find k so that x^2+2x+k is a factor of 2x^4+x^3-14x^2+5x+6.also find all zeroes of the two polynomials?
Answers
Step-by-step explanation:
Given:-
x^2+2x+k is a factor of 2x^4+x^3-14x^2+5x+6.
To find:-
Find the value of k ?
Find all zeroes of the two polynomials?
Solution:-
see the above attachment for division
Given that
Polynomial = P(x) = 2x^4+x^3-14x^2+5x+6.
Given factor = x^2+2x+k
Remainder = 7kx+21x+2k^2-8k+6
Quotient = 2x^2-3x-8-2k
We know that
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 then P(a) = 0,
vice -versa.
So the remainder must be equal to zero
=>7kx+21x+2k^2-8k+6 = 0
=> (7k+21)x + (2k^2-8k+6) = 0
=> 7k+21 = 0 and 2k^2-8k+6 = 0
=> 7k = -21
=> k = -21/7
=> k = -3
Now the divisor = x^2+2x-3
Quotient = 2x^2-3x-8-2(-3)=2x^2-3x+2
We know that
Dividend = Divisor×Quotient+Remainder
=> 2x^4+x^3-14x^2+5x+6
=>[ x^2+2x-3][2x^2-3x+2]+0
=> [ x^2+2x-3][2x^2-3x+2]
To get the zeroes , we write P(x) = 0
=> [ x^2+2x-3][2x^2-3x+2] = 0
=> [ x^2+2x-3] = 0 and [2x^2-3x+2] = 0
=> x^2-x+3x-3=0 and 2x^2-4x+x+2 = 0
=>x(x-1)+3(x-1)=0 and 2x(x-2)+1(x-2) = 0
=> (x-1)(x+3) = 0 and (2x+1)(x-2) = 0
=> x-1 = 0 and x+3 = 0 and 2x+1 = 0 and x-2= 0
=> x = 1 and -3 and 1/2 and 2
The zeroes = 1,-3,1/2,2
Answer:-
The value of k = -3
The zeroes of the given bi-quadratic Polynomial are 1,-3,1/2,2
Zeroes of the Polynomial 2x^2+2x-3 are 1 and -3
Used formulae:-
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 then P(a) = 0,
vice -versa.
- Dividend = Divisor×Quotient+ Remainder
- Since the degree of the bi-quadratic Polynomial is 4 ,It has at most 4 zeroes.
- Since the degree of the quadratic Polynomial is 2 ,It has at most 2 zeroes.
