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Zorluk: ZorFactoring Polynomials

When the algebraic expression 12y3+12y2−45y12y^3 + 12y^2 - 45y is factored completely, which of the following expressions is produced?

  1. 3y(2y−3)(2y+5)3y(2y - 3)(2y + 5)Cevap
  2. B
    3y(2y+3)(2y−5)3y(2y + 3)(2y - 5)
  3. C
    3(2y2−3y)(2y+5)3(2y^2 - 3y)(2y + 5)
  4. D
    −3y(2y−3)(2y+5)-3y(2y - 3)(2y + 5)

Cevap

The completely factored form is 3y(2y−3)(2y+5)3y(2y - 3)(2y + 5).
The correct answer accurately extracts the full GCF of 3y3y and correctly factors the remaining trinomial into (2y−3)(2y+5)(2y - 3)(2y + 5). Expanding this result confirms it equals 12y3+12y2−45y12y^3 + 12y^2 - 45y.

Adım Adım Çözüm

1
Identify and extract the Greatest Common Factor (GCF) from all terms in 12y3+12y2−45y12y^3 + 12y^2 - 45y.
The GCF is 3y3y. Factoring it out leaves 3y(4y2+4y−15)3y(4y^2 + 4y - 15).
Extracting the GCF first simplifies the polynomial and is a required step for complete factorization.
2
Factor the remaining non-monic trinomial, 4y2+4y−154y^2 + 4y - 15, by finding two numbers that multiply to a×ca \times c and add to bb.
We need numbers that multiply to (4)(−15)=−60(4)(-15) = -60 and add to 44. The numbers are 1010 and −6-6.
This prepares the trinomial for factoring by grouping.
3
Rewrite the middle term (4y4y) using these two numbers and factor by grouping.
4y2+10y−6y−15=2y(2y+5)−3(2y+5)=(2y−3)(2y+5)4y^2 + 10y - 6y - 15 = 2y(2y + 5) - 3(2y + 5) = (2y - 3)(2y + 5).
Grouping extracts common binomial factors, breaking down the quadratic into its simplest linear binomials.
4
Combine the original GCF with the newly factored binomials.
3y(2y−3)(2y+5)3y(2y - 3)(2y + 5)
To represent the full and complete factorization of the original polynomial.

Anahtar Kavram

Factoring polynomials by first extracting the GCF and subsequently using the ac-method (or grouping) to factor a non-monic quadratic trinomial.
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