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Unit 6: Quadratic and Polynomial Functions

Topic 2

Factoring and Solving Quadratics

A quadratic equation is an equation of the form
ax2+bx+c=0,a≠0.ax^2 + bx + c = 0, \quad a \ne 0.
One of the most efficient ways to solve quadratics is by factoring—rewriting the equation as a product of linear factors set equal to zero.
Factoring relies on the Zero Product Property: If A⋅B=0A \cdot B = 0, then A=0A = 0 or B=0.B = 0.
For example:
x2−5x+6=0⇒(x−2)(x−3)=0⇒x=2 or x=3.x^2 - 5x + 6 = 0 \quad \Rightarrow \quad (x - 2)(x - 3) = 0 \quad \Rightarrow \quad x = 2 \text{ or } x = 3.
Not all quadratics factor neatly, but recognizing common structures makes factoring fast:
  • Difference of squares: a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)
  • Perfect square trinomials: a2±2ab+b2=(a±b)2a^2 \pm 2ab + b^2 = (a \pm b)^2
  • Common factors: factor out the greatest common factor (GCF) first.
Factoring is most useful when the coefficients are integers and the quadratic can be decomposed into two binomials easily.

Core Skills

  • Factor out any GCF before proceeding.
  • Recognize patterns: difference of squares and perfect square trinomials.
  • Use the zero product property to find roots.
  • Check solutions by substituting back into the original equation.
  • Identify when factoring is not possible (then use quadratic formula).

Example 1: Simple Factoring

Solve x2−7x+10=0.x^2 - 7x + 10 = 0. Find two numbers that multiply to 1010 and add to −7-7: −5-5 and −2.-2.
x2−7x+10=(x−5)(x−2)=0x^2 - 7x + 10 = (x - 5)(x - 2) = 0
Solutions: x=5,2.x = 5, 2.

Example 2: Leading Coefficient Not 1

Solve 2x2+7x+3=0.2x^2 + 7x + 3 = 0. Multiply a⋅c=2×3=6.a \cdot c = 2 \times 3 = 6. Find two numbers that multiply to 6 and add to 7: 66 and 1.1.
2x2+6x+x+3=02x^2 + 6x + x + 3 = 0
Group:
(2x2+6x)+(x+3)=0⇒2x(x+3)+1(x+3)=0(2x^2 + 6x) + (x + 3) = 0 \Rightarrow 2x(x + 3) + 1(x + 3) = 0
(2x+1)(x+3)=0(2x + 1)(x + 3) = 0
Solutions: x=−12,−3.x = -\tfrac{1}{2}, -3.

Example 3: Difference of Squares

Solve x2−49=0.x^2 - 49 = 0.
(x−7)(x+7)=0(x - 7)(x + 7) = 0
Solutions: x=7,−7.x = 7, -7.

Example 4: Perfect Square Trinomial

Solve x2+10x+25=0.x^2 + 10x + 25 = 0.
(x+5)2=0(x + 5)^2 = 0
Solution: x=−5x = -5 (double root).

Example 5: Factoring Out a Common Factor

Solve 3x2−9x=0.3x^2 - 9x = 0.
3x(x−3)=03x(x - 3) = 0
Solutions: x=0,3.x = 0, 3.

Example 6: Application Problem

The product of two consecutive integers is 72. Let the smaller integer be xx, so:
x(x+1)=72x(x + 1) = 72
x2+x−72=0x^2 + x - 72 = 0
(x+9)(x−8)=0(x + 9)(x - 8) = 0
Solutions: x=−9x = -9 or x=8.x = 8. The integers are (−9,−8)(-9, -8) or (8,9).(8, 9).

Key Takeaways

  • Always factor out the GCF first.
  • Use pattern recognition to save time: difference of squares and perfect square trinomials.
  • Apply the zero product property after factoring.
  • When factoring fails, use the quadratic formula.
  • Verify by substitution to ensure no extraneous solutions.