Question

p(x)=2x⁴ − 3x³ − 5x² + 9x − 3 and two of its zeros are $\sqrt{3}$ and $-\sqrt{3}$. Find the other two zeros.

Answer 1

The given polynomial is f(x) = 2x⁴ - 3x³ - 5x² + 9x - 3.
Since, $\sqrt{3}\mathrm{and}-\sqrt{3}$ are are two zeros of f(x), it follows that each one of $\left(x-\sqrt{3}\right)\mathrm{and}\left(\mathrm{x}+\sqrt{3}\right)$ is a factor of f(x).
Consequently, $\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=\left(x^2-3\right)$ is a factor of f(x).
On dividing f(x) by (x² - 3), we get:


∴ f(x) = 0 ⇒ (x² - 3)(2x² - 3x + 1)
⇒ (x² - 3)(2x² - 3x + 1) = 0
⇒ (x² - 3)(2x - 1)(x - 1) = 0
⇒ $\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(2x-1\right)\left(x-1\right)=0$
⇒ $\left(\mathrm{x}-\sqrt{3}\right)=0,\mathrm{or}\left(\mathrm{x}+\sqrt{3}\right)=0,\mathrm{or}\left(2\mathrm{x}-1\right)=0,\mathrm{or}\left(\mathrm{x}-1\right)=0$
⇒ $x=\sqrt{3}\mathrm{or}\mathrm{x}=-\sqrt{3},\mathrm{or}\mathrm{x}=\frac{1}{2},\mathrm{or}\mathrm{x}=1$
Hence, the other two zeros are $\frac{1}{2}\mathrm{and}1$.