dividing polynomal help qn (1 Viewer)

[ronaldinho]

x^3 - x^2 + x + 1 divide by 2x -3

how i do this?

 

[Trebla]

You use fraction (i.e. non integer) co-efficents. For example the first term of the quotient is 0.5x². An alternative method is to divide both 2x - 3 and x³ - x² + x + 1 by 2 and then perform the long division. In other words
(x³ - x² + x + 1)/(2x - 3)
= 2(x³/2 - x²/2 + x/2 + 1/2)/2(x - 3/2)
= (x³/2 - x²/2 + x/2 + 1/2)/(x - 3/2)
So divide x³/2 - x²/2 + x/2 + 1/2 by x - 3/2 should give the same answer.
You should get x²/2 + x/4 + 7/8 with remainder of 29/8.

 

[pLuvia]

This is pretty much like the basic long division on normal integers but you just have to take into account the variables

 

[ronaldinho]

Why do you have to do this?

 

[zeek]

You divide a polynomial to reduce it to its factors

 

[pLuvia]

Sometimes you can reduce the amount of working for long division if you know the factor theorem which finds more zeros to the polynomial hence you have less zeros to find

 

[haque]

since ur dividing by a linear function the degree of the remainder will be 0, thus it is a constant so u only use the factor theorem for this case

 

[jyu]

Why do you have to do this?

This is pretty much like changing an improper fraction to a mixed number:

e.g. 5/3 = 1 + 2/3

Your example:

(x³ - x² + x + 1)/ (2x - 3) = x²/2 + x/4 + 7/8 + (29/8)/(2x - 3)

The mixed form on the RHS is a useful form in some situations,

e.g. in finding integral

integral[(x³ - x² + x + 1)/ (2x - 3)]dx
= integral[x²/2 + x/4 + 7/8 + (29/8)/(2x - 3)]dx


e.g. in sketching graph

Sketching (x³ - x² + x + 1)/ (2x - 3) is the same as sketching
x²/2 + x/4 + 7/8 + (29/8)/(2x - 3), which is the addition of a parabola and a hyperbola.


:wave:

 

[Trebla]

Using the factor or remainder theorem only helps you determine the remainder. You need to apply long division or other means of manipulation to find the factor/quotient.

 

[haque]

Sorry bout that trebla, just a misunderstanding-i thought he only wanted to find the remainder-but of course ur right

 

[bos1234]

but why do u need to make it in..

x+b/a

y cant u divide by ax+b??

 

[Riviet]

but why do u need to make it in..

x+b/a

y cant u divide by ax+b??

Because you can't divide 1x by ax (where a>1) and get an integer out of it.

 

[jyu]

Because you can't divide 1x by ax (where a>1) and get an integer out of it.

But if you divide e.g. x^3 + ..... by (ax + b), you 'll get (1/a)x^2 + ....

:wave:

 

[Slidey]

I suggest that you learn long division but don't use it.

Use synthetic division instead. I stickied a thread about this somewhere. Let me find it.

Hmm. I can't find it. Maybe I accidentally unstickied it. Let me make one, then!

Anyway, here is the link: http://www.purplemath.com/modules/synthdiv.htm

It's a great time saver, as it provides both the dividend and remainder, as with long division. It works for all divisors, however it's only a time saver for linear divisors and you won't find anybody to teach it for quadratic and higher divisors anyway. That is: use normal long division for quadratic divisors.