Math  /  Algebra

QuestionPRIMERA PARTE:
1. A) Sea M(x)(m+k)x3+(3mk)x2+3kx+35M(x) \equiv(m+k) x^{3}+(3 m-k) x^{2}+3 k x+35, hallar mm y kk sabiendo que x=5x=-5 es raíz de M(x)M(x) y que M(3)=80M(-3)=80.

Studdy Solution

STEP 1

STEP 2

Sustituir x=5 x = -5 en M(x) M(x) :
M(5)=(m+k)(5)3+(3mk)(5)2+3k(5)+35 M(-5) = (m+k)(-5)^3 + (3m-k)(-5)^2 + 3k(-5) + 35

STEP 3

Simplificar la expresión sustituida:
M(5)=(m+k)(125)+(3mk)(25)15k+35 M(-5) = (m+k)(-125) + (3m-k)(25) - 15k + 35 M(5)=125(m+k)+25(3mk)15k+35 M(-5) = -125(m+k) + 25(3m-k) - 15k + 35
Dado que x=5 x = -5 es una raíz, M(5)=0 M(-5) = 0 .
0=125(m+k)+75m25k15k+35 0 = -125(m+k) + 75m - 25k - 15k + 35

STEP 4

Simplificar y reorganizar la ecuación:
0=125m125k+75m40k+35 0 = -125m - 125k + 75m - 40k + 35 0=50m165k+35 0 = -50m - 165k + 35
Reorganizar para obtener una relación:
50m+165k=35 50m + 165k = 35

STEP 5

Sustituir x=3 x = -3 en M(x) M(x) :
M(3)=(m+k)(3)3+(3mk)(3)2+3k(3)+35 M(-3) = (m+k)(-3)^3 + (3m-k)(-3)^2 + 3k(-3) + 35

STEP 6

Simplificar la expresión sustituida:
M(3)=(m+k)(27)+(3mk)(9)9k+35 M(-3) = (m+k)(-27) + (3m-k)(9) - 9k + 35 M(3)=27(m+k)+27m9k+35 M(-3) = -27(m+k) + 27m - 9k + 35
Dado que M(3)=80 M(-3) = 80 :
80=27(m+k)+27m9k+35 80 = -27(m+k) + 27m - 9k + 35

STEP 7

Simplificar y reorganizar la ecuación:
80=27m27k+27m9k+35 80 = -27m - 27k + 27m - 9k + 35 80=36k+35 80 = -36k + 35
Reorganizar para obtener una ecuación:
36k=3580 36k = 35 - 80 36k=45 36k = -45 k=4536 k = -\frac{45}{36} k=54 k = -\frac{5}{4}

STEP 8

Sustituir k=54 k = -\frac{5}{4} en la ecuación 50m+165k=35 50m + 165k = 35 :
50m+165(54)=35 50m + 165\left(-\frac{5}{4}\right) = 35 50m8254=35 50m - \frac{825}{4} = 35
Multiplicar toda la ecuación por 4 para eliminar el denominador:
200m825=140 200m - 825 = 140 200m=965 200m = 965 m=965200 m = \frac{965}{200} m=19340 m = \frac{193}{40}
Los valores de m m y k k son:
m=19340,k=54 m = \frac{193}{40}, \quad k = -\frac{5}{4}

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