Marlon's copy of Finding the full factorisation of a polynomial, using the Factor Theorem and long division

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Marlon Arcila

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Marlon Arcila 7 years, 2 months ago

Created this as a copy of Finding the full factorisation of a polynomial, using the Factor Theorem and long division.

Name	Status	Author	Last Modified
Finding the full factorisation of a polynomial, using the Factor Theorem and long division	Ready to use	Elliott Fletcher	20/11/2019 14:39
Marlon's copy of Finding the full factorisation of a polynomial, using the Factor Theorem and long division	draft	Marlon Arcila	03/02/2018 16:18
JD's copy of Finding the full factorisation of a polynomial, using the Factor Theorem and long division	draft	JD Ichwan	21/05/2019 05:43
Finding the full factorization of a polynomial, using the Factor Theorem and long division	draft	Xiaodan Leng	10/07/2019 21:43

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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

y	integer	-2
u	integer	3
z	integer	-1

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xxxxxxxxxx
 
random(-2..3 except 0)

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Description

Factor 3.

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Generated value: `integer`

-2

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For this question, we are given that $(\simplify{x+{z}})$ is a factor of the polynomial

$p(x) = \simplify{x^3+({y}+{u}+{z})*x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}},$

and we are then asked to find the full factorisation of $p(x)$ .

We know that $(\simplify{x+{z}})$ is a factor of $p(x)$ , so we can calculate the other factors of $p(x)$ through long division.

$\begin{align} &\simplify{x^2+({u}+{y})x+{u}{y}}\\ \simplify{x+{z}} \; &\overline{)\simplify{x^3+({y}+{u}+{z})*x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}}}\\ &\;\, \simplify{x^3+{z}x^2}\\ &\qquad\quad \overline{\simplify[all,noLeadingMinus]{({u}+{y})x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}}}\\ &\qquad\quad \simplify[all,noLeadingMinus]{({u}+{y})x^2+({u}{z}+{z}{y})x}\\ &\qquad\quad\quad\quad\quad \overline{\simplify[all,noLeadingMinus]{{y}{u}x+{y}{u}{z}}}\\ &\qquad\quad\quad\quad\quad \simplify[all,noLeadingMinus]{{y}{u}x+{y}{u}{z}}\\ &\qquad\qquad\quad\quad\quad \overline{0.} \end{align}$

We can then factorise $\simplify{x^2+({u}+{y})x+{u}{y}}$ into

$\simplify{x^2+({u}+{y})x+{u}{y}} =(\simplify{x+{y}})(\simplify{x+{u}}).$

Therefore, the full factorisation of $p(x)$ is

$\begin{align} p(x) &= \simplify{x^3+({y}+{u}+{z})*x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}},\\ &= (\simplify{x+{y}})(\simplify{x+{z}})(\simplify{x+{u}}). \end{align}$

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Marlon's copy of Finding the full factorisation of a polynomial, using the Factor Theorem and long division

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Marlon Arcila 7 years, 2 months ago

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Marlon's copy of Finding the full factorisation of a polynomial, using the Factor Theorem and long division

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Marlon Arcila 7 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Mathematical expression

Advanced settings

Checking accuracy

Variable value generators

Pattern restriction

Variables

String restrictions

Generated value: `integer`