Anonymous

Answer

Anonymous

werty

Anonymous

Your question implies that there is only one fourth-degree polynomial having those three roots. That is not so. The complement of 3 + i is 3 - i. (3 + i)(3 - i) = 10 Here is your polynomial: (x² + 10)[x - √(7)]p(x), where p(x) is any linear polynomial Although it is not necessary, many people insist that polynomials should have rational coefficients. If you are one of them, let p be defined like this: p(x) = a[x + √(7)], where a is any rational number Notice that this still leaves you with infinitely many solutions.

Anonymous

Before starting, assuming this equation: x² - 6x + 10 = 0 x² - 6x = - 10 x² - 6x + 9 = - 10 + 9 (x - 3)² = - 1 (x - 3)² = i² (x - 3)² = (± i)² x - 3 = ± i x = 3 ± i x₁ = 3 + i ← first root x₂ = 3 - i ← second root Conclusion: When an equation has (a + ib) has a complex root, you can say that the other complex root is (a - ib), i.e. the conjugate. In your case, (3 + i) is a root, so you can factorize: [x - (3 + i)] According the conclusion, as (3 + i) is a root, so can say that (3 - i) is a root too, so you can factorize: [x - (3 - i)] √7 is a root, so you can factorize: x - √7 …and it gives us: [x - (3 + i)].[x - (3 - i)].(x - √7) = 0 (x - 3 - i).(x - 3 + i).(x - √7) = 0 (x² - 3x + xi - 3x + 9 - 3i - xi + 3i - i²).(x - √7) = 0 (x² - 6x + 9 - i²).(x - √7) = 0 → recall: i² = - 1 (x² - 6x + 10).(x - √7) = 0 x³ - x√7 - 6x² + 6x√7 + 10x - 10√7 = 0 x³ - 6x² + 5x√7 + 10x - 10√7 = 0 ← but you need fourth degree, so you multiply by x x⁴ - 6x³ + 5x²√7 + 10x² - 10x√7 = 0 x⁴ - 6x³ + 5x².(2 + √7) - 10x√7 = 0 ← this is the required equation

Anonymous

3+i => 3 - i is another root √7 => -√7 is another root The roots are 3 + i, 3 - i, -√7 and √7: f(x) = (x - (3 + i))(x - (3 - i))(x - (-√7))(x - (√7)) f(x) = ((x - 3)^2 - i^2) (x^2 - √7^2) f(x) = (x^2 - 6x + 9 - (-1)) (x^2 - 7) f(x) = (x^2 - 6x + 10)(x^2 - 7) f(x) = x^4 - 6x^3 + 3x^2 + 42x - 70

Anonymous

Form the fourth degree whose roots are 3+i and √7? f(x) = (x - (3 + i))^2 (x - (√7))^2 meets these conditions, but is by no means the ONLY fourth degree polynomial that does. f(x) = x^4 - (2√7 + 6 + 2i) x^3 + ((12√7 + 15) + (4√7 + 6)i) x^2 - ((16√7 + 42) + (12√7 + 14)i) x + (56 + 42 i). Nothing in the question stated the coefficients had to be Real or even Integer. ...but assuming it had specified Integer coefficients, we need the other roots to be the conjugate of 0+√7 which is 0-√7, and the complex conjugate of 3+i which is 3-i. f(x) = (x - (3+i))(x - (3-i))(x - √7)(x + √7) which you are welcome to expand... ...or just visit http://www.wolframalpha.com/input/?i=exp...