Example 7.34 from Elementary Algebra, 7.3 Factor Quadratic Trinomials with Leading Coefficient Other than 1

Factor Quadratic Trinomials with Leading Coefficient Other than 1: example 7.34

Factor completely: $6 b^{2} - 13 b + 5$ .

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 the last term, 5 is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.

Consider all the combinations of factors.

$6 b^{2} - 13 b + 5$
Possible factors	Product
$(b - 1) (6 b - 5)$	$6 b^{2} - 11 b + 5$
$(b - 5) (6 b - 1)$	$6 b^{2} - 31 b + 5$
$(2 b - 1) (3 b - 5)$	$6 b^{2} - 13 b + 5$ *
$(2 b - 5) (3 b - 1)$	$6 b^{2} - 17 b + 5$

$\begin{matrix} The correct factors are those whose product \\ is the original trinomial. & (2 b - 1) (3 b - 5) \\ Check by multiplying. \\ (2 b - 1) (3 b - 5) \\ 6 b^{2} - 10 b - 3 b + 5 \\ 6 b^{2} - 13 b + 5 ✓ \end{matrix}$

Your turn

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

Factor completely: $8 x^{2} - 13 x + 3$ .

How did it go?