Question

Find all the zeros of the polynomial x4-3x3-5x2+21x-14 if the given zeros are +√7, -√7

Answer 1

Given, +√7 and -√7 are the zeros of x⁴ - 3x³ - 5x² + 21x - 14

so,x⁴ - 3x³ - 5x² + 21x - 14 is divisible by {(x +√7)(x - √7)}.
e.g., x⁴ - 3x³ - 5x² + 21x - 14 is divisible by x² - 7
now, x⁴ - 7x² + 2x² - 14 - 3x³ + 21x
= x²(x² - 7) + 2(x² - 7) - 3x(x² - 7)
= (x² - 7)(x² - 3x + 2)
hence, next two zeros include in (x²-3x + 2)
so, x²- 3x + 2 = (x - 2)(x - 1)

hence, x = 2 and 1 are other two zeros of given polynomial.

Answer 2

Answer:

Given, +√7 and -√7 are the zeros of x⁴ - 3x³ - 5x² + 21x - 14

so,x⁴ - 3x³ - 5x² + 21x - 14 is divisible by {(x +√7)(x - √7)}.

e.g., x⁴ - 3x³ - 5x² + 21x - 14 is divisible by x² - 7

now, x⁴ - 7x² + 2x² - 14 - 3x³ + 21x

= x²(x² - 7) + 2(x² - 7) - 3x(x² - 7)

= (x² - 7)(x² - 3x + 2)

hence, next two zeros include in (x²-3x + 2)

so, x²- 3x + 2 = (x - 2)(x - 1)

hence, x = 2 and 1 are other two zeros of given polynomial.