynomial p(x)=3x^2+2x+1, find the polynomial whose zeroes are 1-A/1+A and 1-B/1+B.
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As A and B are zeroes,
A+B= -b/a = -2/3
AxB= c/a = 1/3
For new polynomial,
1-A/1+A + 1-B/1+B= (1-A+B-AB+1+A-B-AB) / (1+A+B+AB). [Taking LCM]
= (2 - 2AB) / (1 + A + B + AB)
By substituting values,
{2 - 2 (1/3) }/ (1 -2/3 + 1/3)
= 2
Now
1-A/1+A * 1-B/ 1+B = (1 - A - B + AB)/( 1 + A + B + AB)
By substituting values,
(1 + 2/3 + 1/3)/ (1 -2/3 + 1/3)
= 2/(2/3)=3
Therefore required polynomial is:- k(x^2 - 2x + 3)
A+B= -b/a = -2/3
AxB= c/a = 1/3
For new polynomial,
1-A/1+A + 1-B/1+B= (1-A+B-AB+1+A-B-AB) / (1+A+B+AB). [Taking LCM]
= (2 - 2AB) / (1 + A + B + AB)
By substituting values,
{2 - 2 (1/3) }/ (1 -2/3 + 1/3)
= 2
Now
1-A/1+A * 1-B/ 1+B = (1 - A - B + AB)/( 1 + A + B + AB)
By substituting values,
(1 + 2/3 + 1/3)/ (1 -2/3 + 1/3)
= 2/(2/3)=3
Therefore required polynomial is:- k(x^2 - 2x + 3)
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