Example 7.35 from Elementary Algebra, 7.3 Factor Quadratic Trinomials with Leading Coefficient Other than 1

Factor Quadratic Trinomials with Leading Coefficient Other than 1: example 7.35

Factor completely: $14 x^{2} - 47 x - 7$ .

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution

The trinomial is already in descending order.
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since it is negative, one factor must be positive and one negative.

Consider all the combinations of factors. We use each pair of the factors of $14 x^{2}$ with each pair of factors of $−7 .$

Factors of $14 x^{2}$	Pair with	Factors of $−7$
$x$ , $14 x$		$1$ , $−7$ $−7$ , $1$ (reverse order)
$x$ , $14 x$		$−1$ , $7$ $7$ , $−1$ (reverse order)
$2 x, 7 x$		$1$ , $−7$ $−7$ , $1$ (reverse order)
$2 x, 7 x$		$−1$ , $7$ $7$ , $−1$ (reverse order)

These pairings lead to the following eight combinations.

This table has the heading 14 x ^ 2 – 47 x minus 7. This table has two columns. The first column is labeled “possible factors” and the second column is labeled “product”. The first column lists all the combinations of possible factors and the second column has the products. In the first row under “possible factors” it reads (x+1) and (14 x minus 7). Under product, in the next column, it says “not an option”. In the next row down, it shows (x minus 7) and (14 x plus 1). In the next row down, it shows (x minus 1) and (14 x plus 7). Next to this in the product column, it says “not an option.” The next row down under “possible factors”, it has the equation (x plus 7 and 14 x minus 1. Next to this in the product column it has 14 x ^2 plus 97 x minus 7. The next row down under possible factors, it has 2 x plus 1 and 7 x minus 7. Next to this under the product column, is says “not an option”. The next row down reads 2 x minus 7 and 7x plus 1. Next to this under the product column, it has 14 x ^2 minus 47 x minus 7 with the asterisk following the 7. The next row down reads 2 x minus 1 and 7 x plus 7. Next to this in the product column it reads “not an option”. The final row reads 2 x plus 7 and 7 x minus 1. Next to this in the product column it reads 14, x, ^ 2 plus 47 x minus 7. Next to the table is a box with four arrows point to each “not an option” row. The reason given in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor. That means that each of these combinations is not an option.”

$\begin{matrix} The correct factors are those whose product is the & (2 x - 7) (7 x + 1) \\ original trinomial. \\ Check by multiplying. \\ (2 x - 7) (7 x + 1) \\ 14 x^{2} + 2 x - 49 x - 7 \\ 14 x^{2} - 47 x - 7 ✓ \end{matrix}$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Factor completely: $8 a^{2} - 3 a - 5$ .

How did it go?