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Question

Difficulty: HardFactoring Polynomials

Match each non-monic quadratic polynomial to its completely factored form. Pay close attention to the signs in each binomial factor.

  • 2m2−11m+152m^2 - 11m + 15(2m−5)(m−3)(2m - 5)(m - 3)
  • 2m2+m−152m^2 + m - 15(2m−5)(m+3)(2m - 5)(m + 3)
  • 2m2+11m+152m^2 + 11m + 15(2m+5)(m+3)(2m + 5)(m + 3)
  • 2m2−m−152m^2 - m - 15(2m+5)(m−3)(2m + 5)(m - 3)

Answer

The correct pairings are: 2m2−11m+152m^2 - 11m + 15 with (2m−5)(m−3)(2m - 5)(m - 3); 2m2+m−152m^2 + m - 15 with (2m−5)(m+3)(2m - 5)(m + 3); 2m2+11m+152m^2 + 11m + 15 with (2m+5)(m+3)(2m + 5)(m + 3); and 2m2−m−152m^2 - m - 15 with (2m+5)(m−3)(2m + 5)(m - 3).
Each polynomial requires finding two numbers that multiply to the product of the leading coefficient and the constant term, and add to the middle coefficient. The correct pairings map the expanded polynomials directly to their unique, fully factored equivalents.

Step-by-Step Solution

1
Analyze the signs of the constants in each polynomial.
Polynomials with a positive constant term (+15) must have binomial factors with identical signs (both positive or both negative). Polynomials with a negative constant term (-15) must have binomial factors with opposite signs.
The product of the constant terms in the binomials must equal the constant term of the polynomial.
2
Factor 2m2−11m+152m^2 - 11m + 15.
Since the constant is positive and the middle term is negative, both factors must be negative. 2m2−6m−5m+15=2m(m−3)−5(m−3)=(2m−5)(m−3)2m^2 - 6m - 5m + 15 = 2m(m - 3) - 5(m - 3) = (2m - 5)(m - 3).
Applying factoring by grouping.
3
Factor 2m2+m−152m^2 + m - 15.
The constant is negative, so signs differ. The middle term is positive, so the larger product in FOIL must be positive. 2m2+6m−5m−15=2m(m+3)−5(m+3)=(2m−5)(m+3)2m^2 + 6m - 5m - 15 = 2m(m + 3) - 5(m + 3) = (2m - 5)(m + 3).
Applying factoring by grouping.
4
Factor 2m2+11m+152m^2 + 11m + 15.
Since both the constant and middle term are positive, both factors must be positive. 2m2+6m+5m+15=2m(m+3)+5(m+3)=(2m+5)(m+3)2m^2 + 6m + 5m + 15 = 2m(m + 3) + 5(m + 3) = (2m + 5)(m + 3).
Applying factoring by grouping.
5
Factor 2m2−m−152m^2 - m - 15.
The constant is negative, so signs differ. The middle term is negative, so the larger product in FOIL must be negative. 2m2−6m+5m−15=2m(m−3)+5(m−3)=(2m+5)(m−3)2m^2 - 6m + 5m - 15 = 2m(m - 3) + 5(m - 3) = (2m + 5)(m - 3).
Applying factoring by grouping.

Key Concept

Factoring Non-Monic Trinomials and Sign Analysis
Estimated Time:2m 30s
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Topics

Factoring Polynomials

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