Question

Use suitable identities to find the following products.
$\left( {{y}^{2}}+\dfrac{3}{2} \right)\left( {{y}^{2}}-\dfrac{3}{2} \right)$.

Answer 1

Hint: We equate the equation $\left( {{y}^{2}}+\dfrac{3}{2} \right)\left( {{y}^{2}}-\dfrac{3}{2} \right)$ with the identity formula of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. We then form a square for the left side of the new equation. Then we take square values of the terms. From that we form the simplification of $\left( {{y}^{2}}+\dfrac{3}{2} \right)\left( {{y}^{2}}-\dfrac{3}{2} \right)$.

Complete step-by-step answer:
We need to find the simplified form of the given equation
$\left( {{y}^{2}}+\dfrac{3}{2} \right)\left( {{y}^{2}}-\dfrac{3}{2} \right)$.
We equate with ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
For the simplification of the given quadratic polynomials
$\left( {{y}^{2}}+\dfrac{3}{2} \right)\left( {{y}^{2}}-\dfrac{3}{2} \right)$,
we apply the factorisation identity of difference of two squares as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We put the value of \[a={{y}^{2}};b=\dfrac{3}{2}\].
Simplification of the polynomial gives us
\[\left( {{y}^{2}}+\dfrac{3}{2} \right)\left( {{y}^{2}}-\dfrac{3}{2} \right)={{\left( {{y}^{2}} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}\].
Simplified form gives
\[\left( {{y}^{2}}+\dfrac{3}{2} \right)\left( {{y}^{2}}-\dfrac{3}{2} \right)={{y}^{4}}-\dfrac{9}{4}\].
Therefore, the final simplification of
$\left( {{y}^{2}}+\dfrac{3}{2} \right)\left( {{y}^{2}}-\dfrac{3}{2} \right)$ is \[{{y}^{4}}-\dfrac{9}{4}\].
So, the correct answer is “ \[{{y}^{4}}-\dfrac{9}{4}\]”.

Note: The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have two roots. Cubic polynomials have three. It can be both real and imaginary roots.
In the given polynomial the values will be dependent on the second variable.