Exponentiation is an operation of assigning an exponent to a base. Sometimes a polynomial can be assigned an exponent. Two different situations are possible.
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If the polynomial consists only of multiplications and divisions, the properties of exponents can be used to reduce the algebraic expression.
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If the polynomial contains additions or subtractions, it will have to be expanded to reduce it.
Be careful with the exponentiation in polynomials. In an algebraic expression, the properties of exponents can only be used if the expression consists of multiplications and divisions.
Two properties of exponents that can be used on polynomials are:
||\begin{align}(a^m)^n&=a^{mn}\\ \\
\left(\displaystyle \frac{a^m}{b^n}\right)^c&=\frac{a^{mc}}{b^{nc}}\end{align}||
Situation where there is only multiplication and division
Consider the following algebraic expression: ||\left(\displaystyle \frac{3^2a^3f^2}{27ac^2f}\right)^3|| Before including the exponent inside the brackets, we can simplify – if possible – inside the brackets. ||\begin{align} \left(\displaystyle \frac{3^2a^3f^2}{27ac^2f}\right)^3=\left(\displaystyle \frac{3^2a^3f^2}{3^3ac^2f}\right)^3&\Rightarrow (3^{2-3}a^{3-1}c^{0-2}f^{2-1})^3\\ \\
&=(3^{-1}a^2c^{-2}f^1)^3\end{align}|| We apply one of the properties of the exponents to distribute the exponent. ||(3^{-1}a^2c^{-2}f^1)^3\quad \Rightarrow \quad \left(3^{(-1\times 3)}a^{(2\times 3)}c^{(-2\times 3)}f^{(1\times 3)}\right)\quad =\quad (3^{-3}a^6c^{-6}f^3)|| We write the final answer with positive exponents. ||(3^{-3}a^6c^{-6}f^3)=\displaystyle \frac{a^6f^3}{3^3c^6}||
Situations where there is addition and subtraction
Consider the following algebraic expression: ||(a-3)^3|| We cannot apply the exponent property stated above, because there is a subtraction in the expression.
Cubing a binomial is like multiplying the binomial by itself three times: ||(a-3)(a-3)(a-3)||
We start by multiplying the first two binomials together. ||\begin{align}&\color{white}{=}\color{red}{(a-3)(a-3)}(a-3)\\
&=\color{red}{(a^2-3a-3a+9)}(a-3)\\
&=\color{red}{(a^2-6a+9)}(a-3)\end{align}||Next, we multiply the trinomial obtained previously with the last binomial. ||\begin{align}&\color{white}{=}(a^2-6a+9)(a-3)\\ &=a^3-6a^2+9a-3a^2+18a-27\\ &=a^3-9a^2+27a-27\end{align}|| The answer is: |a^3-9a^2+27a-27| .
Special case of squaring a binomial
Consider the following algebraic expression: ||(3x+4)^2|| We cannot apply the exponent property stated above, because there is an addition in the expression, as follows: ||(3x+4)^2\neq 3^2x^2+4^2|| Squaring the binomial amounts to a multiplication of two identical binomials. The result is the following expression: ||\begin{align} (3x+4)^2 &= (3x+4)(3x+4) \\ &= 9x^2+12x+12x+16 \\ &= 9x^2+24x+16 \end{align}||
By simplifying the square of a binomial, we always obtain a trinomial formed with the following terms:
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The first term is the square of the binomial’s first term.
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The second term is double the product of the binomial’s two terms.
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The third term is the square of the binomial’s second term.
In the example, the first term of the answer |(9x^2)| is the square of the binomial’s first term |(3x)| . The second term of the answer |(24x)| is double the product of the binomial’s two terms. Thus, |2(3x)(4)=24x.| The third term of the answer |(16)| is the square of the binomial’s second term |(4).| ||\begin{align} (\color{blue}{3x} + \color{green}{4})^2 &= (\color{blue}{3x})^2 + 2(\color{blue}{3x})(\color{green}{4}) + (\color{green}{4})^2 \\ &= 9x^2+24x+16 \end{align}||