Find all the zeroes of x² + x³-9x² - 3x + 18,if you know that two of its zeroes are √3 and -√3.
Answers
Step-by-step explanation:
Given :-
√3 and -√3 are the zeroes of
x²+x³-9x² -3x +18
Correction :-
x⁴+ x³-9x² -3x + 18
Reason :-
Bi-quadratic Polynomial has at most four zeroes
To find :-
Find all the zeroes of x⁴ + x³-9x² -3x + 18 ?
Solution :-
Given bi-quadratic polynomial is x⁴+x³-9x²-3x+18
Let P(x) = x⁴+x³-9x²-3x+18
Given zeroes are = √3 and -√3
We know that
If √3 is a zero of P(x) then (x-√3) is a factor of P(x)
If -√3 is a zero of P(x) then (x+√3) is a factor of P(x)
If √3 and -√3 are the zeroes of P(x) then (x-√3) and (x+√3) are the factors
=> (x-√3)(x+√3) is a factor of P(x)
=> x²-(√3)² is a factor of P (x)
Since (a+b)(a-b) = a²-b²
Where, a = x and b = √3
=> x²-3 is a factor of P(x)
To get remaining zeroes,divide P(x) by (x²-3)
x²-3 ) x⁴+x³-9x²-3x+18 (x²+x-6
x⁴ -3x²
(-) (+)
______________
x³ -6x² -3x
x³ -3x
(-) (+)
_______________
-6x²+ 18
-6x²+ 18
(+) (-)
_______________
0
_______________
We get Remainder = 0
Quotient = x²+x-6
For obtaining zeroes we write P(x) as P(x) = 0
=> x⁴+x³-9x²-3x+18 = 0
=> (x²-3)(x²+x-6) = 0
=> (x+3)(x-3)(x²+3x-2x-6) = 0
=> (x+3/(x-3)[x(x+3)-2(x+3)] = 0
=> (x+3)(x-3)(x+3)(x-2) = 0
=> x+3 = 0 or x-3 = 0 or x+3 = 0 or x-2 = 0
=> x = -3 or x = 3 or x = -3 or x = 2
Therefore the zeroes are -3,-3,2,3
Answer :-
The four zeroes of the given bi-quadratic polynomial are -3, -3, 2 and 3
Used Concept :-
For obtaining zeroes we write P(x) as P(x) = 0
Used formulae:-
- (a+b)(a-b) = a²-b²
- Dividend = Divisor × Quotient + Remainder