Question

Difficulty: HardFactoring Polynomials

To optimize a packaging design, a technician models the surface area using the algebraic expression 18xy2−24xy−10x18xy^2 - 24xy - 10x. What is the fully factored equivalent of this expression?

  1. 2x(3y−5)(3y+1)2x(3y - 5)(3y + 1)Answer
  2. B
    2x(3y+5)(3y−1)2x(3y + 5)(3y - 1)
  3. C
    x(6y−10)(3y+1)x(6y - 10)(3y + 1)
  4. D
    2xy(9y−17)2xy(9y - 17)

Answer

The fully factored equivalent is 2x(3y−5)(3y+1)2x(3y - 5)(3y + 1).
The expression is fully factored by first extracting the greatest common factor of 2x2x, resulting in 2x(9y2−12y−5)2x(9y^2 - 12y - 5). Then, using the ac-method, the remaining non-monic quadratic trinomial is factored into (3y−5)(3y+1)(3y - 5)(3y + 1), yielding the final correct product.

Step-by-Step Solution

1
Identify and extract the greatest common factor (GCF) from all three terms.
2x(9y2−12y−5)2x(9y^2 - 12y - 5)
Extracting the GCF simplifies the polynomial and prepares the remaining quadratic trinomial for factoring.
2
Identify the components of the remaining quadratic trinomial ay2+by+cay^2 + by + c.
Here, a=9a = 9, b=−12b = -12, and c=−5c = -5.
This establishes the values needed to use the ac-method for factoring non-monic quadratics.
3
Find two numbers that multiply to acac and add to bb.
ac=9×−5=−45ac = 9 \times -5 = -45. The two numbers that multiply to −45-45 and add to −12-12 are −15-15 and 33.
These numbers allow the middle term to be split, enabling factoring by grouping.
4
Rewrite the middle term and factor the trinomial by grouping.
9y2−15y+3y−5=3y(3y−5)+1(3y−5)=(3y−5)(3y+1)9y^2 - 15y + 3y - 5 = 3y(3y - 5) + 1(3y - 5) = (3y - 5)(3y + 1)
This completes the factorization of the trinomial portion of the expression.
5
Combine the originally extracted GCF with the factored binomials.
2x(3y−5)(3y+1)2x(3y - 5)(3y + 1)
This yields the complete and fully factored form of the original polynomial.

Key Concept

Factoring Polynomials with a GCF
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