Factor Theorem and polynomials

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Adrian Jannetta

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Adrian Jannetta 7 years, 7 months ago

Created this as a copy of Finding the missing value of a constant in a polynomial, using the Factor Theorem .

Name	Status	Author	Last Modified
Finding the missing value of a constant in a polynomial, using the Factor Theorem	Ready to use	Elliott Fletcher	20/11/2019 14:39
Factor Theorem and polynomials	draft	Adrian Jannetta	16/08/2023 14:56
Finding the missing value of a constant in a polynomial, using the Factor Theorem	draft	Adrian Jannetta	16/08/2023 14:56
Marlon's copy of Finding the missing value of a constant in a polynomial, using the Factor Theorem	draft	Marlon Arcila	03/02/2018 16:11
Marlon's copy of Finding the missing value of a constant in a polynomial, using the Factor Theorem	draft	Marlon Arcila	03/02/2018 16:19
Factor Theorem	draft	Mark Hodds	24/03/2018 15:58
Dado que $({x+{d}})$ es un factor de $g(x) = ax^3+{b}x^2+{x}+m$ , Encontrar el valor de $m$ .	Ready to use	Luis Hernandez	19/06/2023 13:36
JD's copy of Finding the missing value of a constant in a polynomial, using the Factor Theorem	draft	JD Ichwan	21/05/2019 05:42
Finding the missing value of a constant in a polynomial, using the Factor Theorem	draft	Xiaodan Leng	11/07/2019 01:38

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Question statement

The factor theorem states that if $f(x)$ is a polynomial and $f(p) = 0$ , then $(x-p)$ is a factor of $f(x)$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

w	integer	2
a	integer	-2
b	integer	-2
d	integer	2
coef_x3	number	-16
coef_x2	number	-8
coef_x	integer	16

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Generated value: `integer`

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Given that $(\simplify{x+{d}})$ is a factor of $g(x) = \simplify{{w}*x^3+({w}{d}+{a}+{w}{b})*x^2+({a}{d}+{w}{b}{d}+{a}{b})*x}+m$ , find the value of $m$ .

$m =$ Unnamed gap.

Advice

Using the factor theorem, we know that if $(x-a)$ is a factor of a polynomial $f(x)$ , then $f(a)=0$ .

We are given that $(\simplify{x+{d}})$ is a factor of $g(x) = \simplify{{w}*x^3+({w}{d}+{a}+{w}{b})*x^2+({a}{d}+{w}{b}{d}+{a}{b})*x+m}$ .

By the factor theorem, this means that $g(\simplify{-{d}}) = 0$ .

Substituting $x=\simplify{-{d}}$ into $g(x)$ gives

$\begin{align} g(\simplify{-{d}}) &= \simplify[all,!collectNumbers]{{coef_x3}+{coef_x2}+{coef_x}+m}\\ &=\simplify{{coef_x3}+{coef_x2}+{coef_x}+m}. \end{align}$

Therefore, as $g(\simplify{-{d}}) = 0$ , we have

$\begin{align} \simplify{{coef_x3}+{coef_x2}+{coef_x}+m}&=0\\ m&=\simplify{-({coef_x3}+{coef_x2}+{coef_x})}. \end{align}$

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This question is used in the following exam:

Maths 1 Assignment #2 by Adrian Jannetta in Foundation Maths.

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Adrian Jannetta 7 years, 7 months ago

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Generated value: `integer`

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Factor Theorem and polynomials

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Adrian Jannetta 7 years, 7 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`