# Which are the numbers m and n if f=x^4-4x^3+4x^2+mx+n is dividede by g=x^2-4x+3?

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f(x)x^4-4x^3+4x+mx+n

g(x)= x^2-4x+3

f is divided by g ==> the roots of g should verify f

g(x) = x^2-4x +3

= (x-3)(x-1)

==> x = {1, 3} ==> f(1) =f(3)=0

f(1) = 1 -4(1)+4(1)+m(1)+n= 0

1+m+n=0 .............(1)

f(3) = 81-108 +36+3m +n = 0

= 9 +3m+n=0 .........(2)

subtract (1) from (2)

==> 8 +2m = 0

==> 2m = -8

==> m = -4

substitute in (1)

1-4+n = 0

==> n = 3

==> f(x) = x^4-4x^3 +4x^2-4x +3

f(x) = x^4-4x^3+4x^2+mx+n is divided by x^2-4x+3. To find m and n.

Solution:

x^2-4x+3 = (x-3)(x-1).

Since f(x) is divisible by x^2-4x+3, f(x) is divisible by x-3 and x-1.

Therefore the remainder when divided ny x-3 is f(3) = 0. Or

f(3) = 3^4-4*3^3+4*3^2+m(3)+n = 0. Or

81-108+36+3m+n = 0. Or

3m+n = 9.......................(1).

Also the remainder when f(x) divided by x-1 is f(1) is zero. Or

f(1) = 1^4-4*1^3+4*1^2+m(1)+n = 0. Or

1-4+4+m+n = 0. Or

m+n = -1.......................(2).

Solving for m and n from the equations (1) and (2):

(1)-(2) gives: 2m = 9-(-1) = 10. Or m =5 and substituting m =5 in (1), we get, 3*5+n=9. Or n = 9-15 = -6.

We'll write the rule of division with reminder.

f=g*C+R

If the polynomial f is divided by g, that means that the roots of g are verifying the polynomial f, so that:

f(x1)=0 and f(x2)=0, where x1 and x2 are the solutions of g(x).

We'll find the roots of g=x^2-4x+3

x^2-4x+3=0

We'll write the quadratic formula:

x1=[4+sqrt(16-12)]/2

x1=(4+2)/2

x1=3

x2=(4-2)/2

x2=1

Now, we'll apply the rule:

f(x1)=0

f(3)=81-108+36+3m+n

81-108+36+3m+n=0

3m+n=-9 (1)

f(1)=1-4+4+m+n

m+n=-1 (2)

We'll multiply the relation (2) by (-1) and we'll add it to (1).

3m-m+n-n=-9+1

2m=-8

**m=-4**

Now, we'll substitute m into the relation (2).

-4+n=-1

n=4-1

**n=3**

**f(x)=x^4-4x^3+4x^2-4x+3**