pls answer the above question
Answers
Step-by-step explanation:
Given :-
i) k²-3x+2k
ii) 3kx²-√2x+1 are completely divisible by x-1
To find:-
Find the value of k for the polynomials ?
Solution:-
I)
Given polynomial P(x) = k²-3x-2k
If P(x) is completely divisible by (x-1) then it satisfies the given equation.i.e. P(1) = 0 (by Factor Theorem)
=> P(1) = k²-3(1)+2k = 0
=> k²-3+2k = 0
=> k²+2k-3 = 0
=> k²-k+3k-3 = 0
=> k(k-1)+3(k-1) = 0
=> (k-1)(k+3) = 0
=>k-1 = 0 or k+3 = 0
=> k = 1 or k = -3
Therefore, k = 1 and -3
ii)Given polynomial P(x) = 3kx²-√2x+1
If P(x) is completely divisible by (x-1) then it satisfies the given equation.i.e. P(1) = 0 (by Factor Theorem)
=> P(1) = 3k(1)²-√2(1)+1= 0
=> 3k-√2+1 = 0
=>3k = 1-√2
=> k = (1-√2)/3
Therefore, k = (1-√2)/3
Answer:-
i)The values of k for the given problem is 1 and -3
ii) The value of k for the given problem is
(1-√2)/3
Used formulae:-
Factor Theorem:-
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 then P(1) = 0 vice versa
hopefully its helped u dear :)
