Question

$\text{If 2 is one of the zeros of}{x}^3-3x^2-4x-12,\text{then find the remaining}\text{two zeroes.}$

• A. $\text{-3 and 2}$
• B. $\text{-3 and -2}$
• C. $\text{3 and 2}$
• D. $\text{3 and -2}$

Answer 1

The correct option is A $\text{-3 and 2}$

$\text{Given,}p\left(x\right)=x^3-3x^2-4x-12\text{and}p\left(2\right)=0$

$x=2\Rightarrow x-2=0\therefore x-2\text{is the factor of p(x).}$

$\text{So, if}{x}^3-3x^2-4x-12\text{is divided by x - 2 the}\text{remainder will be zero.}$

$\text{By long division,}x-2\stackrel{x^2-x-6}{)\overline{x^3-3x^2-4x-12}}\underset{–}{{}_{-}{x}^3\mp 2x^2}\underset{–}{-x^2-4x\mp x^2\pm 2x}\underset{–}{-6x-12\mp 6x\mp 12}00$


$\therefore x^3-3x^2-4x-12=(x-2)(x^2-x-6)$

$\text{On factorising}{x}^2-x-6=x^2+3x-2x-6=x(x+3)-2(x+3)=(x+3)(x-2)$

$x+3=0;x-2=0\therefore x=-3;2$

$\text{So, the remaining two Zeroes are}\text{-3 and 2.}$