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Zorluk: OrtaFactoring Polynomials

If the expression 6x211x106x^2 - 11x - 10 is factored completely into the form (ax+b)(cxd)(ax + b)(cx - d), where a,b,c,a, b, c, and dd are positive integers, what is the value of a+b+c+da + b + c + d?

Cevap: 12

Cevap

The value of the sum of the coefficients and constants a+b+c+da + b + c + d is 1212.
The factored form of 6x211x106x^2 - 11x - 10 is (3x+2)(2x5)(3x + 2)(2x - 5). Matching this with (ax+b)(cxd)(ax + b)(cx - d) where a,b,c,a, b, c, and dd are positive integers results in a=3a = 3, b=2b = 2, c=2c = 2, and d=5d = 5. The sum of these values is 3+2+2+5=123 + 2 + 2 + 5 = 12.

Adım Adım Çözüm

1
Factor the quadratic trinomial 6x211x106x^2 - 11x - 10 using the grouping method.
(3x+2)(2x5)(3x + 2)(2x - 5)
Factoring splits the quadratic expression into its constituent linear binomial factors.
2
Equate the factored expression to the given form (ax+b)(cxd)(ax + b)(cx - d) to find the values of a,b,c,a, b, c, and dd.
a=3a = 3, b=2b = 2, c=2c = 2, and d=5d = 5
Since the variables represent positive integers, we match the positive constant term to bb and the negative constant term to d-d.
3
Sum the values of a,b,c,a, b, c, and dd.
1212
Calculating the final sum answers the target mathematical question.

Anahtar Kavram

Factoring quadratic polynomials with a leading coefficient greater than 1 using the grouping method.
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