Considere el polinomio $p(x)$ con parametros $t$ y $s$. $ p(x) = ax^3+sx^2+t$, halle los valores de $s$ y $t$.

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Created this as a copy of Finding unknown coefficients of a polynomial, using the remainder theorem.

Name	Status	Author	Last Modified
Finding unknown coefficients of a polynomial, using the remainder theorem	Ready to use	Elliott Fletcher	20/11/2019 14:39
Considere el polinomio $p(x)$ con parametros $t$ y $s$ . $p(x) = ax^3+sx^2+t$ , halle los valores de $s$ y $t$ .	Ready to use	Luis Hernandez	19/06/2023 13:40
Simon's copy of Finding unknown coefficients of a polynomial, using the remainder theorem	draft	Simon Thomas	28/02/2019 12:15
Christian's copy of Finding unknown coefficients of a polynomial, using the remainder theorem	draft	Christian Lawson-Perfect	20/05/2019 13:55
Finding unknown coefficients of a polynomial, using the remainder theorem	draft	Xiaodan Leng	10/07/2019 21:44
Finding unknown coefficients of a polynomial by the remainder theorem	Ready to use	Wan Mekwi	11/10/2020 21:25

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Question statement

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

El polinomio $p(x)$ :

tiene resto $\var{rem1}$ cuando se divide por $(\simplify{x+{c}})$
tiene resto $\var{rem2}$ cuando se divide por $(\simplify{x+{d}})$

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			Name	Type	Generated Value

c	integer	2
d	integer	-1
coef2_x3	integer	2
coef2_x	integer	1
rem1	integer	1
rem2	integer	3
numerator	number	19
denominator	number	3
x	number	4
y	number	1

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Dividing term 1.

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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 $t$ .

Gap 0. $s + t$ = Gap 1..

Advice

Nos dicen que el polinomio:

tiene resto $\var{rem1}$ cuando se divide por $(\simplify{x+{c}})$
tiene resto $\var{rem2}$ cuando se divide por $(\simplify{x+{d}})$

a)

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

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

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

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

b)

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

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

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

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

c)

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

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

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

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 $s$ .

Restar las dos ecuaciones da:

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

Entonces, podemos reorganizar esta ecuación para que

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

d)

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

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

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

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

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

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Considere el polinomio $p(x)$ con parametros $t$ y $s$ . $p(x) = ax^3+sx^2+t$ , halle los valores de $s$ y $t$ .

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Luis Hernandez 6 years, 3 months ago

Luis Hernandez 6 years, 3 months ago

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Considere el polinomio p(x)p(x) con parametros tt y ss. p(x)=ax3+sx2+t p(x) = ax^3+sx^2+t, halle los valores de ss y tt.

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Luis Hernandez 6 years, 3 months ago

Luis Hernandez 6 years, 3 months ago

Error in variable testing condition

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Generated value: integer

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a)

b)

c)

d)

Considere el polinomio $p(x)$ con parametros $t$ y $s$ . $p(x) = ax^3+sx^2+t$ , halle los valores de $s$ y $t$ .

Generated value: `integer`