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Zorluk: ZorFactoring Polynomials

Which of the following represents the completely factored form of the algebraic expression 30v3−35v2−15v30v^3 - 35v^2 - 15v?

  1. 5v(3v+1)(2v−3)5v(3v + 1)(2v - 3)Cevap
  2. B
    5v(3v−1)(2v+3)5v(3v - 1)(2v + 3)
  3. C
    v(15v+5)(2v−3)v(15v + 5)(2v - 3)
  4. D
    5v(v−9)(v+2)5v(v - 9)(v + 2)

Cevap

The completely factored form is 5v(3v+1)(2v−3)5v(3v + 1)(2v - 3).
The correct answer accurately extracts the Greatest Common Factor of 5v from the original expression, leaving the non-monic trinomial 6v^2 - 7v - 3. Factoring this remaining trinomial by grouping correctly yields the binomials (3v + 1) and (2v - 3). Expanding these prime factors back confirms they produce the exact original expression.

Adım Adım Çözüm

1
Identify and factor out the greatest common factor (GCF) from all three terms in the polynomial 30v3−35v2−15v30v^3 - 35v^2 - 15v.
5v(6v2−7v−3)5v(6v^2 - 7v - 3)
Factoring out the GCF simplifies the remaining polynomial and is a necessary first step for complete factorization.
2
Analyze the remaining quadratic trinomial 6v2−7v−36v^2 - 7v - 3 to factor it by grouping. Find two numbers that multiply to a×ca \times c (6×−3=−186 \times -3 = -18) and add to bb (−7-7).
The two numbers are −9-9 and 22.
These numbers allow us to split the middle term (−7v-7v) into two separate terms to enable factoring by grouping.
3
Rewrite the trinomial by splitting the middle term using the numbers found, then factor by grouping.
6v2−9v+2v−3=3v(2v−3)+1(2v−3)=(3v+1)(2v−3)6v^2 - 9v + 2v - 3 = 3v(2v - 3) + 1(2v - 3) = (3v + 1)(2v - 3)
Grouping the terms reveals the common binomial factor of (2v−3)(2v - 3).
4
Combine the GCF extracted in step 1 with the binomial factors found in step 3.
5v(3v+1)(2v−3)5v(3v + 1)(2v - 3)
This represents the original expression as a product of its simplest irreducible factors.

Anahtar Kavram

Factoring a polynomial by first extracting the greatest common factor (GCF), followed by factoring a non-monic quadratic trinomial using grouping.
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