Thread: finding polynomials

For the following, find polynomials q(x) and r(x) such that b(x) = q(x)a(x)+r(x), where r(x) = 0 or deg r(x) < deg a(x).

(a) ,

any help would be much appreciated.

Originally Posted by jvignacio

For the following, find polynomials q(x) and r(x) such that b(x) = q(x)a(x)+r(x), where r(x) = 0 or deg r(x) < deg a(x).

(a) ,

any help would be much appreciated.

deg r(x) < deg a(x).
so r(x) must be of degree 1.

b(x) is of degree 5, so q(x)a(x) must be of degree 5.
so we get q(x) of degree 3.

let q(x) = ax^3 + bx^2 + cx + d
and
r(x) = ex + f.

so we have

2x^5 - x^4 + 3x^3 - 2x + 1 = (x^2 - 2x + 4)(ax^3 + bx^2 + cx + d) + ex +f

Now equate the co-efficient of each power of x from LHS and RHS.
You will get 6 equations (linear0 and 6 variables. solve for a,b,c,d,e and f to get both polynomials.

Originally Posted by jvignacio

For the following, find polynomials q(x) and r(x) such that b(x) = q(x)a(x)+r(x), where r(x) = 0 or deg r(x) < deg a(x).

(a) ,

This is just asking you to use do the polynomial long division, and then solve for the needed information. This is because [bthe Remainder Theorem[/b] relates the division to the factoring, etc, etc, as:

. . . . ."b(x) = q(x)a(x) + r(x)"
. . . . .is the same as
. . . . ."b(x)/a(x) = q(x) + r(x)/a(x)"

(This is just like "13 = 2*5 + 3" meaning the same as "13/5 = 2 + 3/5", by the way.)

[HTML] 2x^3 + 3x^2 + 1x - 10
--------------------------------------
x^2 - 2x + 4 )2x^5 - 1x^4 + 3x^3 + 0x^2 - 2x + 1
2x^5 - 4x^4 + 8x^3
--------------------------------------
3x^4 - 5x^3 + 0x^2 - 2x + 1
3x^4 - 6x^3 + 12x^2
--------------------------------------
1x^3 - 12x^2 - 2x + 1
1x^3 - 2x^2 + 4x
--------------------------------------
-10x^2 - 6x + 1
-10x^2 + 20x - 40
--------------------------------------[/HTML]
Since the remainder will obviously be linear (once you complete the division), r(x) will fit the requirements.