Functions and algebra: Work with algebraic expressions using the remainder and the factor theorems

Unit 1: Use the remainder and factor theorem to factorise third degree polynomials

Natashia Bearam-Edmunds

Unit 1 outcomes

By the end of this unit you will be able to:

• Find the factors of a cubic polynomial.
• Find the remainder of cubic polynomial.
• Use division by inspection, synthetic division or long division to factorise and solve cubic polynomials.

What you should know

Before you start this unit, make sure you can:

Introduction

You have already seen examples of polynomial expressions in level 2 subject outcome 2.1, unit 1. Remember that a polynomial is an expression with one or more variables with different coefficients and non-negative integer powers.

We define a polynomial as: $\scriptsize {{a}_{n}}{{x}^{n}}+{{a}_{{n-1}}}{{x}^{{n-1}}}+...+{{a}_{2}}{{x}^{2}}+{{a}_{1}}x+{{a}_{0}}{{x}^{0}}$ where $\scriptsize n\in {{N}_{0}}$. $\scriptsize {{a}_{n}},{{a}_{{n-1}}},{{a}_{2}},{{a}_{1}}$ and $\scriptsize {{a}_{0}}$ are the coefficients of each term and are usually constants.

There is an unlimited variety in the number of terms and the powers of the variable. While the order of the terms in a polynomial is not important for performing operations, we generally arrange the terms in decreasing powers of the variable. This is called the general form.

Notice that the definition of a polynomial states that all exponents of the variables must be elements of the set of whole numbers. If an expression contains terms with exponents that are not whole numbers, then it is not a polynomial.

For example, $\scriptsize \displaystyle \frac{2}{x}+2{{x}^{2}}+1,\text{ }3\sqrt{y}+y$ and $\scriptsize {{k}^{2}}-6k+3{{k}^{{-\tfrac{1}{2}}}}$ are not polynomials. Look at each expression and make sure you understand why these are not polynomials. Can you see that in each case the exponents are not whole numbers?

The degree of a polynomial is the highest power of the variable. If the expression has been written in general form then it is the power of the first variable. The leading term is the term with the highest power of the variable, or the term with the highest degree. The leading coefficient is the coefficient of the leading term.

The polynomial $\scriptsize 2{{x}^{3}}-3{{x}^{2}}+x-1$ has a degree of $\scriptsize 3$, the leading term is $\scriptsize 2{{x}^{3}}$ and the leading coefficient is $\scriptsize 2$.

Exercise 1.1

State whether the following statements are true or false:

1. $\scriptsize 2{{x}^{{-1}}}$ is a monomial because it only has one term.
2. $\scriptsize 45$ is a constant polynomial of degree $\scriptsize 0$.
3. $\scriptsize 2{{x}^{2}}-3x+1$ is a quadratic polynomial of degree $\scriptsize 2$ with a leading coefficient of $\scriptsize 2$.
4. A cubic polynomial always has three terms and all the exponents are natural numbers.

The full solutions are at the end of the unit.

Dividing cubic polynomials

In level 3 subject outcome 2.3 we looked at various methods to factorise and solve quadratic equations. In this unit we will factorise cubic polynomials with one variable. Remember that a cubic polynomial is an algebraic expression with a highest power of $\scriptsize 3$. The standard form of a cubic polynomial is $\scriptsize a{{x}^{3}}+b{{x}^{2}}+cx+d$.

The following activity is useful to remind ourselves of the basics of long division, which can be applied to polynomials too and is needed to factorise cubic polynomials.

Activity 1.1: Investigate simple division

Time required: 10 minutes

What you need:

• a pen and paper

What to do

Consider the following situation and answer the questions that follow.

Six learners are at a product promotion and there are $\scriptsize 15$ free gifts to be given away. Each learner must receive the same number of gifts.

1. How many gifts does each learner get?
2. How many gifts will be left over?
3. Use the following variables to express the above situation as a mathematical equation:
$\scriptsize a=$ total number of gifts
$\scriptsize b=$ total number of learners
$\scriptsize \displaystyle q=$ number of gifts for each learner
$\scriptsize r=$ number of gifts remaining
4. What is each part of the division expression called?
5. What does your equation say in words?

What did you find?

1. Since each learner has to a get the same whole number of gifts, each learner will get $\scriptsize 2$ gifts.
2. In total, the learners will get $\scriptsize 12$ gifts. Therefore, there will be $\scriptsize 3$ left over.
3. Using numerical values we can write this as $\scriptsize \displaystyle \frac{{15}}{6}=2$ remainder $\scriptsize 3$. This is the same as$\scriptsize 15=6\times 2+3$. Using the variables this becomes $\scriptsize a=b\times q+r,\text{ }b\ne 0$.
4. $\scriptsize 15$ is the dividend
$\scriptsize 6$ is the divisor
$\scriptsize 2$ is the quotient
$\scriptsize 3$ is the remainder
5. The dividend is equal to the divisor multiplied by the quotient, plus the remainder.

The activity reminded us that if an integer $\scriptsize a$ is divided by an integer $\scriptsize b$, then the answer is $\scriptsize q$ with a remainder of $\scriptsize r$. Sometimes $\scriptsize r=0$.

For example, $\scriptsize 8$ divided by $\scriptsize 3$ gives a whole number answer of $\scriptsize 2$ with a remainder of $\scriptsize 2$.

We can write this as $\scriptsize \displaystyle \frac{8}{3}=2+\displaystyle \frac{2}{3}$ which is the same as saying that $\scriptsize 8=3\times 2+2$.

Take note!

This rule can be extended to include the division of polynomials; if a polynomial $\scriptsize f(x)$ is divided by a polynomial $\scriptsize g(x)$, then the answer is $\scriptsize Q(x)$ with a remainder $\scriptsize R(x)$.

$\scriptsize f(x)=g(x)\cdot Q(x)+R(x)$ where $\scriptsize g(x)\ne 0$.

We are familiar with the process of long division in ordinary arithmetic. Let’s have a look at the process again to remind us how this is done.

\scriptsize \displaystyle \begin{align*} &3\overset{{59}}{\overline{\left){{178}}\right.}}\\ &\;\underline{{-15}}\downarrow && \text{ Step 1: }5\times 3=15\text{ and }17-15=2\\ &\;\;\;\;\;28 && \text{ Step }2\text{: Bring down the }8\\ &\;\;\underline{{-27}} &&\text{ Step 3: }9\times 3=27 \text{ and }28-27=1\\ &\;\;\;\;\;\;1\text{ }\end{align*}

$\scriptsize \displaystyle \frac{{178}}{3}=59$ remainder $\scriptsize 1$ or $\scriptsize 59\displaystyle \frac{1}{3}\text{ }$.

Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.

\scriptsize \begin{align*}\text{Dividend } &= \text{ divisor}\cdot \text{quotient + remainder}\\178&=3\cdot 59+1\\&=177+1\\&=178\end{align*}

You can use long division and synthetic division, which is explained in detail shortly, to find the quotient and remainder when a cubic polynomial is divided by another polynomial.

Example 1.1

Use the method of long division to find the quotient and remainder when $\scriptsize f(x)=2{{x}^{3}}+3{{x}^{2}}+x-5$ is divided by $\scriptsize x-1$.

Solution

Write down the known and unknown expressions.

$\scriptsize f(x)=g(x)\cdot Q(x)+R(x)$

$\scriptsize 2{{x}^{3}}+3{{x}^{2}}+x-5=(x-1)\cdot Q(x)+R(x)$

Use long division to find $\scriptsize Q(x)$ and$\scriptsize R(x)$.

Make sure that $\scriptsize f(x)$ and $\scriptsize g(x)$ are written in descending order of the exponents. If a term of a certain degree is missing from$\scriptsize f(x)$, then write the term with a coefficient of $\scriptsize 0$.

\scriptsize \displaystyle \begin{align*}&x-1\overset{{2{{x}^{2}}+5x+4}}{\overline{\left){{2{{x}^{3}}+3{{x}^{2}}+x-5\text{ }}}\right.}} && \text{ Divide }2{{x}^{3}}\text{ by }x\text{ and put the answer in the quotient}\\&\text{ }-\underline{{(2{{x}^{3}}-2{{x}^{2}})}} && \text{ Multiply the quotient by the divisor and subtract from }f(x)\\ & \text{ }5{{x}^{2}}+x && \text{ Bring down the next term from }f(x)\text{ and repeat the process }\\ &\text{ }-\underline{{(5{{x}^{2}}-5x)}}\\ & \text{ 6}x-5\\ &\text{ }-(\underline{{6x-6}})\\ & \text{ }1\\&\text{ }\end{align*}

\scriptsize \begin{align*}Q(x)&=2{{x}^{2}}+5x+6\\R(x)&=1\\ &\therefore 2{{x}^{3}}+3{{x}^{2}}+x-5=(x-1)\cdot (2{{x}^{2}}+5x+6)+1\end{align*}

You can multiply the brackets and simplify to check that the answer is correct.

Exercise 1.2

1. Find the quotient and remainder of the following cubic polynomials by using long division:
1. $\scriptsize 6{{x}^{3}}+{{x}^{2}}-4x+5$ divided by $\scriptsize 2x-1$
2. $\scriptsize {{x}^{3}}+3{{x}^{2}}-x-3$ divided by $\scriptsize x-1$
2. Hence or otherwise, factorise 1b) completely.

The full solutions are at the end of the unit.

As you have seen, long division of polynomials involves many steps and can be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is $\scriptsize 1$.

Example 1.2

Use the method of synthetic division to find the quotient and remainder when:

1. $\scriptsize f(x)=2{{x}^{3}}+3{{x}^{2}}+x-5$ is divided by $\scriptsize x-1$.
2. $\scriptsize g(x)=6{{x}^{3}}+{{x}^{2}}-4x+5$ is divided by $\scriptsize 2x-1$.

Solutions

1. Recall the long division solution from Example 1.1 looked this this:
\scriptsize \displaystyle \begin{align*}x-1\overset{{2{{x}^{2}}+5x+6}}{\overline{\left){{2{{x}^{3}}+3{{x}^{2}}+x-5\text{ }}}\right.}}\text{ }\\\text{ }-\underline{{(2{{x}^{3}}-2{{x}^{2}})}}\text{ }\\\text{ }5{{x}^{2}}+x\text{ }\\\text{ }-\underline{{(5{{x}^{2}}-5x)}}\\\text{ 6}x-5\\\text{ }-(6x-6)\\\text{ }1\\\text{ }\end{align*}
.
There is a lot of repetition in the long division. Let’s see how we can simplify the process.
.
Synthetic division allows us to collapse the ‘table’ of long division by moving each of the rows up to fill any vacant spots. Also, instead of dividing by $\scriptsize -1$, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the ‘divisor’ to $\scriptsize +1$, multiply and add.
.
This will become clearer by completing the example $\scriptsize f(x)=2{{x}^{3}}+3{{x}^{2}}+x-5$ divided by $\scriptsize x-1$ using synthetic division.
.
The process starts by writing the constant term of the divisor with the opposite sign and the coefficients of the polynomial. So the $\scriptsize -1$ constant term in the divisor becomes $\scriptsize +1$.
.
$\scriptsize 1\left| \!{\underline {\, {2\text{ }3\text{ }1\text{ }-5} \,}} \right.$
.
Bring down the leading coefficient then multiply this leading coefficient by the divisor and write that answer in column $\scriptsize 2$.
.
\scriptsize \begin{align*}1\left| \!{\underline {\, \begin{align*}2\text{ }3\text{ }1\text{ }-5\\\text{ }2\end{align*} \,}} \right. \\\text{ }2\end{align*}
.
Next, add the numbers in the second column and write this answer down. Next, multiply this answer by the divisor and write the result in the third column and add the numbers in the third column.
\scriptsize \begin{align*}1\left| \!{\underline {\, \begin{align*}2\text{ }3\text{ }1\text{ }-5\\\text{ }2\text{ }5\text{ }6\end{align*} \,}} \right. \\\text{ }2\text{ }5\text{ }6\text{ 1}\end{align*}
.
Once again, multiply this result of adding the numbers in the third column by the divisor and write this number in the fourth column. Add the numbers in the fourth column. This final entry in the last column is the remainder.
.
Using synthetic division the quotient is $\scriptsize 2{{x}^{2}}+5x+6$ and the remainder is $\scriptsize 1$ just as it was in Example 1.1.
2. In this case we see that the coefficient of the divisor is not $\scriptsize 1$. To use synthetic division, the coefficient of the divisor must be $\scriptsize 1$. So, we make the leading coefficient of the divisor equal to $\scriptsize 1$ by dividing by a factor of $\scriptsize 2$ to get $\scriptsize 2(x-\displaystyle \frac{1}{2})$.
Now the $\scriptsize -\displaystyle \frac{1}{2}$ constant term in the divisor becomes $\scriptsize +\displaystyle \frac{1}{2}$. The process for synthetic division will follow the same steps as before.
.
$\scriptsize \tfrac{1}{2}\left| \!{\underline {\, {\text{6 1 }-\text{4 }5} \,}} \right.$
.
\scriptsize \displaystyle \begin{align*}\displaystyle \frac{1}{2}\left| \!{\underline {\, \begin{align*}\text{6 1 }-4\text{ }5\\\text{ 3 2 }-1\end{align*} \,}} \right. \\\text{ }6\text{ }4\text{ }-2\text{ }4\end{align*}
.
Using synthetic division the quotient is $\scriptsize \displaystyle 6{{x}^{2}}+4x-2=2(3{{x}^{2}}+2x-1)$ and the remainder is $\scriptsize 4$. But, we are not done yet. Remember, we wrote the divisor as $\scriptsize 2(x-\displaystyle \frac{1}{2})$. Now we need to rewrite the expression to take it back to its original form.
.
\scriptsize \begin{align*}6{{x}^{3}}+{{x}^{2}}-4x+5&=\displaystyle \frac{1}{2}\cdot 2(x-\displaystyle \frac{1}{2})\cdot 2(3{{x}^{2}}+2x-1)+4\\&=2(x-\displaystyle \frac{1}{2})\cdot (3{{x}^{2}}+2x-1)+4\\&=(2x-1)(3{{x}^{2}}+2x-1)+4\end{align*}

Note

For a demonstration of the synthetic division method watch the video “Synthetic Division”.

Synthetic Division (Duration: 05.20)

Synthetic division is an alternative that can only be used when the divisor is a binomial in the form $\scriptsize x-k$ where $\scriptsize k$ is a real number. To use synthetic division, the coefficient of the divisor must be $\scriptsize 1$. Say, for example, the divisor was $\scriptsize 2x-1$. In this case, we make the leading coefficient of the divisor equal to $\scriptsize 1$ by rewriting it as $\scriptsize 2(x-\displaystyle \frac{1}{2})$. In synthetic division, only the coefficients are used in the division process.

Take note!

To divide two polynomials using synthetic division do the following:

1. Write $\scriptsize k$ for the divisor.
2. Write the coefficients of the dividend.
3. Bring the lead coefficient down.
4. Multiply the lead coefficient by $\scriptsize k$. Write the product in the next column.
5. Add the terms of the second column.
6. Multiply the result by $\scriptsize k$. Write the product in the next column.
7. Repeat steps 5 and 6 for the remaining columns.
8. Use the last row of numbers to write the quotient. The number in the last column is the remainder and has degree $\scriptsize 0$. The next number from the right has degree $\scriptsize 1$, the next number from the right has degree $\scriptsize 2$ and so on.

Exercise 1.3

Find the quotient and remainder of the following by using synthetic division:

1. $\scriptsize 2{{x}^{3}}-3{{x}^{2}}+4x+5$ divided by $\scriptsize x+2$
2. $\scriptsize 4{{x}^{3}}+10{{x}^{2}}-6x-20$ divided by $\scriptsize x-2$
3. $\scriptsize 2{{x}^{3}}+5x-4$ divided by $\scriptsize x-1$

The full solutions are at the end of the unit.

Remainder theorem

Now that we know how to divide polynomials, we can use polynomial division to find the value of polynomials using the remainder theorem.

Activity 1.2: Evaluate polynomials using the remainder theorem

Time required: 15 minutes

What you need:

• a pen and paper

What to do:

Given the following functions:

\scriptsize \begin{align*}f(x)&={{x}^{3}}+3{{x}^{2}}+4x+12\\d(x)&=x-1\\g(x)&=4{{x}^{3}}-2{{x}^{2}}+2x-4\\h(x)&=2x+1\end{align*}

Determine $\scriptsize \displaystyle \frac{{f(x)}}{{d(x)}}$ and $\scriptsize \displaystyle \frac{{g(x)}}{{h(x)}}$.

Write your answers in the form $\scriptsize a(x)=b(x)\cdot Q(x)+R(x)$.

Calculate $\scriptsize f(1)$ and $\scriptsize g(-\displaystyle \frac{1}{2})$.

Compare your answers to question 1 and 3. What do you notice?

Write a mathematical equation to describe your conclusions.

Complete the following sentence: a cubic function divided by a linear expression gives a quotient with a degree of ________ and a remainder with a degree of ______, which is called a constant.

What did you find?

1. Using synthetic division or long division we get:
\scriptsize \displaystyle \begin{align*}&\displaystyle \frac{{f(x)}}{{d(x)}}=\displaystyle \frac{{{{x}^{3}}+3{{x}^{2}}+4x+12}}{{x-1}}\\\ &1\left| \!{\underline {\, \begin{align*}1\text{ }3\text{ }4\text{ }12\\\text{ }1\text{ }4\text{ }8\end{align*} \,}} \right. \\& \text{ }1\text{ }4\text{ }8\text{ }20\\&\text{Quotient}={{x}^{2}}+4x+8\\&\text{Remainder}=20\end{align*}
.
\scriptsize \displaystyle \begin{align*}\displaystyle \frac{{g(x)}}{{h(x)}}=\displaystyle \frac{{4{{x}^{3}}-2{{x}^{2}}+2x-4}}{{2x+1}}\\2x+1\overset{{2{{x}^{2}}-2x+2}}{\overline{\left){\begin{align*}4{{x}^{3}}-2{{x}^{2}}+2x-4\\\underline{{-(4{{x}^{3}}+2{{x}^{2}})}}\\\text{ }-4{{x}^{2}}+2x\\\text{ }\underline{{-\text{(}-4{{x}^{2}}-2x)}}\\\text{ 4}x-4\\\text{ }\underline{{\text{ }-\text{(4}x+2)}}\\\text{ }-6\end{align*}}\right.}}\\\text{Quotient}=2{{x}^{2}}-2x+2\\\text{Remainder}=-6\end{align*}
2. $\scriptsize \displaystyle {{x}^{3}}+3{{x}^{2}}+4x+12=(x-1)({{x}^{2}}+4x+8)+20$
$\scriptsize \displaystyle 4{{x}^{3}}-2{{x}^{2}}+2x-4=(2x+1)(2{{x}^{2}}-2x+2)-6$
3. .
\scriptsize \begin{align*}f(1)&={{(1)}^{3}}+3{{(1)}^{2}}+4(1)+12\\&=20\end{align*}
.
\scriptsize \begin{align*}g(-\displaystyle \frac{1}{2})&=4{{(-\displaystyle \frac{1}{2})}^{3}}-2{{(-\displaystyle \frac{1}{2})}^{2}}+2(-\displaystyle \frac{1}{2})-4\\&=-6\end{align*}
4. The remainders using division of the polynomials are the same as finding the function values using the divisor.
5. If a polynomial is divided by $\scriptsize x-a$, the remainder may be found quickly by evaluating the polynomial function at $\scriptsize a$.
So we can say $\scriptsize R=p(a)$.
6. A cubic function divided by a linear expression gives a quotient with a degree of $\scriptsize \underline{2}$ and a remainder with a degree of $\scriptsize \underline{0}$, which is called a constant.

The remainder theorem:

If a polynomial $\scriptsize f(x)$ is divided by $\scriptsize ax-k$, then the remainder is the value $\scriptsize f\left( {\displaystyle \frac{k}{a}} \right)$.

You can also use the remainder theorem to solve for an unknown variable. This is shown in the next example.

Example 1.3

Determine the value of $\scriptsize t$ if $\scriptsize f(x)={{x}^{3}}+t{{x}^{2}}+4x+2$ gives a remainder of $\scriptsize 16$ when divided by $\scriptsize 2x+1$.

Solution

The remainder theorem tells us that the remainder when $\scriptsize f(x)$ is divided by $\scriptsize 2x+1$ can be found using $\scriptsize f(-\displaystyle \frac{1}{2})$. We are also told that $\scriptsize f(-\displaystyle \frac{1}{2})=16$.

$\scriptsize f(x)={{x}^{3}}+t{{x}^{2}}+4x+2$
\scriptsize \begin{align*}f(-\displaystyle \frac{1}{2})&={{(-\displaystyle \frac{1}{2})}^{3}}+t{{(-\displaystyle \frac{1}{2})}^{2}}+4(-\displaystyle \frac{1}{2})+2\\&=-\displaystyle \frac{1}{8}+\displaystyle \frac{1}{4}t-2+2\\&=\displaystyle \frac{1}{4}t-\displaystyle \frac{1}{8}\end{align*}

But $\scriptsize f(-\displaystyle \frac{1}{2})=16$

\scriptsize \displaystyle \begin{align*}&\displaystyle \frac{1}{4}t-\displaystyle \frac{1}{8}=16\\&\therefore 2t-1=128\\&\therefore 2t=129\\&\therefore t=\displaystyle \frac{{129}}{2}\end{align*}

Exercise 1.4

1. Use the remainder theorem to determine the remainder when $\scriptsize f(x)=3{{x}^{3}}+5{{x}^{2}}-x+1$ is divided by:
1. $\scriptsize x+2$
2. $\scriptsize 2x-1$
3. $\scriptsize 3x+1$
2. Calculate the value of $\scriptsize m$ if $\scriptsize 2{{x}^{3}}-7{{x}^{2}}+mx-26$ is divided by $\scriptsize x-2$ and gives a remainder of$\scriptsize -24$.

The full solutions are at the end of the unit.

Using the factor theorem to solve cubic polynomials

You can use the factor theorem to solve for the zeros or roots (x-intercepts) of a cubic function. The factor theorem describes the relationship between the root of a polynomial and a factor of the polynomial.

You have seen in arithmetic division that if an integer $\scriptsize a$ is divided by an integer $\scriptsize b$ and the answer is $\scriptsize q$ with remainder $\scriptsize r=0$ then $\scriptsize b$ is a factor of $\scriptsize a$. For example, $\scriptsize 27$ divided by $\scriptsize 9$ is $\scriptsize 3$ with remainder $\scriptsize 0$. Therefore, $\scriptsize 9$ is a factor of $\scriptsize 27$. This is also true of polynomials.

Recall that if $\scriptsize f(x)\div (x-k)$ then $\scriptsize f(x)=(x-k)\cdot Q(x)+R(x)$.

If $\scriptsize k$ is a root of the function, then the remainder $\scriptsize R$ is zero and $\scriptsize f\left( k \right)=0$. Therefore, $\scriptsize f(x)=(x-k)\cdot Q(x)+0$ or $\scriptsize f(x)=(x-k)\cdot Q(x)$.

Written in this form, where the remainder is zero, we can say that $\scriptsize x-k$ is a factor of $\scriptsize f(x)$. We can conclude if $\scriptsize k$ is a zero of $\scriptsize f(x)$, then $\scriptsize x-k$ is a factor of $\scriptsize f(x)$.

Conversely, if $\scriptsize x-k$ is a factor of $\scriptsize f(x)$ then the remainder is $\scriptsize 0$.

Factor theorem:

If $\scriptsize f(x)$ is divided by $\scriptsize ax-k$ and the remainder, given by $\scriptsize f\left( {\displaystyle \frac{k}{a}} \right)$ is equal to $\scriptsize 0$, then $\scriptsize ax-k$ is a factor of $\scriptsize f(x)$.

Converse: If $\scriptsize ax-k$ is a factor of $\scriptsize f(x)$ then $\scriptsize f\left( {\displaystyle \frac{k}{a}} \right)=0$.

Cubic polynomials are factorised by using long division, division by inspection or synthetic division. You may use whichever method you are most comfortable with to factorise cubic polynomials. Remember that the degree of the polynomial tells you how many roots the function will have at most. So, a cubic polynomial will have at most three roots, a quadratic function at most two roots and so on.

Example 1.4

Show that $\scriptsize x+2$ is a factor of $\scriptsize g(x)={{x}^{3}}-6{{x}^{2}}-x+30$. Find the remaining factors of $\scriptsize g(x)$. Use the factors to determine the zeros of the cubic polynomial.

Solution

Use the remainder theorem to show that $\scriptsize g(-2)=0$. If$\scriptsize g(-2)=0$ then $\scriptsize x+2$ is a factor of $\scriptsize g(x)$.

\scriptsize \begin{align*}g(-2)&={{(-2)}^{3}}-6{{(-2)}^{2}}-(-2)+30\\&=-8-24+2+30\\&=0\end{align*}

Since $\scriptsize g(-2)=0$ then $\scriptsize x+2$ is a factor of $\scriptsize {{x}^{3}}-6{{x}^{2}}-x+30$.

Now, that we have a linear factor we can divide $\scriptsize g(x)$ by $\scriptsize x+2$ to find the quadratic factor.

You can use long division or synthetic division to factorise further but one of the quickest methods to factorise a cubic polynomial is division by inspection.

We know that $\scriptsize {{x}^{3}}-6{{x}^{2}}-x+30=(x+2)(...)$.

The first term in the second bracket must be $\scriptsize {{x}^{2}}$ to give $\scriptsize {{x}^{3}}$ when we multiply the brackets to make the polynomial a cubic.

The last term in the second bracket must be $\scriptsize 15$ because $\scriptsize 15\times 2=30$. The middle term must be some $\scriptsize ax$ term.

So far we have $\scriptsize {{x}^{3}}-6{{x}^{2}}-x+30=(x+2)({{x}^{2}}+ax+15)$.

Now, we must find the coefficient of the middle term in the second bracket. When you multiply the brackets out, the terms that you will multiply together and collect to get the $\scriptsize -6{{x}^{2}}$ term are: $\scriptsize 2\times {{x}^{2}}$ and $\scriptsize x\times a{{x}^{2}}$.

$\scriptsize (x+2)({{x}^{2}}+ax+15)$

$\scriptsize 2\times {{x}^{2}}+a\times {{x}^{2}}=-6{{x}^{2}}$

Solve this simple equation (which can be done in your head without showing working) to find $\scriptsize a$.

\scriptsize \displaystyle \begin{align*}2{{x}^{2}}+a{{x}^{2}} & =-6{{x}^{2}}\\a{{x}^{2}} & =-8{{x}^{2}}\\\therefore a & =-8\end{align*}

So the coefficient of the x-term is $\scriptsize -8$.

$\scriptsize \therefore {{x}^{3}}-6{{x}^{2}}-x+30=(x+2)({{x}^{2}}-8x+15)$

We must factorise the quadratic factor further to fully factorise the cubic polynomial.

\scriptsize \begin{align*}{{x}^{3}}-6{{x}^{2}}-x+30&=(x+2)({{x}^{2}}-8x+15)\\&=(x+2)(x-3)(x-5)\end{align*}

The zeros of the function are found by solving the previous equation.

\scriptsize \begin{align*}&(x+2)(x-3)(x-5)=0\\&\therefore x=-2,3\text{ or }5\end{align*}

Take note!

To factorise a cubic polynomial we do the following:

1. Find one linear factor by trial and error. Note: generally only try factors that would divide into the constant term. For example, if the constant term is $\scriptsize 3$ then the only factors you need to try are numbers that divide into $\scriptsize 3:-3,-1,1\text{ and }3$.
2. Use the factor theorem to confirm that your answer in question 1 is a factor by showing that $\scriptsize f(k)=0$.
3. Divide the given cubic polynomial by your linear factor to get the quadratic factor.
4. Factorise the quadratic trinomial to find the other two factors of the cubic polynomial.

Exercise 1.5

1. Factorise:
1. $\scriptsize f(x)={{x}^{3}}+{{x}^{2}}-9x-9$
2. $\scriptsize g(x)={{x}^{3}}-3{{x}^{2}}+4$
2. $\scriptsize f(x)=2{{x}^{3}}+{{x}^{2}}-5x+2$
1. Find $\scriptsize f(1)$.
2. Factorise $\scriptsize f(x)$ and list the zeros of the function.

The full solutions are at the end of the unit.

Sometimes it is not possible to factorise a quadratic expression using inspection, in which case we use the quadratic formula to fully factorise and solve the cubic equation.

Summary

In this unit you have learnt the following:

• How to divide polynomials using long division to find the quotient and remainder.
• How to divide polynomials using synthetic division to find the quotient and remainder.
• How to use the remainder theorem to find the remainder of a polynomial.
• How to use the remainder theorem to find an unknown value.
• How to find a linear factor of a cubic polynomial using the factor theorem.
• How to fully factorise a cubic polynomial using division by inspection.
• How to solve equations with cubic polynomials.

Unit 1: Assessment

Suggested time to complete: 45 minutes

1. The volume of a rectangular solid is given by the polynomial $\scriptsize 3{{x}^{4}}-3{{x}^{3}}-33{{x}^{2}}+54x$. The length is $\scriptsize 3x$ and the width is $\scriptsize x-2$. Find the height, h.
2. When $\scriptsize f(x)=3{{x}^{5}}+p{{x}^{4}}+10{{x}^{2}}-21x+12$ is divided by $\scriptsize x-2$ it leaves a remainder of $\scriptsize 10$. Find the value of $\scriptsize p$.
3. Solve $\scriptsize {{x}^{3}}+{{x}^{2}}-16x=16$.
4. Let $\scriptsize g(x)$ be a polynomial in $\scriptsize x$.
1. If $\scriptsize g(x)$ is divided by $\scriptsize x-k$, what does $\scriptsize g(k)$ represent?
2. If $\scriptsize g(k)=0$ what can be said about $\scriptsize x-k$?
5. The polynomial $\scriptsize f(x)={{x}^{3}}+p{{x}^{2}}-qx-6$ is exactly divisible by $\scriptsize {{x}^{2}}+x-2$.
1. Factorise $\scriptsize {{x}^{2}}+x-2$.
2. Calculate the values of $\scriptsize p$ and $\scriptsize q$.

The full solutions are at the end of the unit.

Unit 1: Solutions

Exercise 1.1

1. False. It is not a polynomial since the exponent is not a whole number.
2. True.
3. True.
4. False. A cubic polynomial is an expression with a degree of $\scriptsize 3$ and does not always have three terms.

Back to Exercise 1.1

Exercise 1.2

1. .
1. .
\scriptsize \displaystyle \begin{align*}2x-1\overset{{3{{x}^{2}}+2x-1}}{\overline{\left){\begin{align*}6{{x}^{3}}+{{x}^{2}}-4x+5\\-\underline{{(6{{x}^{3}}-3{{x}^{2}})}}\end{align*}}\right.}}\\\text{ }4{{x}^{2}}-4x\\\text{ }-\underline{{(\text{ }4{{x}^{2}}-2x)}}\text{ }\\\text{ }-2x+5\\\text{ }-\underline{{(2x+1)}}\\\text{ }4\end{align*}
\scriptsize \begin{align*}Q(x)&=3{{x}^{2}}+2x-1\\R(x)&=4\\\therefore 6{{x}^{3}}&+{{x}^{2}}-4x+5=(2x-1)\cdot (3{{x}^{2}}+2x-1)+4\end{align*}
2. .
\scriptsize \begin{align*}x-1\overset{{{{x}^{2}}+4x+3}}{\overline{\left){{{{x}^{3}}+3{{x}^{2}}-x-3}}\right.}}\\\text{ }-\underline{{({{x}^{3}}-{{x}^{2}})}}\\\text{ }4{{x}^{2}}-x\\\text{ }-\underline{{(4{{x}^{2}}-4x)\text{ }}}\text{ }\\\text{ }3x-3\\\text{ }-\underline{{(3x-3)}}\\\text{ 0}\end{align*}
\scriptsize \begin{align*}&Q(x)={{x}^{2}}+4x+3\\&R(x)=0\end{align*}
\scriptsize \begin{align*}\therefore {{x}^{3}}+3{{x}^{2}}-x-3&=(x-1)\cdot ({{x}^{2}}+4x+3)+0\\ &=(x-1)\cdot ({{x}^{2}}+4x+3)\end{align*}
2. Since there is no remainder when $\scriptsize \displaystyle {{x}^{3}}+3{{x}^{2}}-x-3$ is divided by $\scriptsize \displaystyle (x-1)$, we can factorise the divisor multiplied by the quotient completely and we will get the dividend.
\scriptsize \displaystyle \begin{align*}{{x}^{3}}+3{{x}^{2}}-x-3=(x-1)\cdot ({{x}^{2}}+4x+3)\\\text{ }=(x-1)(x+1)(x+3)\end{align*}

Back to Exercise 1.2

Exercise 1.3

1. .
\scriptsize \begin{align*}-2\left| \!{\underline {\, \begin{align*}2\text{ }-3\text{ }4\text{ }5\\\text{ }-4\text{ 14 }-36\end{align*} \,}} \right. \\\text{ }2\text{ }-7\text{ 18 }-31\end{align*}
\scriptsize \begin{align*}&\text{Quotient}=2{{x}^{2}}-7x+18\\&\text{Remainder}=-31\end{align*}
2. .
\scriptsize \begin{align*}2\left| \!{\underline {\, \begin{align*}4\text{ }10\text{ }-6\text{ }-20\\\text{ }8\text{ }36\text{ }60\end{align*} \,}} \right. \\\text{ 4 }18\text{ }30\text{ }40\end{align*}
\scriptsize \begin{align*}\text{Quotient}=4{{x}^{2}}+18x+30\\\text{Remainder}=40\end{align*}
3. .
Remember if a term is missing in the dividend we write the coefficient as $\scriptsize 0$.
\scriptsize \begin{align*}1\left| \!{\underline {\, \begin{align*}2\text{ }0\text{ }5\text{ }-4\\\text{ }2\text{ 2 7}\end{align*} \,}} \right. \\\text{ }2\text{ }2\text{ }7\text{ 3}\end{align*}
\scriptsize \begin{align*}&\text{Quotient}=2{{x}^{2}}+2x+7\\&\text{Remainder}=3\end{align*}

Back to Exercise 1.3

Exercise 1.4

1. .
1. .
\scriptsize \begin{align*}f(x)&=3{{x}^{3}}+5{{x}^{2}}-x+1\\f(-2)&=3{{(-2)}^{3}}+5{{(-2)}^{2}}-(-2)+1\\&=-24+20+2+1\\&=-1\end{align*}
2. .
\scriptsize \begin{align*}f(x)&=3{{x}^{3}}+5{{x}^{2}}-x+1\\f(\displaystyle \frac{1}{2})&=3{{\left( {\displaystyle \frac{1}{2}} \right)}^{3}}+5{{\left( {\displaystyle \frac{1}{2}} \right)}^{2}}-\left( {\displaystyle \frac{1}{2}} \right)+1\\&=\displaystyle \frac{3}{8}+\displaystyle \frac{5}{4}-\displaystyle \frac{1}{2}+1\\&=\displaystyle \frac{{17}}{8}\end{align*}
3. .
\scriptsize \begin{align*}f(x)&=3{{x}^{3}}+5{{x}^{2}}-x+1\\f(-\displaystyle \frac{1}{3})&=3{{(-\displaystyle \frac{1}{3})}^{3}}+5{{(-\displaystyle \frac{1}{3})}^{2}}-(-\displaystyle \frac{1}{3})+1\\&=-\displaystyle \frac{1}{9}+\displaystyle \frac{5}{9}+\displaystyle \frac{1}{3}+1\\&=\displaystyle \frac{{16}}{9}\end{align*}
2. .
$\scriptsize \text{Let }g(x)=2{{x}^{3}}-7{{x}^{2}}+mx-26$
Then $\scriptsize g(2)=-24$
\scriptsize \begin{align*}&g(2)=2{{(2)}^{3}}-7{{(2)}^{2}}+m(2)-2\\&\therefore -24=16-28+2m-2\\&\therefore -2m=10\\&\therefore m=-5\end{align*}

Back to Exercise 1.4

Exercise 1.5

1. .
1. Find a factor by trial and error.
\scriptsize \begin{align*}f(3)&={{(3)}^{3}}+{{(3)}^{2}}-9(3)-9\\&=0\end{align*}
\scriptsize \begin{align*}&f(3)=0\\&\therefore x-3\text{ is a factor of }f(x)\end{align*}
Divide to find the other factors. Use division by inspection, long division or synthetic division.
$\scriptsize {{x}^{3}}+{{x}^{2}}-9x-9=(x-3)({{x}^{2}}+ax+3)$
\scriptsize \begin{align*}{{x}^{2}}&=-3{{x}^{2}}+a{{x}^{2}}\\4{{x}^{2}}&=a{{x}^{2}}\\4&=a\end{align*}
.
\scriptsize \begin{align*}\therefore {{x}^{3}}+{{x}^{2}}-9x-9&=(x-3)({{x}^{2}}+4x+3)\\\text{ }&=(x-3)(x+1)(x+3)\end{align*}
To solve, let $\scriptsize (x-3)({{x}^{2}}+1)(x+3)=0$
$\scriptsize \therefore x=3,-1$ or $\scriptsize -3$
2. .
\scriptsize \begin{align*}g(x)&={{x}^{3}}-3{{x}^{2}}+4\\g(-1)&=-1-3+4\\&=0\end{align*}
$\scriptsize \therefore (x+1)\text{ is a factor of }g(x)$
Remember to include the $\scriptsize 0x$ term especially if using long division or synthetic division.
\scriptsize \begin{align*}g(x)&={{x}^{3}}-3{{x}^{2}}+0x+4\\&=(x+1)({{x}^{2}}+ax+4)\end{align*}
To find coefficient of the middle term:
\scriptsize \begin{align*}{{x}^{2}}+a{{x}^{2}}&=-3{{x}^{2}}\\a{{x}^{2}}&=-4{{x}^{2}}\\a&=-4\end{align*}
\scriptsize \begin{align*}{{x}^{3}}-3{{x}^{2}}+4&=(x+1)({{x}^{2}}-4x+4)\\&=(x+1)(x-2)(x-2)\end{align*}
Solve:
\scriptsize \begin{align*}&(x+1)(x-2)(x-2)=0\\&\therefore x=-1\text{ or }2\end{align*}
2. .
1. .
\scriptsize \begin{align*}f(1)&=2{{(1)}^{3}}+{{(1)}^{2}}-5(1)+2\\&=2+1-5+2\\&=0\end{align*}
2. .
$\scriptsize x-1$ is a factor of $\scriptsize f(x)$
\scriptsize \displaystyle \begin{align*}2{{x}^{3}}+{{x}^{2}}-5x+2 & =(x-1)(2{{x}^{2}}+ax-2)\\-2{{x}^{2}}+a{{x}^{2}} & ={{x}^{2}}\\a{{x}^{2}} & =3{{x}^{2}}\\\therefore a & =3\\2{{x}^{3}}+{{x}^{2}}-5x+2 & =(x-1)(2{{x}^{2}}+3x-2)\\ & =(x-1)(2x-1)(x+2)\end{align*}
The zeros of the function are $\scriptsize \displaystyle 1,\displaystyle \frac{1}{2}\text{ or}-2$

Back to Exercise 1.5

Unit 1: Assessment

1. .
\scriptsize \displaystyle \begin{align*}v&=l\cdot w\cdot h\\&\therefore 3{{x}^{4}}-3{{x}^{3}}-33{{x}^{2}}+54x=3x(x-2)h\\&\therefore (x-2)h=\displaystyle \frac{{3{{x}^{4}}-3{{x}^{3}}-33{{x}^{2}}+54x}}{{3x}}=\displaystyle \frac{{3x({{x}^{3}}-{{x}^{2}}-11x+18)}}{{3x}}={{x}^{3}}-{{x}^{2}}-11x+18\\&\therefore h=\displaystyle \frac{{{{x}^{3}}-{{x}^{2}}-11x+18}}{{x-2}}\end{align*}
$\scriptsize (x-2)$ is a factor of $\scriptsize \displaystyle {{x}^{3}}-{{x}^{2}}-11x+18$
.
\scriptsize \begin{align*}&{{x}^{3}}-{{x}^{2}}-11x+18=(x-2)({{x}^{2}}+ax-9)\\&\therefore -2{{x}^{2}}+a{{x}^{2}}=-{{x}^{2}}\\&\therefore a{{x}^{2}}={{x}^{2}}\\&\therefore a=1\end{align*}
$\scriptsize \displaystyle \therefore {{x}^{3}}-{{x}^{2}}-11x+18=(x-2)({{x}^{2}}+x-9)$
\scriptsize \displaystyle \begin{align*}&h=\displaystyle \frac{{(x-2)({{x}^{2}}+x-9)}}{{(x-2)}}\\&\therefore h={{x}^{2}}+x-9\end{align*}
2. .
$\scriptsize f(x)=3{{x}^{5}}+p{{x}^{4}}+10{{x}^{2}}-21x+12$
But $\scriptsize f(2)=10$
.
\scriptsize \begin{align*}&\therefore 3{{(2)}^{5}}+p{{(2)}^{4}}+10{{(2)}^{2}}-21(2)+12=10\\&\therefore 96+16p+40-42+12=10\\&\therefore 16p=10-106=-96\\&\therefore p=-6\end{align*}
3. .
$\scriptsize {{x}^{3}}+{{x}^{2}}-16x-16=0$
Find a factor.
\scriptsize \begin{align*}&{{(4)}^{3}}+{{(4)}^{2}}-16(4)-16=0\\&\therefore x-4\text{ is factor}\end{align*}
.
\scriptsize \begin{align*}{{x}^{3}}+{{x}^{2}}-16x-16&=(x-4)({{x}^{2}}+ax+4)\\&=(x-4)({{x}^{2}}+5x+4)\\&=(x-4)(x+1)(x+4)\end{align*}
\scriptsize \begin{align*}&(x-4)(x+1)(x+4)=0\\&\therefore x=4;-1\text{ and }-4\end{align*}
4. Let $\scriptsize g(x)$ be a polynomial in $\scriptsize x$.
1. $\scriptsize g(k)$ represents the remainder.
2. If then $\scriptsize x-k$ is a factor of $\scriptsize g(x)$.
5. The polynomial $\scriptsize f(x)={{x}^{3}}+p{{x}^{2}}-qx-6$ is exactly divisible by $\scriptsize {{x}^{2}}+x-2$.
1. .
$\scriptsize {{x}^{2}}+x-2=(x+2)(x-1)$
2. .
$\scriptsize f(-2)=0$
\scriptsize \displaystyle \begin{align*}&\therefore {{(-2)}^{3}}+p{{(-2)}^{2}}-q(-2)-6=0\\&\therefore -8+4p+2q-6=0\\&\therefore 4p+2q-14=0\\&\therefore 2p+q-7=0\end{align*}
.
$\scriptsize \displaystyle f(1)=0$
\scriptsize \displaystyle \begin{align*}&\therefore {{(1)}^{3}}+p{{(1)}^{2}}-q(1)-6=0\\&\therefore 1+p-q-6=0\\&\therefore p-q-5=0\end{align*}
Solve simultaneously
$\scriptsize \displaystyle 2p+q-7=0\text{ (}1)$
$\scriptsize \displaystyle p-q-5=0\text{ (2)}$
.
From (2):$\scriptsize p=q+5(3)$
Substitute into (1):
\scriptsize \displaystyle \begin{align*}&\therefore 2(q+5)+q-7=0\\&\therefore 2q+10+q-7=0\\&\therefore 3q+3=0\\&\therefore q=-1\end{align*}
Substitute into (3):
$\scriptsize p=-1+5=4\text{ }$

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