Question

If two zeroes of the polynomial
X^4 - 9x^3 + 24x^2 – 24x + 8 are 3 +✓ 5 and 3-√5, then find
the other two zeroes.​

Answer 1

Answer: ∴ The zeroes are: $3 +2 \sqrt 2$ and $3 - \sqrt 2$

Step-by-step explanation:

Let the roots of $x^4 - 9x^3 + 24x^2 - 24x + 8$ be $\alpha , \beta, \gamma, \delta$

The roots, $3+ \sqrt{5}$ and $3- \sqrt 5$ are divisible by it (remainder theorem)

∴ Equation with roots as $3+ \sqrt{5}$ and $3- \sqrt 5$ can divide $x^4 - 9x^3 + 24x^2 - 24x + 8$

∴ $x^2 - (3+ \sqrt{5} + 3- \sqrt{5}) + (3+ \sqrt{5})*(3- \sqrt{5})$.... Theorem

∴ $x^2 - 6x + 4 = 0$ divides $x^4 - 9x^3 + 24x^2 - 24x + 8$

∴ $\frac{(x^4 - 9x^3 + 24x^2 - 24x + 8)}{x^2 - 6x + 4 } = 2x^2 - 12x + 2$

∴ $2x^2 - 12x + 2$ is the root of the equation: $x^4 - 9x^3 + 24x^2 - 24x + 8$

Now, the other two zeroes are the factors of $2x^2 - 12x + 2$

Using quadratic equation formula method:

$x = 3 +/- \sqrt 2$

∴ The zeroes are: $3 + \sqrt 2$ and $3 - \sqrt 2$

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