**FAN AND MEDAL**** What type of polynomial is x(x - 6)(x - 7)? and what is its degree?
You need to expand it by multiplying.
So (x^2-6x)(x-7) Then x^3-6x^2-7x^2+42 Now combine the like terms.
ok, so i got 2x^2 - 13x^2 - 13
you need to do them individually I think, so x(x-6) first. Then that times (x-7)
thats not how you do this you do this: (x - 6)(x-7) then you get: 2x - 13x - 13, then you multiply that by x and get 2x^2 - 13x^2 -13
Okay i screwed up lol.
well if that's the answer, the degree is whatever the highest exponent is. So now you know that.
so the degree is 2, so is it a trinomial?
Well hang on, when you do the two binomials together you got 2x^2 - 13x - 13, right?
yeah
Then you need to multiply them all by x if you haven't yet, so it'll be 2x^3 - 13x^2 - 13x.
So it would be to the 3rd degree.
READ WHERE I EXPLAINED HOW TO DO IT, IT HAS ALREADY BEEN MULTIPLYED BY X
thats not how you do this you do this: (x - 6)(x-7) then you get: 2x - 13x - 13, then you multiply that by x and get 2x^2 - 13x^2 -13. That's what you wrote, but when you do the first ones x times x, that should be x^2.
so it would be 2x^2 - 13x - 13 times x.
a trinomial is a polynomial with 3 terms, so does - 13 count as a term.....Where did the extra x come from?
- 13 does count, and I'm not too sure what you mean by the extra x...
were you trying to say i forgot the x that oes on the -13? because i just noticed that
And i say extra x because I already multiplyed the whole equation by x and then you multiplyed it by x a SECOND time which is why i ask where you got the second x
Well sort of... What I'm saying is that when you were multiplying the first two binomials, (x - 6)(x - 7) it should have equaled x^2-7x-6x+42.
which is x^2-13x+42.
and that's before you multiplied it by x, so it should be x(x^2-13x+42)
that doesnt change that its a second degree trinomial
Yes it does, It would make it to the third degree.
because (x)(x^2) = x^3
oh ok
So is that it?
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