Am I expanding (2x+3)(x-5) right?

When I expand brackets I feel like I’m handing out coupons to both terms, but negatives trip me up – is (2x+3)(x-5) = 2x^2 -10x + 3x -15, or am I missing something?

3 Responses

  1. You’re expanding it correctly; the only step left is to combine like terms. Distribute 2x across (x − 5) to get 2x·x = 2x^2 and 2x·(−5) = −10x, then distribute +3 across (x − 5) to get 3·x = 3x and 3·(−5) = −15. Putting these together gives 2x^2 − 10x + 3x − 15, and combining the x-terms yields 2x^2 − 7x − 15. Quick sign check: the x^2 term is positive (positive times positive) and the constant is negative (positive times negative), which matches. Want to try one more, maybe with both binomials having negatives, to reinforce the sign handling?

  2. You’re right on track with the “handing out coupons” idea! Distribute each term in the first bracket to both terms in the second. So (2x+3)(x−5) becomes 2x·x = 2x^2, 2x·(−5) = −10x, 3·x = 3x, and 3·(−5) = −15. Put those together: 2x^2 −10x + 3x −15, and then combine the like terms in the middle to get 2x^2 − 7x − 15. The only place people usually stumble is the signs-just remember the little multiplication rules: plus×minus = minus, minus×minus = plus.

    Quick extra example to lock it in: (x−4)(x−2). Multiply across: x·x = x^2, x·(−2) = −2x, (−4)·x = −4x, and (−4)·(−2) = +8. Add them up: x^2 − 2x − 4x + 8 = x^2 − 6x + 8. Same “coupon” idea, and notice how the last term turned positive because a negative times a negative is a positive. You’ve got it!

  3. Your coupon metaphor is adorable and totally on the right track! Think of x − 5 as x + (−5), then “hand out” each part of 2x + 3 to each part of x − 5. So you get:
    – 2x·x = 2x^2
    – 2x·(−5) = −10x
    – 3·x = 3x
    – 3·(−5) = −15
    Putting them together: 2x^2 − 10x + 3x − 15, and then combine like terms to get 2x^2 − 7x − 15. So you weren’t missing anything-just the final tidy-up step!

    Some folks call this FOIL (First, Outer, Inner, Last), and others call it the commutative property-okay, I always mix those names up-but the idea is simply “distribute each term to each term.” If you want a quick gut-check, plug in a value like x = 1: (2+3)(1−5) = 5·(−4) = −20, and 2(1)^2 − 7(1) − 15 = 2 − 7 − 15 = −20. Match achieved. Hope this helps!

Leave a Reply

Your email address will not be published. Required fields are marked *

Join Our Community

Ready to make maths more enjoyable, accessible, and fun? Join a friendly community where you can explore puzzles, ask questions, track your progress, and learn at your own pace.

By becoming a member, you unlock:

  • Access to all community puzzles
  • The Forum for asking and answering questions
  • Your personal dashboard with points & achievements
  • A supportive space built for every level of learner
  • New features and updates as the Hub grows