Chapter Notes: Quadratics: Multiplying & Factoring
Quadratic expressions are polynomials where the highest power of the variable is 2. You've already worked with linear expressions like \( 2x + 5 \), where the variable has an exponent of 1. Now we'll explore expressions like \( x^2 + 5x + 6 \), where the squared term makes the expression quadratic. Learning to multiply binomials to create quadratic expressions and to factor quadratic expressions back into binomials is a fundamental skill in algebra. These techniques will help you solve quadratic equations, graph parabolas, and understand many real-world relationships involving area, motion, and optimization.
Understanding Quadratic Expressions
A quadratic expression is a polynomial of degree 2, which means the highest exponent on the variable is 2. The standard form of a quadratic expression is:
\[ ax^2 + bx + c \]In this form, \( a \), \( b \), and \( c \) are constants (numbers), and \( a \) cannot equal zero. The term \( ax^2 \) is called the quadratic term, \( bx \) is the linear term, and \( c \) is the constant term.
Examples of quadratic expressions include:
- \( x^2 + 7x + 10 \) where \( a = 1 \), \( b = 7 \), and \( c = 10 \)
- \( 3x^2 - 5x + 2 \) where \( a = 3 \), \( b = -5 \), and \( c = 2 \)
- \( -2x^2 + 8x \) where \( a = -2 \), \( b = 8 \), and \( c = 0 \)
- \( x^2 - 9 \) where \( a = 1 \), \( b = 0 \), and \( c = -9 \)
Quadratic expressions often arise when we multiply two linear expressions called binomials together. Understanding this multiplication process is the foundation for working with quadratics.
Multiplying Binomials
A binomial is an algebraic expression containing exactly two terms, such as \( (x + 3) \) or \( (2x - 5) \). When we multiply two binomials together, we create a quadratic expression. The key principle is the distributive property, which states that every term in the first binomial must be multiplied by every term in the second binomial.
The FOIL Method
The FOIL method is a systematic way to multiply two binomials. FOIL is an acronym that stands for:
- First: Multiply the first terms in each binomial
- Outer: Multiply the outer terms in the product
- Inner: Multiply the inner terms in the product
- Last: Multiply the last terms in each binomial
After applying FOIL, we combine any like terms to simplify the result.
Example: Multiply \( (x + 4)(x + 3) \).
Solution:
First: Multiply the first terms: \( x \cdot x = x^2 \)
Outer: Multiply the outer terms: \( x \cdot 3 = 3x \)
Inner: Multiply the inner terms: \( 4 \cdot x = 4x \)
Last: Multiply the last terms: \( 4 \cdot 3 = 12 \)
Combine all four products: \( x^2 + 3x + 4x + 12 \)
Combine like terms: \( x^2 + 7x + 12 \)
The product is \( x^2 + 7x + 12 \).
Example: Multiply \( (x - 5)(x + 2) \).
Solution:
First: \( x \cdot x = x^2 \)
Outer: \( x \cdot 2 = 2x \)
Inner: \( -5 \cdot x = -5x \)
Last: \( -5 \cdot 2 = -10 \)
Combine: \( x^2 + 2x - 5x - 10 \)
Combine like terms: \( x^2 - 3x - 10 \)
The product is \( x^2 - 3x - 10 \).
Multiplying Binomials with Coefficients
When binomials contain coefficients on the variable terms, we apply the same FOIL process, being careful with multiplication of all coefficients.
Example: Multiply \( (2x + 3)(3x - 4) \).
Solution:
First: \( 2x \cdot 3x = 6x^2 \)
Outer: \( 2x \cdot (-4) = -8x \)
Inner: \( 3 \cdot 3x = 9x \)
Last: \( 3 \cdot (-4) = -12 \)
Combine: \( 6x^2 - 8x + 9x - 12 \)
Combine like terms: \( 6x^2 + x - 12 \)
The product is \( 6x^2 + x - 12 \).
Special Products
Certain binomial products occur frequently and follow recognizable patterns. Learning these special products helps you multiply and factor more quickly.
Square of a Binomial
When we square a binomial, we multiply it by itself:
\[ (a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2 \] \[ (a - b)^2 = (a - b)(a - b) = a^2 - 2ab + b^2 \]Example: Expand \( (x + 5)^2 \).
Solution:
Using the pattern \( (a + b)^2 = a^2 + 2ab + b^2 \) where \( a = x \) and \( b = 5 \):
\( a^2 = x^2 \)
\( 2ab = 2(x)(5) = 10x \)
\( b^2 = 25 \)
Therefore: \( (x + 5)^2 = x^2 + 10x + 25 \)
The expanded form is \( x^2 + 10x + 25 \).
Difference of Squares
When we multiply the sum and difference of the same two terms, the middle terms cancel:
\[ (a + b)(a - b) = a^2 - b^2 \]This pattern is called the difference of squares because the result is the difference between two perfect squares.
Example: Multiply \( (x + 7)(x - 7) \).
Solution:
Using the pattern \( (a + b)(a - b) = a^2 - b^2 \) where \( a = x \) and \( b = 7 \):
\( a^2 = x^2 \)
\( b^2 = 49 \)
Therefore: \( (x + 7)(x - 7) = x^2 - 49 \)
The product is \( x^2 - 49 \).
Introduction to Factoring
Factoring is the reverse process of multiplying. When we factor a quadratic expression, we rewrite it as a product of two or more simpler expressions, usually binomials. Factoring is essential for solving quadratic equations and simplifying algebraic fractions.
For example, if multiplication gives us \( (x + 2)(x + 3) = x^2 + 5x + 6 \), then factoring reverses this: \( x^2 + 5x + 6 = (x + 2)(x + 3) \).
Greatest Common Factor (GCF)
The first step in factoring any expression is to look for a greatest common factor (GCF). The GCF is the largest expression that divides evenly into all terms. If a GCF exists, factor it out first.
Example: Factor \( 3x^2 + 12x \).
Solution:
Identify the GCF of \( 3x^2 \) and \( 12x \).
Both terms are divisible by \( 3x \).
Factor out \( 3x \): \( 3x^2 + 12x = 3x(x + 4) \)
The factored form is \( 3x(x + 4) \).
Factoring Trinomials
A trinomial is a polynomial with three terms. Many quadratic trinomials can be factored into the product of two binomials. We'll focus on trinomials in the form \( x^2 + bx + c \) where the coefficient of \( x^2 \) is 1.
Factoring When a = 1
To factor \( x^2 + bx + c \), we need to find two numbers that:
- Multiply to give \( c \) (the constant term)
- Add to give \( b \) (the coefficient of the linear term)
If those two numbers are \( m \) and \( n \), then:
\[ x^2 + bx + c = (x + m)(x + n) \]Example: Factor \( x^2 + 8x + 15 \).
Solution:
We need two numbers that multiply to 15 and add to 8.
Pairs that multiply to 15: (1, 15), (3, 5)
Check which pair adds to 8: \( 3 + 5 = 8 \) ✓
Therefore: \( x^2 + 8x + 15 = (x + 3)(x + 5) \)
The factored form is \( (x + 3)(x + 5) \).
Example: Factor \( x^2 - 7x + 12 \).
Solution:
We need two numbers that multiply to 12 and add to -7.
Since the product is positive and the sum is negative, both numbers must be negative.
Pairs that multiply to 12: (-1, -12), (-2, -6), (-3, -4)
Check which pair adds to -7: \( -3 + (-4) = -7 \) ✓
Therefore: \( x^2 - 7x + 12 = (x - 3)(x - 4) \)
The factored form is \( (x - 3)(x - 4) \).
Example: Factor \( x^2 + 2x - 15 \).
Solution:
We need two numbers that multiply to -15 and add to 2.
Since the product is negative, one number is positive and one is negative.
Pairs that multiply to -15: (1, -15), (-1, 15), (3, -5), (-3, 5)
Check which pair adds to 2: \( 5 + (-3) = 2 \) ✓
Therefore: \( x^2 + 2x - 15 = (x + 5)(x - 3) \)
The factored form is \( (x + 5)(x - 3) \).
Factoring When a ≠ 1
When the coefficient of \( x^2 \) is not 1, factoring becomes more challenging. For trinomials in the form \( ax^2 + bx + c \) where \( a \neq 1 \), we can use the AC method or trial and error.
The AC Method
The AC method involves these steps:
- Multiply \( a \) and \( c \) to get the product \( ac \)
- Find two numbers that multiply to \( ac \) and add to \( b \)
- Rewrite the middle term using these two numbers
- Factor by grouping
Example: Factor \( 2x^2 + 7x + 3 \).
Solution:
Here \( a = 2 \), \( b = 7 \), \( c = 3 \)
Calculate \( ac = 2 \times 3 = 6 \)
Find two numbers that multiply to 6 and add to 7: These are 1 and 6.
Rewrite the middle term: \( 2x^2 + x + 6x + 3 \)
Factor by grouping: \( x(2x + 1) + 3(2x + 1) \)
Factor out the common binomial: \( (2x + 1)(x + 3) \)
The factored form is \( (2x + 1)(x + 3) \).
Example: Factor \( 3x^2 - 10x + 8 \).
Solution:
Here \( a = 3 \), \( b = -10 \), \( c = 8 \)
Calculate \( ac = 3 \times 8 = 24 \)
Find two numbers that multiply to 24 and add to -10: These are -4 and -6.
Rewrite: \( 3x^2 - 4x - 6x + 8 \)
Factor by grouping: \( x(3x - 4) - 2(3x - 4) \)
Factor out \( (3x - 4) \): \( (3x - 4)(x - 2) \)
The factored form is \( (3x - 4)(x - 2) \).
Factoring Special Patterns
Recognizing special patterns can make factoring faster and easier.
Perfect Square Trinomials
A perfect square trinomial results from squaring a binomial. These trinomials follow the patterns:
\[ a^2 + 2ab + b^2 = (a + b)^2 \] \[ a^2 - 2ab + b^2 = (a - b)^2 \]To recognize a perfect square trinomial, check if the first and last terms are perfect squares and if the middle term is twice the product of their square roots.
Example: Factor \( x^2 + 14x + 49 \).
Solution:
Check if this is a perfect square trinomial.
First term: \( x^2 = (x)^2 \) ✓
Last term: \( 49 = 7^2 \) ✓
Middle term: \( 14x = 2(x)(7) \) ✓
This matches the pattern \( a^2 + 2ab + b^2 \) with \( a = x \) and \( b = 7 \).
Therefore: \( x^2 + 14x + 49 = (x + 7)^2 \)
The factored form is \( (x + 7)^2 \).
Difference of Squares
The difference of squares pattern allows us to factor expressions in the form \( a^2 - b^2 \):
\[ a^2 - b^2 = (a + b)(a - b) \]Notice that the sum of squares \( a^2 + b^2 \) cannot be factored using real numbers.
Example: Factor \( x^2 - 64 \).
Solution:
Recognize this as a difference of squares where \( a = x \) and \( b = 8 \).
\( x^2 = (x)^2 \) and \( 64 = 8^2 \)
Apply the pattern: \( x^2 - 64 = (x + 8)(x - 8) \)
The factored form is \( (x + 8)(x - 8) \).
Example: Factor \( 9x^2 - 25 \).
Solution:
Recognize this as a difference of squares where \( a = 3x \) and \( b = 5 \).
\( 9x^2 = (3x)^2 \) and \( 25 = 5^2 \)
Apply the pattern: \( 9x^2 - 25 = (3x + 5)(3x - 5) \)
The factored form is \( (3x + 5)(3x - 5) \).
Complete Factoring Strategy
When factoring any quadratic expression, follow this systematic approach:
- Factor out the GCF first. Always check for common factors before attempting other methods.
- Count the terms. Two terms suggest difference of squares. Three terms suggest a trinomial. Four terms suggest grouping.
- Look for special patterns. Check for perfect square trinomials or difference of squares.
- Use appropriate method. Apply the factoring method that fits the structure of the expression.
- Check your answer. Multiply the factors to verify you get the original expression.
Example: Factor \( 2x^2 - 50 \) completely.
Solution:
Step 1: Factor out the GCF. Both terms are divisible by 2.
\( 2x^2 - 50 = 2(x^2 - 25) \)
Step 2: Recognize that \( x^2 - 25 \) is a difference of squares.
\( x^2 - 25 = (x + 5)(x - 5) \)
Step 3: Write the complete factorization: \( 2(x + 5)(x - 5) \)
The completely factored form is \( 2(x + 5)(x - 5) \).
Applications and Connections
Understanding how to multiply and factor quadratics has many practical applications. In geometry, the area of a rectangle with dimensions \( (x + 3) \) and \( (x + 5) \) is \( (x + 3)(x + 5) = x^2 + 8x + 15 \). If you know the area is \( x^2 + 8x + 15 \), factoring reveals the dimensions.
In physics, the equation for the height of a projectile often involves quadratic expressions. Factoring allows us to find when the projectile hits the ground (when height equals zero).
In business, revenue and profit functions are often quadratic. Factoring helps identify break-even points and maximum profit scenarios.
Mastering multiplication and factoring of quadratics prepares you for solving quadratic equations, working with parabolas, and understanding more advanced mathematical concepts in future courses.
