Chapter 8.2, Problem 51PPS

(a)

To determine

To calculate: The factors whose product is $-6$ and their sum is 1.

Answer to Problem 51PPS

The required factors are $-2,3$ .

Explanation of Solution

Given information:

The product of the factors is $-6$ and their sum is 1.

Calculation:

Given that the product of the factors is $-6$ and their sum is 1.

Let the factors be $x\text{and}y$ .

Now,

  $xy=-6$   .... (1)

  $x+y=1$   .... (2)

Solve the equations for $x\text{and}y$ .

In equation (1), write $x$ in terms of $y$ .

  $y=\frac{-6}{x}$   .... (3)

Substitute this value in equation (2).

  $\begin{array}{c}x-\frac{6}{x}=1\\ x^2-6=x\\ x^2-x-6=0\end{array}$

From the above equation the coefficient are $1,-1\text{and}-6$ .

Split the coefficient of $x$ into two parts such that their product is equal to the product of $1\text{and}-6$ . Thus,

  $\begin{array}{c}{x}^2-x-6=0\\ x^2-3x+2x-6=0\\ x\left(x-3\right)+2\left(x-3\right)=0\\ \left(x+2\right)\left(x-3\right)=0\end{array}$

Thus, $x=-2,3$

Substitute $x=-2$ in equation (3).

  $\begin{array}{c}y=\frac{-6}{-2}\\ =3\end{array}$

Hence, the required factors are $-2,3$ .

(b)

To determine

To fill: the factors in each empty square in the box

Answer to Problem 51PPS

The required box is

Explanation of Solution

Given information:

The given factors are $-2,3$ .

From 51a. the factors are $-2,3$ .

Find the equation.

  $\left(x+2\right)\left(x-3\right)=0$

Simplify the above equation.

  $\begin{array}{c}\left(x+2\right)\left(x-3\right)=0\\ x\left(x-3\right)+2\left(x-3\right)=0\\ x^2-3x+2x-6=0\end{array}$

Thus, the empty square box are filled with $-3x\text{and}2x$ .

Hence, the required factors are

(c)

To determine

To calculate: The factors for each row and column of the box.

Answer to Problem 51PPS

The required box is

Explanation of Solution

Given information:

The given equation is $x^2+x-6=0$ .

Calculation:

Consider the given equation $x^2+x-6=0$ .

In the above equation the coefficient are $1,1\text{and}-6$ .

Split the coefficient of $x$ into two parts such that their product is equal to the product of $1\text{and}-6$ . Thus,

  $\begin{array}{c}{x}^2+x-6=0\\ x^2+3x-2x-6=0\\ x\left(x+3\right)-2\left(x+3\right)=0\\ \left(x-2\right)\left(x+3\right)=0\end{array}$

Thus, $x=-3,2$

So the factors are $x^2,-3x,2x\text{and}-6$ .

Hence, the factors are

(d)

To determine

To calculate: The factors of the equation $x^2-3x-40=0$ .

Answer to Problem 51PPS

The required box is

Explanation of Solution

Given information:

The given equation is $x^2-3x-40=0$ .

Calculation:

Consider the given equation $x^2-3x-40=0$ .

In the above equation the coefficient are $1,-3\text{and}-40$ .

Split the coefficient of $x$ into two parts such that their product is equal to the product of $1\text{and}-40$ . Thus,

  $\begin{array}{c}{x}^2-3x-40=0\\ x^2+5x-8x-40=0\\ x\left(x+5\right)-8\left(x+5\right)=0\\ \left(x-8\right)\left(x+5\right)=0\end{array}$

Thus, $x=-5,8$

So the factors are $x^2,5x,-8x\text{and}-40$ .

Hence, the factors are