Question

Find the zeros of the polynomials:
(i) 2x – 3
(ii) ${{x}^{4}}-16$

Answer 1

Hint: The given problem is related to zeros of a polynomial. Equate the expressions to zero and solve for x.

Complete step-by-step answer:

A zero of a polynomial is that number, which when put in place of x, makes the polynomial equal to zero. When the curve of the polynomial is plotted on a cartesian plane, the curve intersects the x-axis at the zeroes.
Let f(x) be a polynomial.
If f(a) = 0, then a is a zero of f(x).
Using the above concept, we find the zeros of the given polynomials.
(i) 2x – 3
Let us assume the zero of the polynomial to be a.
Then,
\[\begin{align}
  & 2a-3=0 \\
 & 2a=3 \\
 & a=\dfrac{3}{2} \\
\end{align}\]
Therefore, \[\dfrac{3}{2}\] is the zero of the polynomial 2x – 3.
(ii) ${{x}^{4}}-16$
Let us assume the zeros of the polynomial to be a.
Then,
$\begin{align}
  & {{a}^{4}}-16=0 \\
 & \Rightarrow {{\left( {{a}^{2}} \right)}^{2}}-{{4}^{2}}=0 \\
 & \Rightarrow \left( {{a}^{2}}+4 \right)\left( {{a}^{2}}-4 \right)=0\ \ \ \ \ \ \ \ \ \ \ \left[ {{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right) \right] \\
 & \Rightarrow \left( {{a}^{2}}+4 \right)\left( {{a}^{2}}-{{2}^{2}} \right)=0 \\
 & \Rightarrow \left( {{a}^{2}}+4 \right)\left( a+2 \right)\left( a-2 \right)=0\ \ \ \left[ \text{Since }{{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right) \right] \\
\end{align}$
We have now factorized the given polynomial. We now solve for a.
$\begin{align}
  & \left( {{a}^{2}}+4 \right)\left( a+2 \right)\left( a-2 \right)=0 \\
 & \Rightarrow \left( a+2 \right)\left( a-2 \right)=0\ \ \ \ \ \ \ \ \ \ \ \left[ As\ {{a}^{2}}+4\ge 4\forall a\in R \right] \\
 & \therefore a=2\ or\ -2. \\
\end{align}$
2 and – 2 are the zeroes of the given polynomial.

Note: While factorising, make sure to use correct identities. Also, while doing calculations, take care of the signs, as sign mistakes are very common and can lead to wrong answers. The identities used in the problem must be remembered perfectly as confusion in the identities can lead to wrong answers.