When the polynomial 12x^2 + 11x - 15 is factored completely into the form (ax + b)(cx + d)… — ACT

When the polynomial $12x^2 + 11x - 15$ is factored completely into the form $(ax + b)(cx + d)$ , where $a$ , $b$ , $c$ , and $d$ are integers such that $a > c > 0$ , what is the value of the constant term $d$ ?

Answer: 5

Answer

The value of the constant term

d

5

Factoring the trinomial

12x^2 + 11x - 15

completely gives

(4x - 3)(3x + 5)

. Applying the constraint

a > c > 0

means the factor with the larger

x

-coefficient must be written first in the template

(ax + b)(cx + d)

. This yields

a = 4

b = -3

c = 3

, and

d = 5

. Thus, the constant term

d

5

Step-by-Step Solution

Find the factor pair for the AC method.

We need two numbers that multiply to

12 \times (-15) = -180

and add up to

11

. The numbers are

20

and

-9

This allows us to split the linear middle term to factor by grouping.

Rewrite the polynomial and factor by grouping.

12x^2 + 20x - 9x - 15 = 4x(3x + 5) - 3(3x + 5) = (4x - 3)(3x + 5)

Grouping the first two terms and the last two terms reveals a common binomial factor of

(3x + 5)

Apply the given inequality constraints to match the template.

Comparing

(4x - 3)(3x + 5)

(ax + b)(cx + d)

with

a > c > 0

yields

a = 4

b = -3

c = 3

, and

d = 5

Since the lead coefficient

4

is greater than

3

, the factor

(4x - 3)

must correspond to

(ax + b)

Key Concept

Factoring quadratic trinomials of the form

Ax^2 + Bx + C

using the grouping (AC) method.

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