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Zorluk: Çok zorFactoring Polynomials

When the polynomial 6x37x216x+126x^3 - 7x^2 - 16x + 12 is factored completely into three linear factors of the form (ax+b)(cx+d)(ex+f)(ax + b)(cx + d)(ex + f), where aa, cc, and ee are positive integers, what is the value of a+b+c+d+e+fa + b + c + d + e + f?

Cevap: 5

Cevap

The value of the sum of the coefficients is 5.
The polynomial factors completely over the integers as (x2)(2x+3)(3x2)(x - 2)(2x + 3)(3x - 2). The sum of the six coefficients is 1+(2)+2+3+3+(2)=51 + (-2) + 2 + 3 + 3 + (-2) = 5.

Adım Adım Çözüm

1
Find one linear factor of the cubic polynomial using the Factor Theorem.
The root x=2x = 2 satisfies the equation, so (x2)(x - 2) is a factor.
Testing integer factors of the constant term 12 reveals that x=2x = 2 evaluates the polynomial to 0.
2
Perform synthetic division or polynomial long division to divide the cubic by the linear factor.
The quotient is the quadratic expression 6x2+5x66x^2 + 5x - 6.
This reduces the degree of the polynomial to allow quadratic factoring techniques.
3
Factor the quadratic quotient into two linear binomials.
The quadratic factors into (2x+3)(3x2)(2x + 3)(3x - 2).
Using the AC method, 6×(6)=366 \times (-6) = -36, and the factors of 36-36 that sum to 55 are 99 and 4-4.
4
Identify the coefficients and sum them.
The sum is 1+(2)+2+3+3+(2)=51 + (-2) + 2 + 3 + 3 + (-2) = 5.
The factors are (1x2)(2x+3)(3x2)(1x - 2)(2x + 3)(3x - 2), corresponding to the coefficients a=1,b=2,c=2,d=3,e=3,f=2a=1, b=-2, c=2, d=3, e=3, f=-2.

Anahtar Kavram

Complete factorization of cubic polynomials with integer coefficients using the Rational Root Theorem and quadratic factoring.
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