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) $\displaystyle a(x) = x^2-2x+4$, $\displaystyle b(x) = 2x^5-x^4+3x^3-2x+1$
any help would be much appreciated.
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) $\displaystyle a(x) = x^2-2x+4$, $\displaystyle b(x) = 2x^5-x^4+3x^3-2x+1$
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.
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
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x^2 - 2x + 4 )2x^5 - 1x^4 + 3x^3 + 0x^2 - 2x + 1
2x^5 - 4x^4 + 8x^3
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3x^4 - 5x^3 + 0x^2 - 2x + 1
3x^4 - 6x^3 + 12x^2
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1x^3 - 12x^2 - 2x + 1
1x^3 - 2x^2 + 4x
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-10x^2 - 6x + 1
-10x^2 + 20x - 40
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Since the remainder will obviously be linear (once you complete the division), r(x) will fit the requirements.