Question

Find the zeros of the equation $ 9{x^4} - 148{x^2} + 64 = 0. $

Answer 1

Hint: Zeros of a polynomial can be defined as the points where the polynomial becomes zero as a whole. A polynomial having value zero is called zero polynomial.
So, to find the zeros of the equation given in the question we can use the factorization method.

Complete step-by-step answer:
To find the zeros of the equation,
 $ 9{x^4} - 148{x^2} + 64 = 0. $
The equation can be written as,
 $ \Rightarrow 9{x^4} - 4{x^2} - 144{x^2} + 64 $
We separated the middle polynomial as the factors of equation,
Now, take common between both,
 $ \Rightarrow {x^2}\left( {9{x^2} - 4} \right) - 16\left( {9{x^2} - 4} \right) $
Now separate them,
 $ \Rightarrow \left( {{x^2} - 16} \right)\left( {9{x^2} - 4} \right) $
So,
  $
   \Rightarrow {x^2} = 16 \\
   \Rightarrow x = \pm 4 \;
  $
And,
 $
   \Rightarrow 9{x^2} = 4 \\
   \Rightarrow {x^2} = \dfrac{4}{9} \\
   \Rightarrow x = \pm \dfrac{2}{3} \;
  $
Hence, zeros of the equation are $ 4, - 4,\dfrac{2}{3}, - \dfrac{2}{3}. $
So, the correct answer is “ $ 4, - 4,\dfrac{2}{3}, - \dfrac{2}{3} $ ”.

Note: $ \Rightarrow $ $ '0' $ could be a zero of polynomials but it is not necessarily a zero that has to be $ '0' $ only.
 $ \Rightarrow $ All the linear polynomials have only one zero.
 $ \Rightarrow $ The zeros of the polynomial, depend on its degree.