The cubic polynomial 6x^3 - 19x^2 + 11x + 6 can be factored completely over the integers i… — ACT

The cubic polynomial $6x^3 - 19x^2 + 11x + 6$ can be factored completely over the integers in the form $(x - a)(bx - c)(dx + e)$ , where $a$ , $b$ , $c$ , $d$ , and $e$ are positive integers. What is the value of $a + b + c + d + e$ ?

Answer: 11

Answer

The sum of the coefficients and constants from the factored form is 11.

By applying the Rational Root Theorem, we find the root

x = 2

, which gives the factor

(x - 2)

. Dividing the original cubic expression by

(x - 2)

yields

6x^2 - 7x - 3

. Factoring this quadratic expression by grouping yields

(2x - 3)(3x + 1)

. Writing the completely factored form as

(x - 2)(2x - 3)(3x + 1)

and comparing it to

(x - a)(bx - c)(dx + e)

where

a, b, c, d, e

are positive integers results in

a = 2

b = 2

c = 3

d = 3

, and

e = 1

. The sum

a + b + c + d + e

is equal to 11.

Step-by-Step Solution

Identify a rational root of

6x^3 - 19x^2 + 11x + 6

x = 2

is a root, meaning

(x - 2)

is a factor.

Applying the Rational Root Theorem and testing potential integer root values.

Divide

6x^3 - 19x^2 + 11x + 6

(x - 2)

The quotient is the quadratic polynomial

6x^2 - 7x - 3

To reduce the cubic polynomial to a quadratic expression that can be factored using standard trinomial methods.

Factor the quadratic trinomial

6x^2 - 7x - 3

(2x - 3)(3x + 1)

Finding two numbers that multiply to

-18

and add to

-7

(which are

-9

and

2

) and factoring by grouping.

Match the factored form

(x - 2)(2x - 3)(3x + 1)

(x - a)(bx - c)(dx + e)

a = 2, b = 2, c = 3, d = 3, e = 1

Since

a, b, c, d,

and

e

must be positive integers, the constant terms and signs uniquely determine the mapping of each factor.

Calculate the sum of

a, b, c, d,

and

e

2 + 2 + 3 + 3 + 1 = 11

To find the final numeric value requested by the question.

Key Concept

Factoring cubic polynomials by finding rational roots and factoring quadratic trinomials by grouping.

Estimated Time:2m 30s

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