Exercise 13 Page 673

Here we have a quadratic trinomial of the form $a{x}^2+bx+c,$ where $|a|\ne 1$ and there are no common factors. To factor this expression, we will rewrite the middle term $bx$ as two terms. The coefficients of these two terms will be factors of $ac$ whose sum must be $b.$ Let's first highlight the coefficients. We know that $a=6,$ $b=\text{-}5,$ and $c=\text{-}4.$ There are now three steps we need to follow in order to rewrite the above expression.

1. Find $\mathbf{a}\mathbf{c}.$ Since we have that $a=6$ and $c=\text{-}4,$ the value of $ac$ is $6(\text{-}4)=\text{-}24.$
2. Find factors of $\mathbf{a}\mathbf{c}.$ Since $ac=\text{-}24,$ which is negative, we need factors of $ac$ to have opposite signs. Since $b=\text{-}5$ is also negative, the absolute value of the negative factor will need to be greater than the absolute value of the positive factor.

$$\begin{array}{cccc}\underline{1^{\text{st}}\text{ Factor}}& \underline{2^{\text{nd}}\text{ Factor}}& \underline{\text{Sum}}& \underline{\text{Result}}\\ 1& \text{-}24& 1+(\text{-}24)& \text{-}23\\ 2& \text{-}12& 2+(\text{-}12)& \text{-}10\\ 3& \text{-}8& 3+(\text{-}8)& \text{-}5\\ 4& \text{-}6& 4+(\text{-}6)& \text{-}2\end{array}$$

3. Rewrite $\mathbf{b}\mathbf{x}$ as two terms. Now that we know which factors are the ones to be used, we can rewrite $bx$ as two terms.
   $$\begin{array}{c}6x^2+(\text{-}5)x-4\\ \Updownarrow \\ 6x^2+3x-8x-4\end{array}$$

Finally, we will factor the last expression obtained. To do that, we first need to change the order of the terms so that the ones with a common factor are next to each other.

$6x^2+3x-8x-4$

CommutativePropAdd

Commutative Property of Addition

$6x^2-8x+3x-4$

FactorOut

Factor out $2x$

$2x(3x-4)+3x-4$

AddPar

Add parentheses

$2x(3x-4)+(3x-4)$

FactorOut

Factor out $(3x-4)$

$(2x+1)(3x-4)$