Question

Find the zero of the polynomial:
(i) p(x) = x − 5
(ii) q(x) = x + 4
(iii) r(x) = 2x + 5
(iv) f(x) = 3x + 1
(v) g(x) = 5 − 4x
(vi) h(x) = 6x − 2
(vii) p(x) = ax, a ≠ 0
(viii) q(x) = 4x

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)p\left(x\right)=0\Rightarrow x-5=0 \\ \Rightarrow x=5 \\ \mathrm{Hence},5\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}p\left(x\right). \\ \left(\mathrm{ii}\right)q\left(x\right)=0\Rightarrow x+4=0 \\ \Rightarrow x=-4 \\ \mathrm{Hence},-4\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}q\left(x\right). \\ \left(\mathrm{iii}\right)r\left(x\right)=0\Rightarrow 2x+5=0 \\ \Rightarrow t=\frac{-5}{2} \\ \mathrm{Hence},\frac{-5}{2}\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}p\left(t\right). \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)f\left(x\right)=0\Rightarrow 3x+1=0 \\ \Rightarrow x=-\frac{1}{3} \\ \mathrm{Hence},-\frac{1}{3}\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}f\left(x\right). \\ \left(\mathrm{v}\right)g\left(x\right)=0\Rightarrow 5-4x=0 \\ \Rightarrow x=\frac{5}{4} \\ \mathrm{Hence},\frac{5}{4}\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}g\left(x\right). \\ \left(\mathrm{vi}\right)h\left(x\right)=0\Rightarrow 6x-2=0 \\ \Rightarrow x=\frac{2}{6}=\frac{1}{3} \\ \mathrm{Hence},\frac{1}{3}\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}h\left(x\right). \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vii}\right)p\left(x\right)=0\Rightarrow ax=0 \\ \Rightarrow x=0 \\ \mathrm{Hence},0\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}p\left(x\right). \\ \left(\mathrm{viii}\right)q\left(x\right)=0\Rightarrow 4x=0 \\ \Rightarrow x=0 \\ \mathrm{Hence},0\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{polynomial}q\left(x\right). \end{gathered}$$