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Considere el polinomio $p(x)$ con parametros $t$ y $s$. $ p(x) = \\simplify{{coef2_x3}x^3+s*x^2+{coef2_x}x+t}\\text{.}$

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El polinomio $p(x)$:

\n", "advice": "

Nos dicen que el polinomio:

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a)

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primeramente, sustituyendo $x = \\simplify{-{c}}$  sobre $p(x)$ obtenemos 

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\\begin{align}
p(\\simplify{-{c}}) &= \\simplify[all,!collectNumbers, fractionnumbers]{{coef2_x3*(-c)^3}+{(-c)^2}s+{coef2_x*(-c)}+t},\\\\
&= \\simplify[all,fractionnumbers]{{coef2_x3*(-c)^3}+{(-c)^2}s+{coef2_x*(-c)}+t}.
\\end{align}

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pero, por el teorema del resto $p(\\simplify{-{c}}) = \\var{rem1}$ (usando la primera viñeta), entonces esto se convierte en

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\\begin{align}
\\simplify[all,fractionnumbers]{{coef2_x3*(-{c})^3}+s*{(-{c})^2}+{coef2_x*(-{c})}+t} &= \\var{rem1},\\\\
\\simplify[all,fractionnumbers]{s*{x}+t} &= \\simplify[all,fractionnumbers]{{rem1}-{coef2_x3*(-{c})^3}-{coef2_x*(-{c})}}.
\\end{align}

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b)

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Similarmente, sustituyendo $x = \\simplify{-{d}}$ sobre  $p(x)$, obtenemos

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\\begin{align}
p(\\simplify{-{d}}) &= \\simplify[all,!collectNumbers, fractionnumbers]{{coef2_x3*(-{d})^3}+{(-{d})^2}s+{coef2_x*(-{d})}+t},\\\\
&= \\simplify[all,fractionnumbers]{{coef2_x3*(-{d})^3}+{(-{d})^2}s+{coef2_x*(-{d})}+t}.
\\end{align}

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pero, por el teorema del resto $p(\\simplify{-{d}}) = \\var{rem2}$ (usando la segunda viñeta), entonces esto se convierte en

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\\begin{align}
\\simplify[all,fractionnumbers]{{coef2_x3*(-{d})^3}+s*{(-{d})^2}+{coef2_x*(-{d})}+t} &= \\var{rem2},\\\\
\\simplify[all,fractionnumbers]{s*{y}+t} &= \\simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}.
\\end{align}

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c)

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Tenemos ahora dos ecuaciones simultaneas para $s$ y $t$:

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\\begin{align}
\\simplify[all,fractionnumbers]{s*{x}+t} = \\simplify[all,fractionnumbers]{{rem1}-{coef2_x3*(-{c})^3}-{coef2_x*(-{c})}} \\\\
\\simplify[all,fractionnumbers]{s*{y}+t} = \\simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}
\\end{align}

\n

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A continuación, restamos la segunda ecuación de la primera ecuación.

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Esto nos permite cancelar los términos que involucran $ t $ y nos da una ecuación solo en términos de $ s $, que luego podemos reorganizar para encontrar el valor de $ s $.

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Restar las dos ecuaciones da:

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\\[\\simplify{s*{(-{c})^2-(-{d})^2}} = \\simplify[all,fractionnumbers]{{rem1 - coef2_x3*(-c)^3-coef2_x*(-c)-rem2+coef2_x3*(-d)^3+coef2_x*(-d)}}.\\]

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Entonces, podemos reorganizar esta ecuación para que

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\\[s = \\simplify[all,fractionnumbers]{{rem1 - coef2_x3*(-c)^3-coef2_x*(-c)-rem2+coef2_x3*(-d)^3+coef2_x*(-d)}/{{(-c)^2-(-d)^2}}}.\\]

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d)

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Podemos calcular $ t $ sustituyendo nuestro valor de $ s $ en una de nuestras ecuaciones simultáneas originales. Por ejemplo, usemos la ecuación

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\\[\\simplify[all,fractionnumbers]{s*{(-{d})^2}+t} = \\simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}.\\]

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Sustituyendo nuestro valor de $ s $ en esta ecuación nos da

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\\[
\\begin{align}
\\simplify[all,fractionnumbers,!noleadingMinus]{{numerator/denominator}+t} &= \\simplify[all,fractionnumbers]{{rem2-coef2_x3*(-d)^3-coef2_x*(-d)}},\\\\
t &= \\simplify[all,fractionnumbers]{{rem2-coef2_x3*(-d)^3-coef2_x*(-d) - numerator/denominator}}.
\\end{align}
\\]

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Esta misma respuesta también se hubiera obtenido si hubiésemos sustituido nuestro valor de $ s $ en la otra ecuación.

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Numerator of s

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Coefficient of x^3.

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Simplifies  first coefficient of s.

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Dividing term 2.

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First remainder.

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Use el teorema del resto para hallar el resto cuando $p(x) $ se divide por $(\\simplify {x + {c}}) $, cree una ecuación que incluya $ s $ y $ t $.

\n


[[0]]$s + t$ = [[1]].

\n

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El teorema del resto establece que si un polinomio $f(x)$ se divide por $ (\\simplify{a*x-b})$, entonces el resto es $ f (\\frac {b}{a}) $.

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Use el teorema del resto para el resto cuando $p(x)$ se divide por $(\\simplify{x+{d}}) $, cree otra ecuación que incluya $ s $ y $ t $.

\n

[[0]]$s+t$ = [[1]].

\n

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Encuentra el valor de $ s $. Reduce tu respuesta a su forma fraccionaria más simple.

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$s =$ [[0]]

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\n

Reste las dos ecuaciones simultáneas para $s$ y $t$, obtenidas en las partes a) y b), una de la otra.Luego reorganice esta nueva ecuación para encontrar el valor de $s$.

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Encuentra el valor de $t$. Reduce tu respuesta a su forma fraccionaria más simple.

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$t =$ [[0]]

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Sustituya el valor de $s$ de la parte c) en una de las ecuaciones simultáneas por $s$ y $t$.

\n

Luego, reorganiza esta ecuación para encontrar el valor de $t$.

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