Answer This
@Mods,Stars and Best Users..
☆ If two zeroes of the polynomial x⁴-6x³-26x²+138x-35 are 2 ±√3, find other zeroes.
Answers
Answer
The two zeroes of the polynomial is 2+
3
,2−
3
Therefore, (x−2+
3
)(x−2−
3
)=x
2
+4−4x−3
= x
2
−4x+1 is a factor of the given polynomial.
Using division algorithm, we get
x
4
−6x
3
−26x
2
+138x−35=(x
2
−4x+1)(x
2
−2x−35)
So, (x
2
−2x−35) is also a factor of the given polynomial.
x
2
−2x−35=x
2
−7x+5x−35
=x(x−7)+5(x−7)
=(x−7)(x+5)
Hence, 7 and −5are the other zeros of this p
Step-by-step explanation:
Given :-
Two zeroes of the polynomial x⁴-6x³-26x²+138x-35 are 2 ±√3
To find :-
Find other zeroes?
Solution :-
Given bi-quadratic polynomial is x⁴-6x³-26x²+138x-35
Let P(x) = x⁴-6x³-26x²+138x-35
Given zeroes = 2±√3 = 2+√3 and 2-√3
We know that by Factor Theorem
If 2+√3 is a zero of P(x) then [x-(2+√3)] is a factor of P(x).
If 2-√3 is a zero of P(x) then [x-(2-√3)] is a factor of P(x).
If both 2+√3 and 2-√3 are the zeroes of P(x) then
[x-(2+√3)][x-(2-√3)] is a factor of P (x)
=> (x-2-√3)(x-2+√3)
=> [(x-2)-√3][(x-2)+√3]
=> (x-2)²-(√3)²
Since (a+b)(a-b) = a²-b²
Where a = x-2 and b = √3
=> x²-4x+4-3
Since (a-b)² = a²-2ab+b²
=> x²-4x+1 is a factor of P (x).-----(1)
To get other zeroes of P(x) , then divide P(x) by x²-4x+1.
x²-4x+1 ) x⁴-6x³-26x²+138x-35 ( x²-2x-35
x⁴-4x³+x²
(-) (+) (-)
_________________
-2x³-27x²+138x
-2x³+8x²-2x
(+) (-) (+)
__________________
-35x² +140x-35
-35x²+140x-35
(+) (-) (+)
__________________
0
__________________
Quotient = x²-2x-35
Remainder = 0
Now,
P(x) = Divisor×Quotient + Remainder
=> P(x) = (x²-4x+1)(x²-2x-35)+0
=> P(x) = (x²-4x+1)(x²-2x-35)
=> P(x) = (x²-4x+1)(x²-7x+5x-35)
=> P(x) = (x²-4x+1)[x(x-7)+5(x-7)]
=> P(x) = [x-(2+√3)][x-(2-√3)](x-7)(x+5)
from (1)
To get zeroes we write P(x) as P(x) = 0
=> x-(2+√3) = 0 or x-(2-√3) = 0 or x-7 = 0 or
x+5 = 0
=> x = 2+√3 or x = 2-√3 or x = 7 or x = -5
Therefore the other zeroes = -5 and 7
Answer:-
The other zeroes of the given bi-quadratic polynomial are -5 and 7
Used formulae:-
- (a+b)(a-b) = a²-b²
- (a-b)² = a²-2ab+b²
Fundamental theorem on Polynomials :-
P(x) = g(x)×q(x) +r(x)
Dividend= Divisor×Quotient + Remainder
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.
Used Concepts:-
- To get zeroes we write P(x) we equate P(x) to zero .
- If a is a factor of b and c is a factor b then ac is a factor of b.
Ex:- 2 is a factor of 20
5 is a factor of 20
2×5 = 10 is also a factor of 20