# Algebraic Division

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Algebraic Division

Algebraic division is the process of dividing a polynomial by a linear expression. It’s useful as it breaks down complex polynomials into easier ones.

There are some key terms that you need to understand first:

• Degree – the highest power in a polynomial.
• Divisor –  the thing you are dividing by.
• Quotient – what you get when you divide by a divisor (it includes the remainder).
• Remainder – what is left over (this will be a constant in A level Maths).

There are 3 methods for algebraic division that you will see.

Make sure you are happy with the following topics before continuing.

A Level

## The Factor Theorem

The Factor Theorem is defined as:

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

or

“If $f \left(\dfrac{b}{a} \right)=0$, then $(ax-b)$ is a factor of $f(x)$

i.e. if you know the roots then you know the factors, and if you know the factors then you know the roots.

A Level

## Method 1: Subtracting Multiples of the Divisor

This method uses the following procedure:

Step 1: Subtract a multiple of $(x-k)$ to cancel the highest power of $x$.

Step 2: Repeat Step 1, until there are no powers of $x$ remaining.

Step 3: Work out how many lots of $(x-k)$ you subtracted, and write this as an expression with the remainder.

Example: Divide $2x^3 - 5x^2 + 4x - 3$ by $\textcolor{limegreen}{(x-3)}$

Step 1: Subtract $2x^2$ lots of $x-3$, so that the $x^3$ term is cancelled:

$(2x^3 - 5x^2 + 4x - 3) - \textcolor{red}{2x^2}\textcolor{limegreen}{(x-3)} = (2x^3 - 5x^2 + 4x - 3) - 2x^3 + 6x^2 = x^2+ 4x - 3$

Step 2: Repeat this process, to remove all powers of $x$.

Start subtracting $x$ lots of $x-3$ to remove the $x^2$ term in $x^2 + 4x - 3$:

$(x^2 + 4x - 3) - \textcolor{red}{x}\textcolor{limegreen}{(x-3)} = (x^2 + 4x - 3) - x^2 + 3x = 7x - 3$

Then, subtract $7$ lots of $(x-3)$ to remove the $x$ term:

$(7x-3) - \textcolor{red}{7}\textcolor{limegreen}{(x-3)} = (7x-3) - 7x + 21 = \textcolor{blue}{18}$

Step 3: In total, we have subtracted $\textcolor{red}{2x^2 + x + 7}$ lots of $\textcolor{limegreen}{(x-3)}$, and there is $\textcolor{blue}{18}$ left over.

So,

$(2x^3 - 5x^2 + 4x - 3) \div \textcolor{limegreen}{(x-3)} = \textcolor{red}{2x^2 + x + 7}$ remainder $\textcolor{blue}{18}$

A Level

## Method 2: Algebraic Long Division

This method uses the same principles as long division for numbers, but for algebraic expressions.

Example: Divide $x^3 + 6x^2 - 11x + 4$ by $\textcolor{limegreen}{(x-2)}$

Step 1: Divide $x^3$ by $x$ to get $\textcolor{red}{x^2}$, and put this at the top

Step 2: Multiply $x^2$ by $\textcolor{limegreen}{(x-2)}$ to get $x^3 - 2x^2$

Step 3: Subtract to get $8x^2$ and bring the $-11x$ down

Step 4: Divide $8x^2$ by $x$ to get $\textcolor{red}{8x}$, and put this at the top

Step 5: Multiply $8x$ by $\textcolor{limegreen}{(x-2)}$ to get $8x^2 - 16x$

Step 6: Subtract to get $5x$ and bring the $+4$ down

Step 7: Divide $5x$ by $x$ to get $\textcolor{red}{5}$, and put this at the top

Step 8: Multiply $5$ by $\textcolor{limegreen}{(x-2)}$ to get $5x-10$

Step 9: Subtract to get $\textcolor{blue}{14}$, which is the remainder – since this term has a degree that’s less than the divisor, therefore it can’t be divided.

Hence,

$(x^3 + 6x^2 - 11x + 4) \div \textcolor{limegreen}{(x-2)} = \textcolor{red}{x^2 + 8x + 5}$ remainder $\textcolor{blue}{14}$

Note: If the polynomial you are dividing doesn’t have an $x^2$ term for example, just put $0x^2$ where the $x^2$ usually goes.

A Level

## Method 3: Using a Formula

This method makes use of the following identity.

“A polynomial, $f(x)$, can be written as

$f(x) \equiv q(x)d(x) + r(x)$

where $q(x)$ is the quotient, $d(x)$ is the divisor and $r(x)$ is the remainder.”

You then use the following procedure:

Step 1: Find the degrees of the quotient and remainder. The degree of the quotient is $\text{deg } f(x) - \text{deg } d(x)$. For the degree of the remainder, $\text{deg } r(x) < \text{deg } d(x)$.

Step 2: Write the division in the form above, replacing $q(x)$ and $r(x)$ with general polynomials (i.e. $Ax^2 + Bx + C$ is a general polynomial of degree $2$)

Step 3: Find the values of the constants $A$, $B$ and $C$ etc., by substituting in values for $x$ and equating coefficients.

Step 4: Replace $A$, $B$ and $C$ etc. in the general polynomial with the values you have just found.

Example: Divide $x^3 + 4x^2 + 6x + 8$ by $\textcolor{orange}{(x+2)}$

Step 1: This polynomial has degree $3$, since the highest power if $x$ is $3$, and the divisor has degree $1$. Therefore the quotient has degree $3-1=2$ (so it is a quadratic). The remainder has degree $0$.

Step 2: Write the division in the form $f(x) \equiv q(x)d(x) + r(x)$:

$x^3 + 4x^2 + 6x + 8 \equiv (Ax^2 + Bx + C)\textcolor{orange}{(x+2)} + D$

Step 3: Substitute $x= -2$ to make $d(x)=0$, therefore the $q(x)d(x)$ part will disappear and will leave the remainder, $D$.

\begin{aligned} (-2)^3 + 4(-2)^2 +6(-2) + 8 &= D \\ 4 &= D \end{aligned}

Now, substitute $D=4$ and $x=0$ into the equation:

\begin{aligned} 8 &= 2C +4 \\ 2 &= C \end{aligned}

So, we have

\begin{aligned} &x^3 + 4x^2 + 6x + 8 \\ &\equiv (Ax^2 + Bx + 2)\textcolor{orange}{(x+2)} + 4 \\ &\equiv Ax^3 + (2A + B)x^2 + (2B+2)x + 8 \end{aligned}

Equating coefficients $x^3$, $x^2$ and $x$ gives:

$A = 1$ and $2A + B = 4$ so $B = 2$

Step 4: Put the values of $\textcolor{purple}{A=1}$, $\textcolor{limegreen}{B=2}$, $\textcolor{blue}{C=2}$ and $\textcolor{red}{D=4}$ into the identity, which gives

$x^3 + 4x^2 + 6x + 8 \equiv (x^2 + \textcolor{limegreen}{2}x + \textcolor{blue}{2})\textcolor{orange}{(x+2)} + \textcolor{red}{4}$

Hence,

$(x^3 + 4x^2 + 6x + 8) \div \textcolor{orange}{(x+2)} = (x^2 + \textcolor{limegreen}{2}x + \textcolor{blue}{2})$ remainder $\textcolor{red}{4}$

Note: For A level maths, you will only see questions involving $\text{deg } d(x) = 1$ and $\text{deg } r(x) = 0$

A Level

## Note:

The Factor Theorem can be combined with the 3 methods for dividing polynomials, which will enable you to factorise cubics and quartics.

A Level

## Example: The Factor Theorem

a) Show that $(x-2)$ is a factor of $f(x) = x^3 + 5x^2 - 19x + 10$

[2 marks]

b) The polynomial $g(x) = x^3 - 8x^2 + 11x + 20$ has roots at $x=4$, $x=5$ and $x=-1$. Factorise $g(x)$ completely.

[2 marks]

a) If $f(2) = 0$, then $(x-2)$ is a factor of $f(x)$ by the Factor Theorem.

$f(2) = (2)^3 +5(2)^2 - 19(2) + 10 = 0$

Hence, by the Factor Theorem, $(x-2)$ is a factor of $f(x)$.

b) By the Factor Theorem, if $a$ is a root of $g(x)$, then $g(a) = 0$. So $(x-a)$ is a factor of $g(x)$.

We’re given all the roots of the cubic, so we can factorise it using the Factor Theorem.

$g(x) = x^3 - 8x^2 + 11x + 20 = (x-4)(x-5)(x+1)$

A Level

## Example Questions

Use the factor theorem with $a =3$ and $b = 1$

If $f \left( \dfrac{1}{3} \right)=0$, then $(3x-1)$ is a factor of $f(x)$

$f \left( \dfrac{1}{3} \right) = 3 \left( \dfrac{1}{3} \right)^3 + 8 \left( \dfrac{1}{3} \right)^2 - 15 \left(\dfrac{1}{3} \right) + 4 = 0$

Hence, $(3x-1)$ is a factor of $f(x)$

a) Use the Factor Theorem with $a = -1$

If $f(-1)=0$ then $(x+1)$ is a factor of $f(x)$.

$f(-1) = 2(-1)^3 + 7(-1)^2 + 2(-1) - 3 = 0$

Hence, by the Factor Theorem, $(x+1)$ is a factor of $f(x)$.

b) $(x+1)$ is a factor of $f(x)$, so divide $f(x) = 2x^3 + 7x^2 + 2x - 3$ by $(x+1)$

$(2x^3 + 7x^2 + 2x - 3) - 2x^2(x+1) \\ = (2x^3 + 7x^2 + 2x - 3) - 2x^3 - 2x^2 \\ = 5x^2 + 2x - 3$

$(5x^2 + 2x - 3) - 5x(x+1) \\ = (5x^2 + 2x - 3) - 5x^2 - 5x \\ = -3x - 3$

$(-3x - 3) - (-3(x+1)) \\ = (-3x - 3) + 3x + 3 = 0$

So,

$2x^3 + 7x^2 + 2x - 3 = (x+1)(2x^2 + 5x - 3)$

$(2x^2 + 5x - 3) = (2x-1)(x+3)$

Hence,

$2x^3 + 7x^2 + 2x - 3 = (x+1)(2x-1)(x+3)$

Put $x = -5$ into both sides of the identity $2x^3 + 17x^2 + 37x + 18 \equiv (Ax^2 + Bx + C)(x+5) + D$:

$2(-5)^3 + 17(-5)^2 + 37(-5) + 18 = D$

$D = 8$

Now, let $x=0$

$18 = 5C + 8$, so $C = 2$

Equate the coefficients of $x^3$ to get $A=2$

Equate the coefficients of $x^2$ to get $5A + B = 17$, so $B = 7$

Hence,

$2x^3 + 17x^2 + 37x + 18 = (2x^2 + 7x + 2)(x+5) + 8$

A Level

A Level

A Level

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