Chapter 8.8, Problem 40PPE

To determine

the two different values that the trinomial can be factored into the product of two binomials

Answer to Problem 40PPE

The two different values of the trinomial form is 48,-7

The trinomial $8r^2+\text{□}r-42$ can be factored into the product of two binomials that are $\left(8r-7\right)\left(r+6\right)$

Explanation of Solution

Given info:

The trinomial form is given as $8r^2+\text{□}r-42$

From the above expression $a=8,b=?,c=-42$

Formula used:

Show the general form of trinomial as follows:

  $a{x}^2+bx+c$

Here, $a,b,c$ are integers, $b$ is the value which depends on the multiplies $\left(x,y\right)$ of the product of $a$ and $c$ .

Condition of $b$ value depends on $c$ value. If the value $c$ is negative, then $b=x-y$ . If the value $c$ is positive, then $b=x+y$ .

Calculation:

Finding the $b$ value as follows:

The product of $a$ and $c$ is $a\times c=\left(8\times -42\right)=-336$

Choose two multiplier of -336 which suits for the $b$ value condition.

Take $x=48,y=-7$ and substitute in trinomial form to find $b$ value as follows:

If the value $c$ is negative, then $b=x-y$ .If $c=-42$ (negative), then $b=48-7=41$ .

Take $b=41$ in trinomial form

  $8r^2+\text{□}r-42$

  $8r^2+48r-7r-42$

  $\left(8r^2+48r\right)-\left(7r+42\right)$

  $\begin{array}{l}8r\left(r+6\right)-7\left(r+6\right)\\ \left(r+6\right)\left(8r-7\right)\\ \left(8r-7\right)\left(r+6\right)\end{array}$

Hence, the two different values of the trinomial form is 48,-7 and the trinomial $8r^2+\text{□}r-42$ can be factored into the product of two binomials that are $\left(8r-7\right)\left(r+6\right)$ .

Conclusion:

Thus, the two different values of the trinomial form is 48,-7 and the trinomial $8r^2+\text{□}r-42$ can be factored into the product of two binomials that are $\left(8r-7\right)\left(r+6\right)$ .