Factor 6x^2 + x - 12 into two binomial terms.

• A. (2x + 3)(3x - 4)
• B. (3x + 4)(2x - 3)
• C. x(6x + 1) - 12
• D. (2x - 4)(3x + 3)

Answer: (A)

To factor 6x^2 + x - 12 into two binomial terms, look for numbers that help match the leading, middle, and constant parts when you form a product (ax + b)(cx + d). The leading term 6x^2 comes from ac, so a and c must multiply to 6. The constant term -12 comes from bd, so b and d must multiply to -12. The cross terms ad + bc must sum to the middle coefficient, 1.

A convenient way is the ac method: multiply a and c to get 6, and then find two numbers that multiply to ac times the constant (-72) and add to the middle coefficient (1). The numbers 9 and -8 fit because 9 + (-8) = 1 and 9(-8) = -72. Rewrite the middle as 9x and -8x:

6x^2 + 9x - 8x - 12.

Now factor by grouping:

(6x^2 + 9x) + (-8x - 12) = 3x(2x + 3) - 4(2x + 3).

Factor out the common binomial (2x + 3):

(2x + 3)(3x - 4).

You can verify by expanding: (2x + 3)(3x - 4) = 6x^2 - 8x + 9x - 12 = 6x^2 + x - 12.

This is the product of two binomials that matches the original expression. The other forms either give the wrong middle term sign or aren’t written as a product of two binomials.