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Answers
Step-by-step explanation:
Given :-
(x-2√2) is a factor of x³-(4+2√2)x²+(8√2+3)x-6√2
To find:-
Find the other zeroes of the polynomial?
Solution :-
Given polynomial is x³-(4+2√2)x²+(8√2+3)x-6√2
Let P(x) = x³-(4+2√2)x²+(8√2+3)x-6√2
Given factor of P(x) = (x-2√2)
To get another zeroes we have to divide P(x) by (x-2√2)
=>[x³-(4+2√2)x²+(8√2+3)x-6√2]/(x-2√2)
=> [x³-4x²-2√2x²+8√2x+3x-6√2]/(x-2√2)
=> [(x³-2√2x²)-(4x²+8√2x)+(3x-6√2)]/(x-2√2)
=> [x²(x-2√2)-4x(x-2√2)+3(x-2√2)]/(x-2√2)
=> [(x-2√2)(x²-4x+3)]/(x-2√2)
On cancelling (x-2√2) then
We get quotient = x²-4x+3
we have ,
P(x) = (x-2√2)(x²-4x+3)
=> P(x)=(x-2√2)[x²-x-3x+3]
=> P(x) = (x-2√2)[x(x-1)-3(x-1)]
=> P(x) = (x-2√2)(x-1)(x-3)
To get zeroes we write P(x) = 0
=> (x-2√2)(x-1)(x-3) = 0
=> (x-2√2) = 0 or (x-1) = 0 or (x-3) = 0
=> x = 2√2 or x = 1 or x = 3
zeroes are 2√2 , 1 and 3
Answer:-
The zeores of the given cubic polynomial are 2√2 ,1 and 3
Used formulae:-
- To get zeores of a Polynomial P(x) then we write P(x) = 0
Points to know :-
Factor Theorem :-
Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if x-a is a factor of P (x) then P(a) = 0 vice-versa.
Division Algorithm on Polynomials :-
P(x) = g(x)×q(x)+r(x)
Where ,
P(x) = Dividend
g(x) = Divisor
q(x) = quotient
r(x) = remainder.
Step-by-step explanation:
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