Chapter 1.2, Problem 1CFU

To determine

$f\left(x\right)\text{ and }g\left(x\right)$ , and explain the process.

Answer to Problem 1CFU

The value of $f\left(x\right)\text{ and }g\left(x\right)$ will be one of these values, $2x+1\text{ and }x+6$ .

They are determined by factoring.

Explanation of Solution

Given information: $\left(f\cdot g\right)\left(x\right)=2x^2+11x-6$

Calculation:

Solve the function as follows:

  $\begin{array}{c}\left(f\cdot g\right)\left(x\right)=2x^2+11x-6\\ \left(f\left(x\right)\right)\cdot \left(g\left(x\right)\right)=2x^2+11x-6\end{array}$

Try to make the factors. The only one possibility is $2x\text{ time }x$ . So, it can will be something as shown below.

  $\left(f\left(x\right)\right)\cdot \left(g\left(x\right)\right)=\left(2x\pm ?\right)\left(x\pm ?\right)$

Next step is to think about the possible factors of $-6$ . The following 4 sets of possible factors are as follows:

  $\left(1,6\right),\left(-1,6\right),\left(2,-3\right),\text{ and }\left(-2,3\right)$

Multiply each factor by $2x,x$ and then multiply each by $x,2x$ as shown below.

Factor	Multiply by $\left(2x,x\right)$	Multiply by $\left(x,2x\right)$
$\left(1,6\right)$	$2x,-6x$	$x,-12$
$\left(-1,6\right)$	$-2x,6x$	$-x,12x$
$\left(2,-3\right)$	$4x,-3x$	$2x,-6x$
$\left(-2,3\right)$	$-4x,3$	$-2x,6x$

Now, add each of the 8 possible sets until you find $+11x$ .

Factor	Possible set	Add 8 to the set
$\left(1,6\right)$	$2x,-6x$	$-4x$
$\left(1,6\right)$	$x,-12$	$-11x$
$\left(-1,6\right)$	$-2x,6x$	$4x$
$\left(-1,6\right)$	$-x,12x$	$11x$
$\left(2,-3\right)$	$4x,-3x$	$x$
$\left(2,-3\right)$	$2x,-6x$	$-4x$
$\left(-2,3\right)$	$-4x,3$	$-x$
$\left(-2,3\right)$	$-2x,6x$	$4x$

Since, you need $2x\text{ times +6}$ , make sure they are not in the same set of parentheses.

  $\left(f\left(x\right)\right)\cdot \left(g\left(x\right)\right)=\left(2x-1\right)\left(x+6\right)$

Therefore the value of $f\left(x\right)$ and $g\left(x\right)$ is either $\left(2x-1\right)$ and $\left(x+6\right)$ respectively, or $\left(x+6\right)$ and $\left(2x-1\right)$