Ninth Grade Mathematics: Factoring Quadratic Trinomials & Grouping

Factoring is the algebraic reverse of expansion, transforming a polynomial sum into a product of simpler linear or irreducible polynomial factors. In this chapter, you will master factoring by grouping terms, examine common monomial extraction, and study the method of splitting the middle term for quadratic trinomials.

Factoring by Extracting Common Monomial Factors

The most fundamental factoring technique is reversing the distributive law by identifying the greatest common factor (GCF) among all terms and factoring it out.

Extracting the GCF:
12*x^3*y^2 - 18*x^2*y^3 + 24*x^2*y

Step 1: Find the GCF of the numerical coefficients (12, 18, 24)  ---> 6
Step 2: Find the lowest power of variable x present (x^3, x^2, x^2) ---> x^2
Step 3: Find the lowest power of variable y present (y^2, y^3, y)   ---> y
The complete GCF is 6*x^2*y.

Factored Form:
6*x^2*y * (2*x*y - 3*y^2 + 4)

Always check for common monomial factors before attempting any other factoring method, as this simplifies the remaining polynomial significantly.

Factoring Four-Term Polynomials by Grouping

When a polynomial has four terms with no single factor common to all four, grouping terms into pairs can reveal a shared binomial factor.

Algorithm for Factoring by Grouping:
1. Group the four terms into two pairs: (Term 1 + Term 2) + (Term 3 + Term 4).
2. Factor out the GCF from each pair separately.
3. If the resulting binomial expressions in parentheses are identical,
   factor out that common binomial.

Example: Factor a*x + b*x + a*y + b*y

Step 1: Group into pairs:
        (a*x + b*x) + (a*y + b*y)

Step 2: Factor out the GCF from each pair:
        x * (a + b) + y * (a + b)

Step 3: Factor out the common binomial (a + b):
        (a + b) * (x + y)

If the binomials in Step 2 do not match, rearrange the original four terms into a different order (for instance, pairing Term 1 with Term 3, and Term 2 with Term 4) and repeat the process.

Factoring Quadratic Trinomials by Splitting the Middle Term

A quadratic trinomial in standard form is written as ax^2 + bx + c. To factor this trinomial into the product of two linear binomials, we split the middle linear term bx into two terms, px and q*x, such that the resulting four-term polynomial can be factored by grouping.

The ac-Method for Splitting the Middle Term:

Quadratic Trinomial:  a*x^2 + b*x + c

Find two real numbers p and q such that:
  1. p + q = b       (Sum equals the middle coefficient)
  2. p * q = a * c   (Product equals leading coefficient times constant)

Example: Factor 6*x^2 + 17*x + 12

Step 1: Identify a, b, and c:
        a = 6,  b = 17,  c = 12.

Step 2: Compute product a * c:
        6 * 12 = 72.

Step 3: Find two numbers p and q that multiply to 72 and add to 17:
        Pairs multiplying to 72: (1, 72), (2, 36), (3, 24), (4, 18), (6, 12), (8, 9).
        Check sums: 8 + 9 = 17.  Thus, p = 8 and q = 9.

Step 4: Split the middle term 17*x into 8*x + 9*x:
        6*x^2 + 8*x + 9*x + 12

Step 5: Factor by grouping:
        2*x * (3*x + 4) + 3 * (3*x + 4)

Step 6: Extract the common binomial:
        (3*x + 4) * (2*x + 3)

Chapter Exercises

Exercise 1: Factor completely by extracting common factors: 15a^3b - 25a^2b^2 + 10ab^3.

Exercise 2: Factor the four-term expression 3x^3 - 6x^2 + 4*x - 8 by grouping.

Exercise 3: Factor each of the following quadratic trinomials by splitting the middle term: x^2 - 11x + 24, and 3x^2 - 10*x - 8.

Solutions and Explanations

Solution 1: Find the greatest common factor of the terms 15a^3b, -25a^2b^2, and 10ab^3. The GCF of coefficients 15, -25, and 10 is 5. The lowest power of a is a^1, and the lowest power of b is b^1. Factoring out 5ab yields: 5ab * (3a^2 - 5ab + 2b^2). The quadratic trinomial inside can be further factored: 3a^2 - 3ab - 2ab + 2b^2 = 3a(a - b) - 2b(a - b) = (a - b)(3a - 2b). Thus, the full factorization is 5ab * (a - b) * (3a - 2*b).

Solution 2: Group the four terms into two pairs: (3x^3 - 6x^2) + (4x - 8). From the first pair, factor out 3x^2 to get 3x^2 * (x - 2). From the second pair, factor out 4 to get 4 * (x - 2). Both terms now share the binomial (x - 2). Factoring out (x - 2) gives (x - 2) * (3x^2 + 4).

Solution 3: For x^2 - 11x + 24, here a = 1, b = -11, and c = 24. The product ac is 24. We need two numbers whose product is 24 and whose sum is -11. These numbers are -8 and -3. Splitting the middle term: x^2 - 8x - 3x + 24 = x(x - 8) - 3(x - 8) = (x - 8)(x - 3). For 3x^2 - 10x - 8, here a = 3, b = -10, and c = -8. The product ac is 3 * (-8) = -24. We need two numbers whose product is -24 and whose sum is -10. These numbers are -12 and +2. Splitting the middle term: 3x^2 - 12x + 2x - 8 = 3x(x - 4) + 2(x - 4) = (x - 4)(3x + 2).