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# MS Algebra

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If x3 and x2 don't exist in the polynomial (x3+mx+n) (x2-3x+4);

What is the value of m, n?

ISmellGood  Aug 28, 2017
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#1
+91038
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If x3 and x2 don't exist in the polynomial (x3+mx+n) (x2-3x+4);

What is the value of m, n?

$$(x^3+mx+n) (x^2-3x+4)\\ x^3\;\;terms: 4x^3+mx*x^2= 4x^3+mx^3= (4+m)x^3 \\ x^2\;\;terms: -mx*3x+nx^2=-3mx^2+nx^2=(-3m+n)x^2\\~\\ 4+m=0\\ m=-4\\ -3m+n=0\\ n=3m\\ n=3*-4\\ n=-12\\~\\ n=-12,\qquad m=-4$$

Melody  Aug 28, 2017
edited by Melody  Aug 28, 2017
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Amazing! Thanks!

ISmellGood  Aug 28, 2017
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(x^3+mx+n) (x^2-3x+4)=

x^5 - 3x^4 + 4x^3

+ mx^3 -3mx^2 + 4mx

+ nx^2   - 3nx   + 4n

_______________________________

x^5  -  3x^4 + (4 + m)x^3 + (-3m + n)x^2 + (4m -3n)x + 4n

4 + m  = 0  →  m = -4

-3m+ n = 0 →  -3(-4) + n  = 0  →  12 + n = 0  →  n  = -12

Proof

(x^3 - 4x - 12) (x^2 - 3x + 4)  =

x^5 - 3 x^4 + 20 x - 48      .....no x^3  or x^2  terms in the product polynomial...!!!

CPhill  Aug 28, 2017

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