**Example 5: **Simplify the given expression by following the proper steps and write the type of the expression.

**a)** $\displaystyle {{(x+3)}^{2}}+2{{x}^{2}}+12-21$

**b)** $\displaystyle 3x-2+{{(2y-3)}^{2}}+{{({{x}^{2}})}^{{-2}}}$

**Solution**

**a)** Remove the parenthesis by expanding the “Square Formula”

$\displaystyle {{(x+3)}^{2}}+2{{x}^{2}}+12-21=$

$\displaystyle {{x}^{2}}+2\cdot 3\cdot x+{{(3)}^{2}}+2{{x}^{2}}+12-21=$

$ \displaystyle {{x}^{2}}+6x+9+2{{x}^{2}}+12-21=$

**Combine the like terms of the expression**

$ \displaystyle ({{x}^{2}}+2{{x}^{2}})+6x+(9+12-21)=$

$ \displaystyle 3{{x}^{2}}+6x+0=$

$\displaystyle 3{{x}^{2}}+6x$

**The type of the expression is Binomial Expression.**

**b) **Remove the parenthesis by expanding “Square Formula”

$\displaystyle 3x-2+{{(2y-3)}^{2}}+{{({{x}^{2}})}^{{-2}}}$

$\displaystyle 3x-2+{{(2y)}^{2}}-2\cdot 2y\cdot 3+{{(-3)}^{2}}+{{({{x}^{2}})}^{{-2}}}=$

$\displaystyle 3x-2+4{{y}^{2}}-12y+9+{{({{x}^{2}})}^{{-2}}}=$

**Secondly **we use the exponent rule by multiplying the exponents.

$\displaystyle 3x-2+4{{y}^{2}}-12y+9+{{x}^{{-4}}}=$

$\displaystyle 3x-2+4{{y}^{2}}-12y+9+\frac{1}{{{{x}^{4}}}}=$

**Combine the like terms of the expression**

$\displaystyle 3x+4{{y}^{2}}-12y+\frac{1}{{{{x}^{4}}}}+(9-2)=$

$ \displaystyle 3x+4{{y}^{2}}-12y+\frac{1}{{{{x}^{4}}}}+7=$

**The type of the expression is a Polynomial Expression.**

See also: **Expanding and Factorising Algebraic Expressions**