Algebraic Division

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.

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 $textcolor{blue}{18}$

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.

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 $text{deg } r(x) = 0$