Question

Solve the polynomial inequality. Express the solution set in in interval notation: $\frac{(x+7)(x-3)}{x-1}\ge 0$

• A. $(-\infty ,-7]\cup [3,\infty )$
• B. $(-\infty ,-7]\cup [1,3)$
• C. $(-\infty ,1)\cup (3,\infty )$
• D. $(-\infty ,-1)\cup [3,\infty )$

Answer 1

The correct option is C $(-\infty ,1)\cup (3,\infty )$

Given, $\frac{(x+7)(x-3)}{x-1}\ge 0$

$\therefore x-1<0$ and $(x+7)(x-3)\ge 0$

$\therefore x<1$ and $x\ge -7,x\ge 3$

Hence, Answer is $(-\infty ,1)\cup (3,\infty )$