Question

$\text{If two zeroes of the polynomial}{x}^4–6x^3–26x^2+138x–35\text{are}2\pm \sqrt{3},\text{find the other zeroes.}$

• A. $-2-\sqrt{3},2+\sqrt{3}$
• B. 7 & -5
• C. -7 & 5
• D. $-2+\sqrt{3},-2-\sqrt{3}$

Answer 1

The correct option is B 7 & -5

$\text{Given that:}2+\sqrt{3}\&2-\sqrt{3}\text{are two}\text{zeros of f(x), where}f\left(x\right)=x^4–6x^3–26x^2+138x–35.$

$\text{Let}\alpha =2+\sqrt{3}\text{and}\beta =2-\sqrt{3}.$

$\text{Using two zeroes let us frame a}\text{quadratic equation.}{x}^2-(\alpha +\beta )x+\alpha \beta =x^2-(2+\sqrt{3}+2-\sqrt{3})x+(2+\sqrt{3})(2-\sqrt{3})=x^2-4x+(2^2-(\sqrt{3}{)}^2)=x^2-4x+4-3=x^2-4x+1$

$\text{So,}{x}^2-4x+1\text{is a factor of f(x)}.$

$\text{Let us now divide f(x) by}{x}^2-4x+1.$

$x^2-2x-35x^2-4x+1)\overline{x^4–6x^3–26x^2+138x–35}\underset{–}{{}_{-}{x}^4\mp 4x^3\pm x^2}\underset{–}{-2x^3-27x^2+138x\mp 2x^3\pm 8x^2\mp 2x}\underset{–}{-35x^2+140x-35\mp 35x^2\pm 140x\mp 35}00$

$f\left(x\right)=x^4–6x^3–26x^2+138x–35=(x^2-4x+1)(x^2-2x-35)$

$\text{Hence, other two zeroes of f(x) are the}\text{zeroes of the polynomial}{x}^2-2x-35.$

$x^2-2x-35=x^2-7x+5x-35=(x-7)(x+5)$
$\Rightarrow x-7=0;x+5=0$

$\text{Hence, other two zeroes of f(x)}\text{are 7 and -5.}$