chemistry1111 Expand then factorising LHS gives (a+b)x + (2a + 3b) = 18x + 8 --> by comparing coefficients, you obtain a+b = 18 and 2a+3b =8, solve simultaneously for a and b, giving a=46 and b=-28

    thank you so much that makes sense

    also this question
    For the polynomial P(x)=(a+1)x3 + (b-7)x2 +c+5, find values for a,b and c if P(x) has a degree 2, a leading coefficient of 3 and the constant term is -1

    when factorizing polynomials in methods unit 3/4, what method do most people use vcaa prefer, long division or equating coefficients

    9 days later

    Do we need to buy the new Methods 3/4 book for the new study design (2023)?
    I feel like the textbook questions are the same...

    4 days later

    PizzaMaster
    Howdy, I did QCE Methods, so if I mess this up, I’m really sorry.

    https://ibb.co/m0Q1xdV

    For b) may I know what the answers are? I may be able to work backwards 🙂

    I didn’t do d) because I wasn’t that confident in doing the question.

    Hope this helps,
    -jinx_58

      20 days later

      Hi jinx_58,

      Thank you so much for solving!
      The answer to part b is Domain = [0, 6]; Range = [0, 9/2]
      The answer to part c is 9/2

      The graph of y = 5/x -6 is reflected in the x-axis and
      then in the y-axis. The equation of the final image is:

      could someone help with this question

        chemistry1111

        1. reflected in the x-axis
        2. reflected in the y-axis

        **I'm not quite sure if y = (5/x) -6 or if it's y = 5/(x-6) so I'll show the working for both

        let f[x] = (5/x) -6

        1. x reflection: -f[x] = f1[x] = -[ (5/x) - 6 ] = -( 5/x ) + 6

        2. y reflection: f1[-x] = f2[x] = -(5/ [-x] ) +6 = -(5/ x ) +6

        let g[x] = 5/(x-6)

        1. x reflection: -g[x] = f1[x] = -[ 5/(x-6) ] = -5/(x-6)

        1. y reflection: g1[-x] = f2[x] = -5/( [-x] - 6) = -5/( -x - 6)

        hope that helps!

          Hi everyone,

          Is (2x-1) the same as (x-1/2)
          I mean they both give x= 1/2
          But let's say if we had to factorise 12x3 + 20x2 − x − 6
          and we get (2x-1)(2x+3)(3x+2)

          Can I also write this as:
          (x-1/2)(2x+3)(3x+2)?

          ALSO can some please help and find the roots of this: 12x3 + 20x2 − x − 6
          using rational root theorem

          much appreciated!!!

            PizzaMaster
            good question! to answer your first one, (2x-1) = 2 (x-1/2). they're not the same but the same x value will make both of them equal (0). I'm not home rn so I'll reply to the second one when I have access to my iPad.

            EDIT
            I think it's easier if you watch a video, here's a good one I found (watching 0:00-2:40 should be enough): https://www.youtube.com/watch?v=4XytYH35AP0

              So I can write (2x-1)(2x+3)(3x+2) as 2(x-1/2)(2x+3)(3x+2) NOT (x-1/2)(2x+3)(3x+2)
              Great, thanks crumblycupcakes!

              With rational root theorem, I find that it is super time consuming... after we find the possible roots, is there a shortcut to know which ones to try first?
              For example with 12x3 + 20x2 − x − 6:
              the possible roots are +- [1/2, 1/3, 1/4, 1/6, 1/12, 2/3, 3/4, 3/2]
              That is a lot! Is there a way to know which one we can try first?

              Also does simplifying work?
              for example with 12x3 + 20x2 − x − 6, we do like factors of 6/ factors of 12. Can we do factors of 1/ factors of 2 (since 6/12 = 1/2?)

              Billzene
              I think that would only be the case in Specialist maths with VCAA so 'no solutions' is usually accepted in methods because complex numbers aren't taught in it. BUT we can't say for sure so it's always good to say 'no real solutions'. 😅

              [ Edit ]
              istg VCAA is so weird. They changed the study design for spesh and they included proofs but didn't specify if it was matrix proofs, number proofs or circle proofs or all of them. Apparently my teacher emailed them a couple of times but they never replied.

                5 days later

                Hi, could someone please help me with this question. I am pretty sure the answer is 150-t but i'm not too sure how I got there and what the working out should look like.

                Towns A and B are 150 km apart along a straight road. A car leaves town A and heads towards town B at 60 km/h. The distance, in kilometres, of the car from town B after t minutes is...

                Thanks

                  :) It sounds like you might have caught on to the right idea. We know the car travels at 60 km/h which equals 1 km/minute. Each minute, the car travels 1 km, therefore in t minutes the car travels t km. Therefore the distance from town B is 150-t km.
                  I feel that the following working should be sufficient, but it might depend on mark allocation.
                  distance = 150 - (60 / 60) * t = 150 - t

                    2 months later

                    hey guys! I was just wondering if any of you know if the methods checkpoints books are designed to be done using a calculator or without one? thank you!!