Question

Solution of $(x-1{)}^2(x+4)<0$ is

• A. $\left(-\infty ,1\right)$
• B. $\left(-\infty ,-4\right)$
• C. $\left(-1,4\right)$
• D. $\left(1,4\right)$

Answer 1

Answer: (B)

$(x-1{)}^2$ is always positive except when $x=1$ (and then it is $0$)
$\therefore$ Solution is when $x+4<0$ and $x\ne 1$
i.e. $x<-4$, $x\ne 1\therefore x\in \left(-\infty ,-4\right)$.