Very quick maths question (1 Viewer)

[hscishard]

I found this question quite interesting. Note this isn't the full question, but the part that made it interesting.

Without using a 3 unit method, Fully factorise:

 

[mirakon]

Does subbing in values for x such as 1 to see if remainder is zero count as 3 unit method?

Because by subbing x=1, the polynomial equals zero and then (x-1) is hence a factor. Use long division to divide the initial polynomial by (x-1) and the quadratic quotient should be easy to factorise from there.

Hope I helped

 

[lolman12567]

2x³-14x +12

=2(x³- x - 6x+6)
=2(x(x+1)(x-1) - 6(x-1))
=2(x-1)(x² + x -6)
=2(x-1)(x-2)(x+3)

without 3unit methods

 

[x jiim]

is it bad that I can't think of a 3u method of doing that? o___o

 

[lolman12567]

i'm guessing 3unit method would be what mirakon used, factor theorem.

 

[x jiim]

..that was 3u? O__O

 

[lolman12567]

yeah im pretty sure long division and factor theorem is 3unit

 

[mirakon]

Yeah i was wondering because i learnt it in year 10 at school but maybe it was just an extension topic because we finished the course early, i don't know.

Anyway, good solution lolman12567. Very elegant

 

[hscishard]

Nice work. Its interesting how that method still works for cubics. Not many people use this method.

 

[thongetsu]

Nice lolman. I found that using the factor theorem made it SO MUCH easier.

 

[johNNyboy772]

wat's a '3 unit' method to do this question?

 

[mioumiou]

yeah im pretty sure long division and factor theorem is 3unit

<HR style="COLOR: #d1d1e1; BACKGROUND-COLOR: #d1d1e1" SIZE=1> wat's a '3 unit' method to do this question?

Using the polynomial methods?

 

[hscishard]

Yep using polynomials