Factor a Quadratic Equation by Grouping

Any quadratic equation with rational roots can be factored by the grouping method. Here are the steps, followed by a few examples.

Steps:

1. Use algebra to put the equation in standard form with all terms on one side and 0 on the other.
2. Divide both sides by the greatest common factor (GCF), if any, to get to lowest terms.
3. Multiply the A coefficient (the squared term) by the C coefficient (the constant), to find AC. Note: "A" and "C" come from Ax²+Bx+C=0, which is the standard form of a quadratic equation.
4. Find the two integers that multiply together to equal AC and also add together to equal B (the coefficient of the x term).
5. Rewrite the B term as two new terms using the pair of numbers just found.
6. Group so that one x term falls into each resultant binomial.
7. Factor each pair of terms.
8. Factor again -- you should now have a fully factored expression.
9. Set each factor equal to zero and solve.
10. Check your answers.

Example: 4x²-13x=x+30

1. Subtract x+30 from both sides to get 4x²-14x - 30 = 0.
2. The GCF is 2, so divide each term by 2. The equation is now 2x²-7x-15=0.
3. 2 * (-15) = -30, so AC is -30.
4. The integer pairs that multiply together to equal -30 are: (-1, 30), (-2, 15), (-3, 10), (-5, 6), (1, -30), (-2, 15), (3, -10), and (5, -6)
   
   We want the pair that adds together to equal -7. Since 3 + (-10) = -7, we choose 3 and -10 as our two numbers.
5. Rewrite the middle term of our equation using the two numbers just found. The middle term was -7x, so substitute 3x - 10x for the -7x, giving the equation as 2x² + 3x - 10x - 15=0.
6. Group to get (2x²+3x)+(-10x-15)=0
7. Factor each pair of terms: x(2x+3) - 5(2x+3)=0
8. Factor out the common factor (2x + 3) to get (2x+3)(x-5)=0
9. Solve each factor
   When 2x + 3 = 0, x = -3/2.
   When x - 5 = 0, x = 5.
   Therefore, x is either -3/2 or 5
10. Check using the original equation:
    4(-3/2)² - 13(-3/2) = (-3/2) + 30, so 9 + 39/2 = 57/2, and 57/2 = 57/2. Therefore x = -3/2 is a solution.
    4(5)² - 13(5) = (5) + 30, so 100 - 65 = 35, and 35 = 35. Therefore x = 5 is a solution and the quadratic has been solved.

Example: x²- 2 = 0

1. This is already in standard form, although you could write it x² + 0x - 2 = 0.
2. The GCF is 1, so no need to divide.
3. AC = -2
4. Possible integer pairs are (-1, 2) and (1, -2). Neither pair sums to 0 (our B coefficient), so we now know that x² - 2 = 0 does not have rational roots.

Result: x² - 2 = 0 is not factorable in the rational number system. (It can be factored in the reals.)

Example: 6y² + 11y - 35 = 0

1. Equation is already in standard form.
2. Terms are relatively prime (GCF = 1), so no need to divide out common factors.
3. AC = -210
4. Possible pairs: (-1, 210), (-2, 105), (-3, 70), (-5, 22), (-6, 35), (-7, 30), (-10, 21) <<< stop here because -10 + 21 = 11, which is what we want!
5. 6y² - 10y + 21y - 35 = 0
6. (6y² - 10y) + (21y - 35) = 0
7. 2y(3y - 5) + 7(3y - 5) = 0
8. (3y - 5)(2y + 7) = 0
9. 3y - 5 = 0 implies that y = 5/3
   2y + 7 = 0 implies that y = -7/2
   Therefore y is 5/3 or -7/2.
   
10. Does 6(5/3)² + 11(5/3) - 35 = 0? Yes, true.
    Does 6(-7/2)² + 11(-7/2) - 35 = 0? Yes, true.