Find all other zeroes of 2x⁴-3x³-3x²+6x-2 , if you know that two of its zeroes are √2 and -√2.
Answers
Answer:
Step-by-step explanation:
Let p ( x ) = 2x⁴ - 3x³ - 3x² + 6x - 2.
Given that, √2 and ( - √2 ) are the zeroes of p ( x )
So, we can divide by ( x - √2 ) and ( x + ≡2 ) = x² - 2, we get
∴ 2x⁴ - 3x³ - 3x² + 6x - 2 = ( x² - 2 ) ( 2x² - 3x + 1 )
Now splitting the middle term,
2x⁴ - 3x³ - 3x² + 6x - 2 = ( x² - 2) [ 2x² - 2 x - x + 1 )]
= ( x² - 2 ) [ 2x ( x - 1 ) - 1 ( x - 1 ) ]
= ( x² - 2 ) ( x - 1 ) ( 2x -1 )
The zeroes are ( x - 1 ) and ( 2x - 1 ) are 1 and 1/2.
Hence all zeroes of p ( x ) are - √2 , √2 , 1 / 2 , 1 is the answer.
Given :-
♦ The zeroes of the biquadratic polynomial 2x⁴-3x³-3x²+6x-2 are √2 and -√2
To find :-
♦ The remaining zeroes of the polynomial.
Solution :-
Given biquadratic polynomial is
P(x) = 2x⁴-3x³-3x²+6x-2
Given zeroes are √2 and -√2
We know that
By Factor Theorem
√2 is a zero of P(x) then (x-√2) is a factor of P(x)
-√2 is a zero of P(x) then (x+√2) is a factor of P(x)
=> (x+√2)(x-√2) is also the factor P(x)
=> x²-(√2)² is also the factor of P(x)
Since , (a+b)(a-b) = a²-b²
Where, a = x and b = √2
=> x²-2 is also the factor of P(x).
Now to find the other two zeroes of P(x) then we have to divide P(x) by x²-2
x²-2 ) 2x⁴-3x³-3x²+6x-2 ( 2x²-3x+1
2x⁴ -4x²
(-)
________________
-3x³ +x²
-3x³ +6x
(-)
_________________
x² -2
x² -2
(-)
_________________
0
_________________
The quotient = 2x²-3x+1
Therefore,
2x⁴-3x³-3x²+6x-2 = (x²-2)(2x²-3x+1)
=> (x²-2)(2x²-2x-x+1)
=> (x²-2)[2x(x-1)-1(x-1)]
=> (x²-2)(x-1)(2x-1)
To get zeroes of P(x) then we have to write
P(x) = 0
=> (x²-2)(x-1)(2x-1) = 0
=> x²-2 = 0 or x-1 = 0 or 2x-1 = 0
=> x² = 2 or x = 1 or 2x = 1
=> x = ±√2 or x = 1 or x = 1/2
Therefore ,x = √2 , -√2 , 1 and 1/2
Answer:-
The other two zeroes of the biquadratic polynomial are 1 and 1/2