Soru

Zorluk: Çok zorFactoring Polynomials

Decompose the quadratic trinomial 6x2−11x−106x^2 - 11x - 10 into a product of two binomial factors structured as (ax−b)(cx+d)(ax - b)(cx + d). Assume all parameters (aa, bb, cc, and dd) are strictly positive integers. To ensure a unique arrangement, the leading coefficient aa must be strictly less than cc. Provide the values of the four parameters.

Cevap:a=a = 【2】, b=b = 【5】, c=c = 【3】, and d=d = 【2】.

Cevap

The parameters are a=2a = 2, b=5b = 5, c=3c = 3, and d=2d = 2.
Factoring the polynomial 6x2−11x−106x^2 - 11x - 10 using the AC method produces (2x−5)(3x+2)(2x - 5)(3x + 2). Matching these two factors against the template (ax−b)(cx+d)(ax - b)(cx + d) under the rule that a<ca < c (since 2<32 < 3) perfectly determines the positive integers as a=2a = 2, b=5b = 5, c=3c = 3, and d=2d = 2.

Adım Adım Çözüm

1
Identify the factoring technique for a non-monic quadratic trinomial in the form Ax2+Bx+CAx^2 + Bx + C.
The AC method (factoring by grouping) is required because the leading coefficient A=6A = 6 is greater than 1.
Directly finding factors of the constant CC that sum to BB only works when A=1A = 1.
2
Multiply the leading coefficient AA by the constant term CC.
6×(−10)=−606 \times (-10) = -60.
This establishes the target product needed to properly split the middle term for grouping.
3
Find two integers that multiply to −60-60 and add to the middle coefficient B=−11B = -11.
The correct pair of integers is −15-15 and 44.
(−15)×4=−60(-15) \times 4 = -60 and −15+4=−11-15 + 4 = -11.
4
Rewrite the middle term using these two integers and factor the resulting four-term expression by grouping.
6x2−15x+4x−10=3x(2x−5)+2(2x−5)=(2x−5)(3x+2)6x^2 - 15x + 4x - 10 = 3x(2x - 5) + 2(2x - 5) = (2x - 5)(3x + 2).
Grouping the first two terms and the last two terms reveals a common binomial factor of (2x−5)(2x - 5).
5
Match the factored expression to the required template (ax−b)(cx+d)(ax - b)(cx + d) while adhering to the condition a<ca < c.
The factors are (2x−5)(2x - 5) and (3x+2)(3x + 2). Since 2<32 < 3, (ax−b)(ax - b) corresponds exactly to (2x−5)(2x - 5), yielding a=2a=2 and b=5b=5. Consequently, (cx+d)(cx + d) corresponds to (3x+2)(3x + 2), yielding c=3c=3 and d=2d=2.
This guarantees a unique, correctly-ordered set of strictly positive parameters.

Anahtar Kavram

Factoring non-monic quadratic trinomials via grouping
Bu soruyu puanla