Question

(7x+3)(5x^(3)-4x^(2)-6x+8)
and (m^(2)+4mn-4n^(2))(3m^(2)-3mn+5n^(2)) =

Answer 1

it says to solve it using synthetic multiplication

Answer 2

Or synthetic division?

Answer 3

no it was multiplication

Answer 4

i know there is only synthetic division that's why i'm confused

Answer 5

Ah interesting!
I had never heard of this either.
But I found a website explaining it.
http://astarmathsandphysics.com/ib-maths-notes/polynomials/ib-maths-notes-polynomials-synthetic-multiplication.html

\[\Large\rm (7x+3)(5x^3-4x^2-6x+8) \]

Answer 6

Similar to synthetic division, it's important that we have our powers of x in `descending order`, with NO powers missing.

So if you had something like:\[\Large\rm x^2+2\]You would want to think of it as:\[\Large\rm x^2+0x+2\]So that way you can see that you have a 0 coefficient.

Answer 7

So for our problem, everything is in order already.
Looks like we want to multiply \(\Large\rm 73\) and \(\Large\rm 5468\), yes?

Ooo I'm not sure how the negatives work actually.. hmmm

Answer 8

i really dont understand this

Answer 9

how did you get 73 and 5468?

Answer 10

`Synthetic multiplication is limited and cannot handle carried digits`
From the website^

So we can't actually use synthetic multiplication :( Hmmmm

Answer 11

is there another way to solve this?

Answer 12

\[\Large\rm (7x+3)(5x^3-4x^2-6x+8)\]
See the coefficients?
7 3 5 4 6 8

Answer 13

Yes of course :) we can do normal people math lol

Answer 14

yes lol

Answer 15

1. (7x+3)(5x^(3)-4x^(2)-6x+8)
2. (m^(2)+4mn-4n^(2))(3m^(2)-3mn+5n^(2)) =

Answer 16

3x^5+9x^4+9x^3+x^2+6x+4

Answer 17

i actually got 35^(4)-13x^(3)-54x^(2)+38x+24

Answer 18

So this would be the first set of distributions, giving the 7x to each term in the second set of brackets.