Polynomials and Factoring Word Problems
The word problems presented in this workbook will help you understand how Mathematics relates to the real world. As you explore the problems presented in the book, try to make connections between Mathematics and the world around you! [color=#1551b5]Reference book: [color=#198f88] Larson R., Boswell L., Kanold T. D., & Stiff L. (2001) [i]Algebra 1[/i] McDougal Little Inc.[/color][/color] [color=#d69210]Images courtesy: Creative Commons[/color]
Table of Contents
- Using Polynomials in Real Life
- Subtracting Polynomials
- Polynomials: Word Problem 1
- Multiplying Binomials
- Polynomials: Word Problem 2
- Special products of Polynomials
- Modeling a Punnett Square
- Punnett square Word Problem 1
- Punnett square Word Problem 2
- Plant Genetics
- Solving Quadratic Equation by Factoring
- Writing a Quadratic Model
- The Taj Mahal
- Vertical Motion models
- Using a Falling Object Model
- Falling Object Word Problem 1
- Using Quadratic Models in Real Life
- Writing a Quadratic Model
- Vertical Motion Model: Word Problems
- Factoring Special Products
- Writing and Using a Quadratic Model
- Safe Working Load: Word Problem
- Hang Time Word Problems
- Pole-Vaulting
Subtracting Polynomials
You are enlarging a [b][i]5[/i][/b]-inch by [b][i]7[/i][/b]-inch photo by a scale factor of [i][b]x[/b][/i] and mounting it on a mat. You want the mat to be twice as wide as the enlarged photo and [b][i]2[/i][/b] inches less than twice as high as the enlarged photo.[br][br][u][b]a.[/b][/u] Draw a diagram to represent the described situation. Label the dimensions.[br][br][b][u]b.[/u][/b] Write a model for the area of the mat around the photograph as a function of the scale factor.
SOLUTION
Use a verbal model. Use the diagram to find expression for the labels.[br][br][b]Area of mat = Total Area - Area of photo[br][/b][br]Area of mat = [b][i]A[/i][/b] (square inches)[br][br]Total area = [math]\left(10x\right)\left(14x-2\right)[/math] (square inches)[br][br]Area of photo = [math]\left(5x\right)\left(7x\right)[/math] (square inches)[br][br]Algebraic Model[br][br][math]A=\left(10x\right)\left(14x-2\right)-\left(5x\right)\left(7x\right)[/math][br][br] [math]=140x^2-20x-35x^2[/math][br][br] [math]=105x^2-20x[/math][br][br]A model for the area of the mat around the photograph as a function of the scale factor [b][i]x[/i][/b] is[br][br][math]A=105x^2-20x[/math]
Modeling a Punnett Square
Science Connection
The Punnett square below is an area model that shows the possible results of crossing two pink snapdragons, each with one red gene [b][i]R[/i] [/b]and one white gene [i][b]W[/b][/i]. Each parent snapdragon passes along only one gene for color to its offspring. Show how the square of a binomial can be used to model the Punnett square.
Solution
Each parent snapdragon has half red and half white genes. You can model the genetic makeup of each parent as [math]0.5R+0.5W[/math]. The genetic makeup of the offspring can be modeled by the product [math]\left(0.5R+0.5W\right)^2[/math].[br][br][math]\left(0.5R+0.5W\right)^2=\left(0.5R\right)^2+2\left(0.5R\right)\left(0.5W\right)+\left(0.5W\right)^2[/math][br][br] [math]=0.25R^2+0.5RW+0.25W^2[/math][br][br] (Red) (Pink) (White)[br][br][br][color=#ff0000]25% of the offspring will be red, 50% will be pink, and 25% will be white.[/color]
Writing a Quadratic Model
Lanscape Design
You are putting a stone border along two sides of a rectangular Japanese garden that measures [b][i]6[/i][/b] yards by [b][i]15[/i][/b] yards. Your budget limits you to only enough stones to cover [b][i]46[/i][/b] square yards. How wide should the border be?
Solution
Begin by drawing and labeling a diagram.[br][br]Area of Border = Total Area - Garden Area[br][br][math]46=\left(x+15\right)\left(x+6\right)-\left(15\right)\left(6\right)[/math] [color=#1155cc]Write quadratic model.[/color][br][br][math]46=x^2+21x+90-90[/math] [color=#1155cc]Multiply[/color][br][br][math]0=x^2+21x-46[/math] [color=#1155cc]Write in standard form[/color][br][br][math]0=\left(x+23\right)\left(x-2\right)[/math] [color=#1155cc]Factor[/color][br][br][math]\left(x+23\right)=0[/math] or [math]\left(x-2\right)=0[/math] [color=#1155cc]Use zero-product property[/color][br][br][math]x+23=0[/math] [color=#1155cc]Set first factor equal to 0[/color][br][br][math]x=-23[/math] [color=#1155cc]Solve for x[/color][br][br][math]x-2=0[/math][color=#1155cc] Set second factor equal to 0[/color][br][br][math]x=2[/math][color=#1155cc] Solve for x[/color][br][br]The solutions are [math]-23[/math] and [math]2[/math]. Only [math]x=2[/math] is a reasonable solution, because negative values for dimensions do not make sense. Mark the border [color=#ff0000]2 yards[/color] wide.
Writing and Using a Quadratic Model
BLOCK AND TACKLE
An object lifted with a rope or wire should not weigh more than the [i]safe working load[/i] for the rope or wire. The safe working load [i]S[/i] (in pounds) for a natural fiber rope is a function of [i]C[/i], the circumference of the rope in inches.[br][br] [b]Safe working load model[/b]: [math]150\cdot C^2=S[/math][br][br]You are setting up a block and tackle to lift a 1350-pound safe. What size natural fiber rope do you need to have a safe working load?
SOLUTION
Use the model to find a safe rope size. Substitute 1350 for [i]S[/i].[br][br] [math]150C^2=S[/math] [color=#0000ff]Write model.[br][br][/color] [math]150C^2=1350[/math] [color=#0000ff]Substitute.[br][br][/color] [math]150C^2-1350=0[/math] [color=#0000ff]Subtract 1350 from each side.[br][/color][br] [math]150\left(C^2-9\right)=0[/math] [color=#0000ff]Factor out common factor.[br][/color][br] [math]150\left(C-3\right)\left(C+3\right)[/math] [color=#0000ff]Factor.[br][/color][br] [math]\left(C-3\right)=0[/math] or [math]\left(C+3\right)=0[/math] [color=#0000ff]Use zero-product property.[/color][br][br] [math]C-3=0[/math] [color=#0000ff]Set first factor equal to 0.olve for [i]C[/i].[br][br][/color] [math]C=3[/math] [color=#0000ff]Solve for [i]C[/i].[/color][br] [br] [math]C+3=0[/math] [color=#0000ff]Set second factor equal to 0.[br][/color][color=#0000ff][br][/color] [math]C=-3[/math] Solve for [i]C[/i].[br][color=#0000ff][br][/color]The negative solution makes no sense. You need a rope with a circumference of at least [b][i][color=#ff0000]3 inches.[br][/color][/i][/b]