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Factor completely 18x2 − 21x −15.Group of answer choices3(2x + 1)(3x − 5)3(2x − 5)(3x + 1)3(2x − 1)(3x + 5)3(6x + 1)(x − 5)

Question

Factor completely 18x² − 21x − 15.

Group of answer choices

  • 3(2x + 1)(3x − 5)
  • 3(2x − 5)(3x + 1)
  • 3(2x − 1)(3x + 5)
  • 3(6x + 1)(x − 5)
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Solution

To factor the quadratic equation 18x^2 - 21x - 15 completely, we first look for the greatest common factor (GCF). The GCF of 18, 21, and 15 is 3. So, we can factor out 3 from the equation:

3(6x^2 - 7x - 5)

Next, we need to factor the quadratic equation 6x^2 - 7x - 5. We look for two numbers that multiply to (6*-5)=-30 and add to -7. The numbers -10 and 3 fit these conditions because (-10)*3=-30 and -10+3=-7.

So, we can write the quadratic equation as:

3(6x^2 - 10x + 3x - 5)

Next, we group the terms and factor by grouping:

3[(6x^2 - 10x) + (3x - 5)]

This simplifies to:

3[2x(3x - 5) + 1(3x - 5)]

We can then factor out (3x - 5) to get:

3[(2x + 1)(3x - 5)]

So, the completely factored form of the equation 18x^2 - 21x - 15 is 3(2x + 1)(3x - 5).

This problem has been solved

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