Question

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

• A. -2, -3
• B. $\frac{3}{2},-5$
• C. $\sqrt{3},-\sqrt{3}$
• D. $\frac{5}{2},-3$

Answer 1

The correct option is B $\frac{3}{2},-5$

$p\left(x\right)=2x^4+7x^3-19x^2-14x+30$

Given that $\sqrt{2}\text{and}-\sqrt{2}$ are zeroes of p(x).

$\Rightarrow (x-\sqrt{2})(x+\sqrt{2})$ is a factor of p(x)

$\Rightarrow (x^2-2)$ is a factor of p(x)



$\therefore 2x^4+7x^3-19x^2-14x+30=(x^2-2)(2x^2+7x-15)$

$=(x^2-2)(2x^2+10x-3x-15)$

$=(x^2-2)\left[2x\right(x+5)-3(x+5\left)\right]$

$=(x^2-2)(x+5)(2x-3)$

So, for other zeroes: x + 5 = 0 and 2x – 3 = 0.

$\Rightarrow x=-5andx=\frac{3}{2}$

Hence, the correct answer is option (b).