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