Determine the values of m and n so that the polynomials 2x^3 + mx^2 +nx -3 AND x^3 - 3mx^2 + 2nx + 4 are both divisible by x -2
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2 m n -3 | 2 1 -3m 2n 4 | 2
4 8+2m 16+4m+2n 2 4-6m 8-12m+4n)
2 (4+m) (8+2m+n) (16+4m+2n-3) 1 (2-3m) (4-6m+2n) (8-12m+4n+4)
16+4m+2n-3=0 and 8-12m+4n+4=0
4m+2n=-13 -12m+4n=-12
- 6m+2n=-6
10m=-7
m=-7/10
n=-51/10
4 8+2m 16+4m+2n 2 4-6m 8-12m+4n)
2 (4+m) (8+2m+n) (16+4m+2n-3) 1 (2-3m) (4-6m+2n) (8-12m+4n+4)
16+4m+2n-3=0 and 8-12m+4n+4=0
4m+2n=-13 -12m+4n=-12
- 6m+2n=-6
10m=-7
m=-7/10
n=-51/10
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Pretty easy way would be to apply long division to both polynomials, and then equal the remainders to zero, we would get the equation from each polynomials.
I hope you are able to do long division, because its impossible to show it in the YA.
So for the first equation I get:
(x-2)( 2x^2 + x(4+m) + n+8+2m) + 2(n+8+2m) = 2x^3 + mx^2 +nx -3
So if the remainder is equal to zero, we got it factored: 2(n+8+2m) = 0
Another equation is from another polynomial :
(x-2)(x^2 + x(2-3m) + 2(2-3m+n) + 4(2-3m+n) = x^3 - 3mx^2 + 2nx + 4
Remainder must once again be equal to zero: 4(2-3m+n) = 0
So we got an linear equation system:
2-3m+n = 0
n+8+2m=0
And if I got it right: m = -6/5, n = -28/5
I hope you are able to do long division, because its impossible to show it in the YA.
So for the first equation I get:
(x-2)( 2x^2 + x(4+m) + n+8+2m) + 2(n+8+2m) = 2x^3 + mx^2 +nx -3
So if the remainder is equal to zero, we got it factored: 2(n+8+2m) = 0
Another equation is from another polynomial :
(x-2)(x^2 + x(2-3m) + 2(2-3m+n) + 4(2-3m+n) = x^3 - 3mx^2 + 2nx + 4
Remainder must once again be equal to zero: 4(2-3m+n) = 0
So we got an linear equation system:
2-3m+n = 0
n+8+2m=0
And if I got it right: m = -6/5, n = -28/5