MathIsimple
Lesson 1.3: Polynomial Factoring (Linear & Quadratic)

Master Polynomial Factoring Techniques

Unlock the power of factoring! Learn systematic techniques to break down polynomials into their simplest forms, discover polynomial zeros, and understand how factoring reveals the structure of algebraic expressions.

Learning Objectives

Extract greatest common factors
Factor quadratic trinomials
Recognize perfect squares and differences
Find polynomial zeros through factoring

Core Concept 1: Greatest Common Factor (GCF) Extraction

Finding the GCF

The GCF is the largest factor that divides all terms in a polynomial. Always look for the GCF first before applying other factoring methods.

Example: 6x2−9x6x^2-9x

Step 1:Identify GCF of coefficients: 6 and 9
GCF(6, 9) = 33
Step 2:Identify GCF of variables: x2x^2 and xx
GCF(x2x^2, xx) = xx
Step 3:6x2−9x=3x(2x−3)6x^2-9x=3x(2x-3)

Example: 12x3y2−8x2y+4xy12x^3y^2-8x^2y+4xy

Step 1:GCF of coefficients: 12, 8, 4 → 4
Step 2:GCF of x: x3,x2,xx^3, x^2, x → xx
Step 3:GCF of y: y2,y,yy^2, y, y → yy
Result:4xy(3x2y−2x+1)4xy(3x^2y-2x+1)

GCF of Numbers

Find largest number dividing all coefficients

GCF(12, 18, 24) = 6

GCF of Variables

Take lowest power of each variable

GCF(x³, x², x) = x

Core Concept 2: Quadratic Trinomial Factoring

Cross-Multiplication Method

For trinomials of the form x² + bx + c, find two numbers that multiply to c and add to b.

Example: x2+5x+6x^2+5x+6

Need two numbers that:
• Multiply to 66: (1,6), (2,3), (-1,-6), (-2,-3)
• Add to 55: 2+3=52+3=5 ✓
Result:x2+5x+6=(x+2)(x+3)x^2+5x+6=(x+2)(x+3)

Example: x2−7x+12x^2-7x+12

Need numbers that multiply to 12 and add to -7:
(−3)imes(−4)=12(-3) imes(-4)=12 ✓
(−3)+(−4)=−7(-3)+(-4)=-7 ✓
Result: (x−3)(x−4)(x-3)(x-4)

Example: 2x2+7x+32x^2+7x+3

For ax2+bx+cax^2+bx+c where ae1a e 1:
• Find factors of ac=2imes3=6ac=2 imes 3=6
• That add to b=7b=7: 6+1=76+1=7
• Split middle term: 2x2+6x+x+32x^2+6x+x+3
• Factor by grouping: 2x(x+3)+1(x+3)2x(x+3)+1(x+3)
• Result: (2x+1)(x+3)(2x+1)(x+3)

Core Concept 3: Perfect Squares & Difference of Squares

Special Factoring Patterns

Recognize these special patterns for quick factoring without trial and error.

Perfect Square Trinomials

Pattern: a² + 2ab + b² = (a + b)²

Example: x² + 6x + 9 = (x + 3)²

Pattern: a² - 2ab + b² = (a - b)²

Example: x² - 8x + 16 = (x - 4)²

How to recognize: First and last terms are perfect squares, middle term is 2 × √(first) × √(last)

Difference of Squares

Pattern: a² - b² = (a - b)(a + b)

Example: x² - 9 = (x - 3)(x + 3)

Example: 4x² - 25 = (2x - 5)(2x + 5)

How to recognize: Two perfect squares separated by a minus sign

Core Concept 4: Finding Polynomial Zeros Through Factoring

Zero Product Property

If the product of factors equals zero, then at least one of the factors must be zero. This allows us to find the zeros (roots) of polynomial functions.

Example: Find zeros of y = x² - 6x + 8

Step 1:Factor the polynomial
x² - 6x + 8 = (x - 2)(x - 4)
Step 2:Set equal to zero: (x - 2)(x - 4) = 0
Step 3:Apply zero product property
x - 2 = 0 or x - 4 = 0
Step 4:Solve: x = 2 or x = 4
Zeros: (2, 0) and (4, 0)

Example: y = 2x² + 5x - 3

Factor: 2x² + 5x - 3 = (2x - 1)(x + 3)
Set to zero: (2x - 1)(x + 3) = 0
Solve: 2x - 1 = 0 → x = 1/2
Or: x + 3 = 0 → x = -3
Zeros: (1/2, 0) and (-3, 0)

Zero Product Property

If ab = 0, then a = 0 or b = 0

Essential for finding polynomial roots

Graphical Connection

Zeros are x-intercepts of the graph

Points where y = 0

Real-World Application: Function Analysis & Optimization

Problem Scenario

A ball is thrown upward with initial velocity. Its height h (in feet) after t seconds is given by h = -16t² + 32t + 48. Find when the ball hits the ground and its maximum height.

Solution Steps:

Step 1:Factor the height function
h=−16t2+32t+48=−16(t2−2t−3)h=-16t^2+32t+48=-16(t^2-2t-3)
=−16(t−3)(t+1)=-16(t-3)(t+1)
Step 2:Find when ball hits ground (h = 0)
−16(t−3)(t+1)=0-16(t-3)(t+1)=0
t−3=0⇒t=3 extst-3=0\Rightarrow t=3\ ext{s}
(t = -1 is not physically meaningful)

Key Insights:

  • • The ball hits the ground after 3 seconds
  • • Maximum height occurs at t = 1 second (vertex of parabola)
  • • Factoring reveals the structure of the motion equation
  • • Zeros represent important physical events (ground contact)

Practice Problems

Problem 1:

Factor: 8x² - 12x

Show Solution

GCF of 8x² and 12x is 4x

8x² - 12x = 4x(2x - 3)

Problem 2:

Factor: x² + 8x + 15

Show Solution

Need numbers that multiply to 15 and add to 8

3 × 5 = 15 and 3 + 5 = 8

x² + 8x + 15 = (x + 3)(x + 5)

Problem 3:

Find zeros of y = x² - 9

Show Solution

x² - 9 = (x - 3)(x + 3) = 0

x - 3 = 0 → x = 3

x + 3 = 0 → x = -3

Zeros: (3, 0) and (-3, 0)

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