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Zorluk: OrtaFactoring Polynomials

During an algebra calculation, an engineer needs to rewrite the expression 2x2+7x−152x^2 + 7x - 15 in its completely factored form. Which of the following is equivalent to this expression?

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

Cevap

The correct factorization is (2x−3)(x+5)(2x - 3)(x + 5).
The correct answer is mathematically equivalent to the original polynomial. Expanding it using the FOIL method confirms this: (2x)(x)+(2x)(5)+(−3)(x)+(−3)(5)=2x2+10x−3x−15=2x2+7x−15(2x)(x) + (2x)(5) + (-3)(x) + (-3)(5) = 2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15.

Adım Adım Çözüm

1
Identify the coefficients of the quadratic expression ax2+bx+cax^2 + bx + c.
a=2a = 2, b=7b = 7, and c=−15c = -15.
Identifying these values is required to determine the target product and sum for factoring by grouping.
2
Find two numbers that multiply to a×ca \times c and add up to bb.
The numbers are 1010 and −3-3 because 10×(−3)=−3010 \times (-3) = -30 and 10+(−3)=710 + (-3) = 7.
These specific numbers allow us to logically split the middle term of the polynomial.
3
Rewrite the middle term 7x7x using these two numbers and factor the polynomial by grouping.
2x2+10x−3x−15=2x(x+5)−3(x+5)=(2x−3)(x+5)2x^2 + 10x - 3x - 15 = 2x(x + 5) - 3(x + 5) = (2x - 3)(x + 5).
Extracting the greatest common factor from each pair reveals the common binomial factor, completing the factorization.

Anahtar Kavram

Factoring non-monic quadratic trinomials by grouping
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