// Numbas version: exam_results_page_options {"name": "Apply the factor and remainder theorems", "metadata": {"description": "

Apply the factor and remainder theorems to manipulate polynomial expressions

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Apply the factor theorem to check which of a list of linear polynomials are factors of another polynomial.

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "variable_groups": [], "advice": "

To find the factors of the polynomial $f(x) = \\simplify{x^3+({a}+{b}+{c})x^2+({a}{b}+{a}{c}+{b}{c})x+{a}{b}{c}}$, we use the factor theorem. 

\n

If $f(x)$ is a polynomial and $f(p) = 0$, then $(x-p)$ is a factor of $f(x)$.

\n

If $(\\simplify{(x+{a})})$ is a factor of $f(x)$ then by the factor theorem, $f(\\simplify{-{a}}) = 0$.

\n

We see that

\n

\\[
\\begin{align}
f(\\simplify{-{a}}) &= \\simplify[all,!collectNumbers]{{coef1_x3}+{coef1_x2}+{coef1_x}+{const}}\\\\
&= \\simplify{{coef1_x3}+{coef1_x2}+{coef1_x}+{const}}.
\\end{align}
\\]

\n

Therefore, $(\\simplify{(x+{a})})$ is a factor of $f(x)$.

\n

Similarly for $(\\simplify{(x+{d})})$,

\n

\\[
\\begin{align}
f(\\simplify{-{d}}) &= \\simplify[all,!collectNumbers]{{coef2_x3}+{coef2_x2}+{coef2_x}+{const}}\\\\
&= \\simplify{{coef2_x3}+{coef2_x2}+{coef2_x}+{const}}\\\\
&\\neq 0.
\\end{align}
\\]

\n

Therefore, $(\\simplify{(x+{d})})$ is not a factor of $f(x)$.

\n

Finally, for $(\\simplify{(x+{c})})$,

\n

\\[
\\begin{align}
f(\\simplify{-{c}}) &= \\simplify[all,!collectNumbers]{{coef3_x3}+{coef3_x2}+{coef3_x}+{const}}\\\\
&= \\simplify{{coef3_x3}+{coef3_x2}+{coef3_x}+{const}}.
\\end{align}
\\]

\n

Therefore, $(\\simplify{(x+{c})})$ is also a factor of $f(x)$.

", "statement": "

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

", "preamble": {"js": "", "css": ""}, "variables": {"coef1_x3": {"templateType": "anything", "description": "

Number obtained from putting x=-a into the first term of the equation.

", "name": "coef1_x3", "group": "Ungrouped variables", "definition": "(-a)^3"}, "coef1_x2": {"templateType": "anything", "description": "

Number obtained from putting x=-a into the second term of the equation.

", "name": "coef1_x2", "group": "Ungrouped variables", "definition": "(a+b+c)*(-a)^2"}, "coef2_x3": {"templateType": "anything", "description": "

Number obtained from putting x=-d into the first term in the equation.

", "name": "coef2_x3", "group": "Ungrouped variables", "definition": "(-d)^3"}, "coef1_x": {"templateType": "anything", "description": "

Number obtained from putting x=-a into the first term of the equation.

", "name": "coef1_x", "group": "Ungrouped variables", "definition": "(a*b+b*c+a*c)*(-a)"}, "coef3_x3": {"templateType": "anything", "description": "

Number obtained for putting x=-c into the first term of the equation.

", "name": "coef3_x3", "group": "Ungrouped variables", "definition": "(-c)^3"}, "const": {"templateType": "anything", "description": "

Constant term in the equation.

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Random number between -2 and 3 except 0 for creating polynomial.

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Incorrect answer for part a.

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Number obtained from putting x=-d into the 3rd term for the equation.

", "name": "coef2_x", "group": "Ungrouped variables", "definition": "(a*b+b*c+a*c)*(-d)"}, "a": {"templateType": "anything", "description": "

Random number between -2 and 3, not including 0 for creating polynomial.

", "name": "a", "group": "Ungrouped variables", "definition": "random(-2..3 except 0 except c)"}, "coef2_x2": {"templateType": "anything", "description": "

Number obtained from putting x=-d into the second term of the equation.

", "name": "coef2_x2", "group": "Ungrouped variables", "definition": "(a+b+c)*(-d)^2"}, "b": {"templateType": "anything", "description": "

Random number between -2 and 3 except 0 for creating polynomial.

", "name": "b", "group": "Ungrouped variables", "definition": "random(-2..3 except 0 except c)"}, "coef3_x": {"templateType": "anything", "description": "

Number obtained by putting x=-c into the third term of the equation.

", "name": "coef3_x", "group": "Ungrouped variables", "definition": "(a*b+b*c+a*c)*(-c)"}, "coef3_x2": {"templateType": "anything", "description": "", "name": "coef3_x2", "group": "Ungrouped variables", "definition": "(a+b+c)*(-c)^2"}}, "parts": [{"variableReplacementStrategy": "originalfirst", "choices": ["

$(\\simplify{x+{a}})$

", "

$(\\simplify{x+{d}})$

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$(\\simplify{x+{c}})$

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Use the factor theorem to find which two of the following are factors of the polynomial

\n

\\[f(x) = \\simplify{x^3+({a}+{b}+{c})x^2+({a}{b}+{a}{c}+{b}{c})x+{a}{b}{c}}.\\]

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Number obtained by putting x=-d into the first term of the equation.

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Random number between -2 and 3 except 0 for creating polynomial.

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Number obtained by putting x=-d into the third term of the equation.

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Used in creation of the polynomial.

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Random number between 2,3,4.

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Number obtained by putting x=-d into the second term of the equation.

", "group": "Ungrouped variables", "templateType": "anything", "name": "coef_x2", "definition": "(w*d+a+w*b)*(-d)^2"}, "a": {"description": "

Random number between -2 and 3, not including 0 for creating polynomial.

", "group": "Ungrouped variables", "templateType": "anything", "name": "a", "definition": "random(-2..3 except 0)"}}, "functions": {}, "statement": "

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

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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 $m$.

\n

$m =$ [[0]].

\n

", "type": "gapfill"}], "ungrouped_variables": ["w", "a", "b", "d", "coef_x3", "coef_x2", "coef_x"], "rulesets": {}, "preamble": {"css": "", "js": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Given a factor of a cubic polynomial, factorise it fully by first dividing by the given factor, then factorising the remaining quadratic.

"}, "advice": "

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

\n

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

\n

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

\n

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

\n

\\[
\\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}
\\]

\n

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

\n

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

"}, {"name": "Finding the full factorisation of a polynomial, using the Factor Theorem and long division", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "type": "question", "statement": "

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

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Factor 3.

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Factor 2.

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Factor 1.

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Given that $(\\simplify{x+{z}})$ is a factor of \\[p(x) = \\simplify{x^3+({y}+{u}+{z})*x^2+({y}*{u}+{z}*{u}+{y}*{z})*x+{y}*{u}*{z}}.\\]

\n

Find the full factorisation of $p(x)$.

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Use a given factor of a polynomial to find the full factorisation of the polynomial through long division.

"}, "preamble": {"css": "", "js": ""}, "advice": "

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

\n

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

\n

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

\n

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.

\n

\\[
\\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}
\\]

\n

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

\n

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

\n

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

\n

\\[
\\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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This question tests the student's knowledge of the remainder theorem and the ways in which it can be applied.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Consider the polynomial

\n

\\[ p(x) = \\simplify{{coef2_x3}x^3+s*x^2+{coef2_x}x+t}\\text{.}\\]

\n

The polynomial:

\n", "advice": "

We are told that the polynomial:

\n\n

a)

\n

Firstly, substituting $x = \\simplify{-{c}}$ into $p(x)$ gives us 

\n

\\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}

\n

But, by the remainder theorem $p(\\simplify{-{c}}) = \\var{rem1}$ (using the first bullet point), so this becomes

\n

\\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}

\n

b)

\n

Similarly, substituting $x = \\simplify{-{d}}$ into $p(x)$, gives us

\n

\\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}

\n

But, by the remainder theorem $p(\\simplify{-{d}}) = \\var{rem2}$ (using the second bullet point), so this becomes

\n

\\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}

\n

c)

\n

We now have two simultaneous equations for $s$ and $t$:

\n

\\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

Next, we subtract the second equation from the first equation.

\n

This allows us to cancel out the terms involving $t$ and gives us an equation only in terms of $s$, which we can then rearrange to find the value of $s$.

\n

Subtracting the two equations gives

\n

\\[\\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)}}.\\]

\n

Then, we can rearrange this equation so that

\n

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

\n

d)

\n

We can calculate $t$ by substituting our value of $s$ into one of our original simultaneous equations. For example, let's use the equation

\n

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

\n

Substituting our value of $s$ into this equation gives us

\n

\\[
\\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}
\\]

\n

This same answer would've also been obtained if we had substituted our value of $s$ into the other equation instead.

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Numerator of s

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Dividing term 1.

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First remainder.

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Coefficient of x.

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Denominator of s.

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Dividing term 2.

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Coefficient of x^3.

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Simplifies first coefficient of s.

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Simplifies second coefficient of s.

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Second remainder.

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Using the remainder theorem for the remainder when $p(x)$ is divided by $(\\simplify{x+{c}})$, create an equation involving $s$ and $t$.

\n

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

\n

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The remainder theorem states that if a polynomial $f(x)$ is divided by $(\\simplify{a*x-b})$ then the remainder is $f(\\frac{b}{a})$.

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Using the remainder theorem for the remainder when $p(x)$ is divided by $(\\simplify{x+{d}})$, create another equation involving $s$ and $t$.

\n

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

\n

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Find the value of $s$. Reduce your answer to its simplest fractional form.

\n

$s =$ [[0]]

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Subtract the two simultaneous equations for $s$ and $t$, obtained in parts a) and b), from each other.

\n

Then rearrange this new equation to find the value of $s$.

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Find the value of $t$. Reduce your answer to its simplest fractional form.

\n

$t =$ [[0]]

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Substitute the value of $s$ from part c) into one of the simultaneous equations for $s$ and $t$.

\n

Then, rearrange this equation to find the value of $t$.

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Constant term

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Coefficient of x^2.

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Coefficient of x.

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Coefficient of x^3

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Correct remainder.

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Free coefficient in the dividing equation.

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Leading coefficient in the dividing equation.

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Find the remainder when $f(x) = \\simplify{{coef_x3}x^3+{coef_x2}x^2+{coef_x}x+{const}}$ is divided by $(\\simplify{{a}x+{k}})$, using the remainder theorem.

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This question tests the student's ability to find remainders using the remainder theorem. 

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The remainder theorem states that if a polynomial $f(x)$ is divided by $(\\simplify{a*x-b})$ then the remainder is $f \\left( \\frac{b}{a} \\right)$.

\n

This means that if we substitute $x = \\frac{b}{a}$ into the equation for $f(x)$, the result will be equal to the remainder when $f(x)$ is divided by $(\\simplify{a*x-b})$.

\n

Therefore, to calculate the remainder when $f(x) = \\simplify{{coef_x3}*x^3+{coef_x2}*x^2+{coef_x}*x+{const}}$ is divided by $(\\simplify{{a}*x+{k}})$, we use this same principle.

\n

As we are dividing $f(x)$ by $(\\simplify{{a}*x+{k}})$, using the remainder theorem tells us that substituting

\n

\\[
\\begin{align}
x &= \\frac{b}{a}\\\\
&= \\simplify{-({k}/{a})} 
\\end{align}
\\]

\n

into our equation for $f(x)$ will give us the remainder when $f(x)$ is divided by $(\\simplify{{a}*x+{k}})$. Substituting this value of $x$ into $f(x)$ gives us

\n

\\[
\\begin{align}
f(\\simplify{-({k}/{a})}) &= \\simplify[all,!collectNumbers, fractionnumbers]{{coef_x3*(-({k}/{a}))^3}+{coef_x2*(-({k}/{a}))^2}+{coef_x*(-({k}/{a}))}+{const}}\\\\
&= \\simplify[all,fractionnumbers]{{coef_x3*(-({k}/{a}))^3}+{coef_x2*(-({k}/{a}))^2}+{coef_x*(-({k}/{a}))}+{const}}.
\\end{align}
\\] 

\n

Therefore, the remainder when $f(x) = \\simplify{{coef_x3}*x^3+{coef_x2}*x^2+{coef_x}*x+{const}}$ is divided by $(\\simplify{{a}*x+{k}})$ is  $\\displaystyle\\simplify[all,fractionnumbers]{{coef_x3*(-({k}/{a}))^3}+{coef_x2*(-({k}/{a}))^2}+{coef_x*(-({k}/{a}))}+{const}}$.

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Quotient and remainder, polynomial division.

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Apply the remainder and factor theorems.

", "advice": "

The polynomial remainder theorem is useful here, it states that the remainder of the division of a polynomial $f(x)$ by a linear polynomial $x-a$ is equal to $f(a)$.

\n

To find the quotient, here are two methods:

\n

1) Perform polynomial long division

\n

2) First find the remainder by the polynomial remainder theorem. Minus the remainder from the initial polynomial, then factorise. 

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Remainder

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By using the Polynomial Remainder Theorem, find the remainder $R$ when $\\simplify{f(x)=x^3+{c_coeff2}x^2+{c_coeff1}x+({c_coeff0}+ {c_r})}$ is divided by $\\simplify{(x-{c_root1})}$

\n

$R = f($[[0]]$)=$[[1]]

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Find the remainder $R$ when $\\simplify{f(x)={n_coeff3}x^3+{n_coeff2}x^2+{n_coeff1}x+({n_coeff0}+ {n_r})}$ is divided by $\\simplify{({n_coeff3}x-{n_root1})}$

\n

$R =$ [[0]] 

\n

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Obtain the quotient $Q(x)$ and remainder $R$ when $\\simplify{f(x)={e_coeff3}x^3+{e_coeff2}x^2+{e_coeff1}x+({e_coeff0}+ {e_r})}$ is divided by $\\simplify{({e_coeff3}x-{e_root1})}$

\n

$Q(x) =$ [[0]]

\n

$R =$ [[1]] 

\n

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Long division of a quartic by a quadratic

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Obtain the quotient and remainder when \\[ \\simplify[all]{x^4+{a1}x^3 +  {a2} x^2+{a3}x+{a4}} \\]

\n

is divided by $\\simplify{x^2+{b1}x+{b2}}$.

", "advice": "

Use long division.

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Quotient $= \\quad$ [[0]]

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Use long division  with the dividend and divisor.

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Remainder $=\\quad$ [[0]]

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Use long division again.

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