Question

If two zeros of the polynomial p(x) = $2x^4+7x^3-19x^2-14x+30\mathrm{are}\sqrt{2}\mathrm{and}-\sqrt{2},$ then find the other two zeroes.

Answer 1

The given polynomial is f(x) = 2x⁴ + 7x³ - 19x² - 14x + 30
Since, $\sqrt{2}\mathrm{and}-\sqrt{2}$ are are two zeros of f(x), it follows that each one of $\left(x-\sqrt{2}\right)\mathrm{and}\left(\mathrm{x}+\sqrt{2}\right)$ is a factor of f(x).
Consequently, $\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=\left(x^2-2\right)$ is a factor of f(x).
On dividing f(x) by (x² - 2), we get:


∴ f(x) = 0
⇒ (x² - 2)(2x² + 7x - 15)=0
⇒ (x² - 2)(2x² + 7x - 15) = 0
⇒ (x² - 2)(2x - 3)(x + 5) = 0
⇒ $\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(2x-3\right)\left(x+5\right)=0$
⇒ $\left(\mathrm{x}-\sqrt{2}\right)=0,\mathrm{or}\left(\mathrm{x}+\sqrt{2}\right)=0,\mathrm{or}\left(2\mathrm{x}-3\right)=0,\mathrm{or}\left(\mathrm{x}+5\right)=0$
⇒ $x=\sqrt{2}\mathrm{or}\mathrm{x}=-\sqrt{2},\mathrm{or}\mathrm{x}=\frac{3}{2},\mathrm{or}\mathrm{x}=-5$
Hence, the other two zeros are $\frac{3}{2}\mathrm{and}-5$.