Chapter 11.3, Problem 50PPE

a.

To determine

To divide the expressions $\left(d^2-d+1\right)÷\left(d+1\right)$ .

Answer to Problem 50PPE

  $d-2+\frac{3}{d+1}$

Explanation of Solution

Given information:

The expression $\left(d^2-d+1\right)÷\left(d+1\right)$

Concept used:

Exponent properties: If a , b are real numbers and m , n are whole numbers, then

• Product Property: $a^m\cdot a^n=a^{m+n}$
• Quotient property: $\frac{a^m}{a^n}=a^{m-n}$
• Power Property: ${\left(a^m\right)}^n=a^{m\cdot n}$
• Product to a Power Property: ${\left(ab\right)}^m=a^m{b}^m$

Calculation:

In order to divide the given expression, use long division as shown below,

  $d+1\begin{array}{c}\hfill d-2\\ \hfill \overline{)\begin{array}{l}{d}^2-d+1\\ \underset{\_}{-\left(d^2+d\right)}\\ -2d+1\\ \underset{\_}{-\left(-2d-2\right)}\\ 3\end{array}}\end{array}$

Thus, on dividing it gives $d-2+\frac{3}{d+1}$ .

b.

To determine

To divide the expressions $\left(d^3-d^2+d-1\right)÷\left(d+1\right)$ .

Answer to Problem 50PPE

  $d^2-2d+3-\frac{4}{d+1}$

Explanation of Solution

Given information:

The expression $\left(d^3-d^2+d-1\right)÷\left(d+1\right)$

Concept used:

Exponent properties: If a , b are real numbers and m , n are whole numbers, then

Product Property: $a^m\cdot a^n=a^{m+n}$

Quotient property: $\frac{a^m}{a^n}=a^{m-n}$

Power Property: ${\left(a^m\right)}^n=a^{m\cdot n}$

Product to a Power Property: ${\left(ab\right)}^m=a^m{b}^m$

Calculation:

In order to divide the given expression, use long division as shown below,

  $d+1\begin{array}{c}\hfill d^2-2d+3\\ \hfill \overline{)\begin{array}{l}{d}^3-d^2+d-1\\ \underset{\_}{-\left(d^3+d^2\right)}\\ -2d^2+d\\ \underset{\_}{-\left(-2d^2-2d\right)}\\ 3d-1\\ \underset{\_}{-\left(3d+3\right)}\\ -4\end{array}}\end{array}$

Thus, on dividing it gives $d^2-2d+3-\frac{4}{d+1}$ .

c.

To determine

To divide the expressions $\left(d^4-d^3+d^2-d+1\right)÷\left(d+1\right)$ .

Answer to Problem 50PPE

  $d^3-2d^2+3d-4+\frac{5}{d+1}$

Explanation of Solution

Given information:

The expression $\left(d^4-d^3+d^2-d+1\right)÷\left(d+1\right)$

Concept used:

Exponent properties: If a , b are real numbers and m , n are whole numbers, then

Product Property: $a^m\cdot a^n=a^{m+n}$

Quotient property: $\frac{a^m}{a^n}=a^{m-n}$

Power Property: ${\left(a^m\right)}^n=a^{m\cdot n}$

Product to a Power Property: ${\left(ab\right)}^m=a^m{b}^m$

Calculation:

In order to divide the given expression, use long division as shown below,

  $d+1\begin{array}{c}\hfill d^3-2d^2+3d-4\\ \hfill \overline{)\begin{array}{l}{d}^4-d^3+d^2-d+1\\ \underset{\_}{-\left(d^4+d^3\right)}\\ -2d^3+d^2\\ \underset{\_}{-\left(-2d^3-2d^2\right)}\\ 3d^2-d\\ \underset{\_}{-\left(3d^2+3d\right)}\\ -4d+1\\ \underset{\_}{-\left(-4d-4\right)}\\ 5\end{array}}\end{array}$

Thus, on dividing it gives $d^3-2d^2+3d-4+\frac{5}{d+1}$ .

d.

To determine

To divide the expressions $\left(d^5-d^4+d^3-d^2+d-1\right)÷\left(d+1\right)$ .

Answer to Problem 50PPE

  $d^4-2d^3+3d^2-4d+5-\frac{6}{d+1}$

Explanation of Solution

Given information:

The expression $\left(d^5-d^4+d^3-d^2+d-1\right)÷\left(d+1\right)$

Concept used:

Exponent properties: If a , b are real numbers and m , n are whole numbers, then

Product Property: $a^m\cdot a^n=a^{m+n}$

Quotient property: $\frac{a^m}{a^n}=a^{m-n}$

Power Property: ${\left(a^m\right)}^n=a^{m\cdot n}$

Product to a Power Property: ${\left(ab\right)}^m=a^m{b}^m$

Calculation:

Following the pattern of quotient from pervious subparts, the solution seems to be,

  $\left(d^5-d^4+d^3-d^2+d-1\right)÷\left(d+1\right)=d^4-2d^3+3d^2-4d+5-\frac{6}{d+1}$

Thus, on dividing it gives $d^4-2d^3+3d^2-4d+5-\frac{6}{d+1}$ .

e.

To determine

To verify the prediction by dividing the polynomials.

Explanation of Solution

The quotient of the division of polynomials is 1 degree less than the degree of dividend. For example any nth degree dividend, the quotient would be

  $d^{n-1}-2d^{n-2}+3d^{n-3}+4d^{n-4}-\mathrm{...}+{\left(-1\right)}^{n+1}n+\frac{{\left(-1\right)}^n\left(n+1\right)}{d+1}$ .