Find all the zeroes of p(x) = 2x4 11x3 + 7x2 + 13x 7, if its two zeroes are (3 + √2) and (3 √2)
Answers
Answer:
Step-by-step explanation:
If α and β are zeroes of the polynomial then x2 -(α+ β)x + αβ .
α = (3+√2) and β = (3 - √2).
α+ β = 6 and αβ = 7.
∴ x2 - 6x + 7.
Given polynomial 2x4-11x3+ 7x2 + 13x - 7
x2 - 6x + 7 ) 2x4-11x3+ 7x2 + 13x - 7 ( 2x2 + x -1
2x4-12x3+14x2 (subtract)
x3 - 7x2 + 13x
x3- 6x2 + 7x (subtract)
- x2 + 6x - 7
- x2 + 6x - 7 (subtract)
∴ The Quotient is 2x2 + x -1
= 2x2 + 2x - x -1
= ( 2x -1)( x +1).
∴ x = 1/2 , -1 are the other zeros of the polynomial.
2)
Using remainder theorem f(x) = g(x) (2x-1) + (x+3)
4x3-8x2 + 8x + 1 = g(x) (2x-1) + (x+3)
4x3-8x2 + 7x - 2 = g(x) (2x-1).
g(x) = (4x3-8x2 + 7x - 2) / (2x -1).
2x -1 ) 4x3-8x2 + 7x - 2 ( 2x2 - 3x + 2
4x3-2x2 (subtract)
- 6x2 + 7x
- 6x2 + 3x (subtract)
4x - 2
4x - 2 (subtract)
∴ g(x) = 2x2 - 3x + 2.