If alfa and bita are zeros of polynomial f(x)=x² -p(x+1)-c , then (alfa+1)(bita+1)
Answers
Solution :-
f(x) = x² + x - 2
To find zeroes equate the given polynomial to 0
x² + x - 2 = 0
Splitting the middle term
x² + 2x - x - 2 = 0
x(x + 2) - 1(x + 2) = 0
(x + 2)(x - 1) = 0
x + 2 = 0 or x - 1 = 0
x = - 2 or x = 1
Zeroes of the polynomial x² + x - 2 are - 2 , 1
So
• α = - 2
• β = 1
Finding a polyinomial whose zeroes are (2α + 1) and (2β + 1)
2α + 1
= 2(-2) + 1
= - 4 + 1
(2α + 1)= - 3
2β + 1
= 2(1) + 1
= 2 + 1
(2β + 1) = 3
The new polynomial zeroes should be - 3 and 3
Here
• α = - 3
• β = 3
Sum of zeroes = α + β = - 3 + 3 = 0
Product of zeroes = αβ = - 3(3) = - 9
Quadratic polynomial ax² + bx + c = k{x² - x(α + β) + αβ}
(Where k ≠ 0)
= k{x² - x(0) + (-9)}
= k(x² - 0 - 9)
= k(x² - 9)
When k = 1
= 1(x² - 9)
= x² - 9
Therefore the polynomial is x² - 9.
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