Apply the factor and remainder theorems

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Apply the factor and remainder theorems

Question 1

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

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

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

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

Expected answer:

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

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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.

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

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

We see that

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

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

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

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

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

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

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

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

Question 2

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

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 $m$.

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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

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

Question 3

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

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

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

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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

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

Question 4

Consider the polynomial

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

The polynomial:

has remainder $\var{rem1}$ when divided by $(\simplify{x+{c}})$
has remainder $\var{rem2}$ when divided by $(\simplify{x+{d}})$

a)

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

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b)

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

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c)

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

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d)

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

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We are told that the polynomial:

has a remainder $\var{rem1}$ when divided by $(\simplify{x+{c}})$
has a remainder $\var{rem2}$ when divided by $(\simplify{x+{d}})$

a)

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

\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}

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

\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)

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

\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}

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

\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)

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

\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}

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

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 $s$.

Subtracting the two equations gives

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

Then, we can rearrange this equation so that

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

d)

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

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

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

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

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

Question 5

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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Advice

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

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 $(\simplify{a*x-b})$.

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.

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

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

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

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

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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Apply the factor and remainder theorems

Apply the factor and remainder theorems

Question 1

Question 1

Advice

Question 2

Question 2

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Question 3

Question 3

Advice

Question 4

Question 4

a)

b)

c)

d)

Advice

a)

b)

c)

d)

Question 5

Question 5

Advice