Let
Since 1 and
Consequently,
On dividing
f(x)=0 \Rightarrow\left(x^{2}+x-2\right)(x-5)=0 \\
\Rightarrow(x-1)(x+2)(x-5)=0 \\
\Rightarrow x=1 \text { or } x=-2 \text { or } x=5
Hence, the third zero is 5 .
Let
Since 1 and
Consequently,
On dividing
Hence, the third zero is 5 .