Math Problem Statement

how to do simple trinomials x^2 + 3x −18

Solution

To solve a simple trinomial such as x2+3x18x^2 + 3x - 18, you typically factorize it into the form (x+p)(x+q)(x + p)(x + q), where pp and qq are numbers that satisfy the following:

  1. Their sum is equal to the coefficient of xx, which is 33 in this case.
  2. Their product is equal to the constant term, which is 18-18.

Steps:

  1. Write down the equation: x2+3x18x^2 + 3x - 18

  2. Identify the sum and product conditions:

    • p+q=3p + q = 3 (sum condition)
    • pq=18p \cdot q = -18 (product condition)
  3. Find two numbers that satisfy these conditions:

    • The numbers 66 and 3-3 satisfy both conditions:
      • 6+(3)=36 + (-3) = 3
      • 6×(3)=186 \times (-3) = -18
  4. Write the factorized form: x2+3x18=(x+6)(x3)x^2 + 3x - 18 = (x + 6)(x - 3)

  5. (Optional) Verify by expanding: (x+6)(x3)=x23x+6x18=x2+3x18(x + 6)(x - 3) = x^2 - 3x + 6x - 18 = x^2 + 3x - 18

Solution:

The factorized form of x2+3x18x^2 + 3x - 18 is: (x+6)(x3)(x + 6)(x - 3)

If solving x2+3x18=0x^2 + 3x - 18 = 0, the solutions are: x=6orx=3x = -6 \quad \text{or} \quad x = 3


Would you like more examples or further clarification? Here are some related questions:

  1. How do you factor trinomials where the coefficient of x2x^2 is greater than 1?
  2. What is the difference between factoring and completing the square?
  3. How can trinomials be solved if they are not easily factorable?
  4. What are some tips for finding pp and qq quickly?
  5. How does this method relate to solving quadratic equations using the quadratic formula?

Tip: Always double-check your factorization by expanding it back to confirm the original expression.

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Math Problem Analysis

Mathematical Concepts

Algebra
Factoring Trinomials
Quadratic Equations

Formulas

Factored form of a trinomial: (x + p)(x + q)
Sum and product conditions for factoring

Theorems

Zero Product Property

Suitable Grade Level

Grades 8-10