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