Elementary Algebra

1. We first write the numerator and denominator in factored form. Then we divide any common factors from the numerator and denominator.

   \begin{equation*} \dfrac{a^5}{a^3} = \dfrac{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot a\cdot a}{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}} = \dfrac{a^2}{1} = a^2 \end{equation*}

   You may observe that the exponent of the quotient can be obtained by subtracting the exponent of the denominator from the exponent of the numerator. In other words,

   \begin{equation*} \dfrac{a^5}{a^3}= a^{5-3} = a^2 \end{equation*}

2. In this quotient, the larger power occurs in the denominator.

   \begin{equation*} \dfrac{a^4}{a^8} = \dfrac{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot \cancel{a}}{\cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot \cancel{a}\cdot a\cdot a\cdot a \cdot a} = \dfrac{1}{a^4} \end{equation*}

   We see that we can subtract the exponent of the numerator from the exponent of the denominator. That is,

   \begin{equation*} \dfrac{a^4}{a^8}=\dfrac{1} {a^{8-4}} = \dfrac{1}{a^4} \end{equation*}

We consider the numerical coefficients and the powers of each base separately. We use the second law of exponents to simplify each quotient of powers.

The largest integer that divides evenly into all three coefficients is 2. The highest power of \(a\) is \(a^1\text{,}\) and the highest power of \(b\) is \(b^2\text{.}\) Thus, the GCF is \(2ab^2\text{.}\) Note that the exponent on each variable of \(2ab^2\) is the smallest exponent that appears on that variable among the terms of the polynomial.

The GCF for this polynomial is \(2ab^2\text{,}\) as we saw in Example 7.22. We factor out the GCF from each term and write the polynomial as a product,

To find the factor inside parentheses, we divide each term of the polynomial by the GCF.

We apply the distributive law to factor \(2ab^2\) from each term.

The given expression has two terms, and \((2x+1)\) is a factor of each. We factor out the entire binomial \((2x+1)\text{.}\)

We begin by factoring the quadratic term, \(2x^2\text{,}\) which can only be factored as \(x\) times \(2x\text{,}\) so we can fill in the First terms in each binomial.

Next, we factor the constant term, \(3\text{,}\) which can only be factored as 3 times 1. Because the linear term is negative, we make both factors negative: \(3=-3(-1)\text{.}\) Finally, we have to decide which number appears as the Last term in each binomial. We check \(O+I\) for each possibility.

The second choice gives the correct middle term, so the factorization is

As usual, we begin by factoring the first term into \(x\) times \(x\text{.}\)

Next we look for factors of the last term, \(6y^2\text{.}\) In order to obtain the \(xy\)-term in the middle, we need a \(y\) in each factor. Thus the possibilities are

We'll check the sum \(O+I\) for each possibility.

The second possibility gives the correct middle term, so the factorization is

We begin by factoring out the greatest common factor, \(2b\text{.}\)

The remaining factor, \(b^2+4b+3\text{,}\) is a quadratic trinomial that can be factored. We look for two numbers \(p\) and \(q\) so that \(pq=3\) and \(p+q=4\text{.}\) You can check that \(p=3\) and \(q=1\) will work.

Thus, \(~~b^2+4b+3 = (b+3)(b+1)\text{,}\) and

1. We construct a rectangle with sides \(3t+2\) and \(t+3\text{,}\) as shown below. We see that the product of the two binomials is

   \(3t^2+9t+2t+6=3t^2+11t+6\)

   	\(~~~~~t~~~~~\)	\(~~~~3~~~~~\)
   \(3t\)	\(3t^2\)	\(9t\)
   \(2\)	\(2t\)	\(6\)

2. Now let's compute the product of the expressions along each diagonal of the rectangle:

   \begin{equation*} 3t^2 \cdot 6 = 18t^2~~~~~~\text{and}~~~~~~9t \cdot 2t = 18t^2 \end{equation*}

   The two products are equal. This is not surprising when you think about it, because each diagonal product is the product of all four terms of the binomials, namely \(3t,~2,~t,\) and \(3\text{,}\) just multiplied in a different order. You can see where the diagonal entries came from in our example:

   \begin{align*} 18t^2 \amp = 3t^2 \cdot 6 =3t \cdot t \cdot 2 \cdot 3\\ 18t^2 \amp = 9t \cdot 2t =3t \cdot 3 \cdot 2 \cdot t \end{align*}

\(\blert{\text{Step 2}~~}\) Compute the diagonal product:

\(\blert{\text{Step 3}~~}\) List all possible factors of \(D\text{,}\) and compute the sum of each pair of factors. (Note that both factors must be negative.)

\(\blert{\text{Step 5}~~}\) Factor \(2x\) from the top row of the rectangle, and write the result, \(x-3\text{,}\) at the top, as shown below.

Finally, factor \(5\) from the bottom row, and write it on the left. The correct factorization is