Algebra Toolkit

We can rationalize the denominator by multiplying mumerator and denominator by \(\blert{\sqrt{3}}\text{:}\)

or we can divide 3 by \(\sqrt{3}\) to get \(\sqrt{3}\text{.}\) (Remember that \(\sqrt{3}~\sqrt{3} = 3\text{.}\))

The LCD for the two fractions is 2, and the building factor for the first fraction is \(\blert{\sqrt{2}}\text{.}\)

We use the fact that \(~a^{-n} = \dfrac{1}{a^n}\text{,}\) and consequently that \(~\dfrac{1}{a^{-n}} = a^n\text{.}\)

1. \(\displaystyle \dfrac{x}{4y^2}\)
2. \(\displaystyle \dfrac{1}{a^{3}b^{2}}\)
3. \(\displaystyle \dfrac{2c^2}{ab}\)
4. \(\displaystyle xy^{2}+\dfrac{1}{x^2y}\)

1. Add the exponents: \(~3x^{-3}x^5 = 3x^2\)
2. Subtract the exponents: \(~\dfrac{4a^{-4}}{8a^{-8}} = \dfrac{1}{2}a^{-4-(-8)} = \dfrac{a^4}{2}\)

3. Raise each factor to the power \(-2\text{.}\) Multiply exponents:

   \begin{equation*} (2bc^{-3})^{-2} = 2^{-2}b^{-2}(c^{-3})^{-2} = \dfrac{c^6}{4b^2} \end{equation*}

4. We cannot add or subtract powers with different exponents.

   \begin{equation*} 3x^{-4}-2x^{-3} = \dfrac{3}{x^4} - \dfrac{2}{x^3} \end{equation*}

The quotient tells us that 15 divides into 536 thirty-five times, with a remainder of 11. This in turn means that if we multiply 15 by 35, and then add 11, we should get 536 back again.

Note the pattern: divisor \(\times\) quotient \(+\) remainder \(=\) starting number

The answer tells us that \(n+2\) divides into \(3n^2+n-6\) to give a quotient of \(3n-5\text{,}\) with a remainder of 4. If we multiply \(n+2\) by \(3n-5\text{,}\) and then add 4, we should get \(3n^2+n-6\) back again.