Algebraic division is used to determine factors to expressions that have a cubic or higher term. It is virtually the same as long division. For algebraic division, a linear term (called a binomial) `(x +n)`, where `n` is an integer, is used as the divisor.

Factorise `x^3 + 7x^2 + 14x + 8` by dividing the cubic by `(x + 1)`.

Set up a long division with `(x+1)` being the divisor

What multiplies `(x + 1)` to obtain an `x^3` term?

The answer is `x^2`: place that in the `x^2` column

Multiply `(x + 1)` by this value of `x^2` to get a value of `x^3 + x^2`: place this in the appropriate columns and subtract

Draw down the `14x` for a new expression of `6x^2 + 14x`

What should multiply `(x+1)` to obtain an `6x^2` term? The answer is `6x`

Carry on repeating until the division is complete

This gives a quadratic `x^2 + 6x + 8` as an answer

Factorise the quadratic to obtain `(x + 4)` and `(x + 2)`.

`x^2` | `+ 6x` | `+ 8` | ||

`(x + 1)` | `x^3` | `+ 7x^2` | `+ 14x` | ` + 8` |

`-` | `x^3` | `+ x^2` | ||

`6x^2` | `+ 14x` | |||

`-` | `6x^2` | `+ 6x` | ||

`8x` | `+ 8` | |||

`-` | `8x` | `+8` |

Answer: `(x + 4)(x + 2)(x + 1)`

A cubic function `x^3 - 19x + 30` crosses the `x`-axis at `x=3`. What are the values of `x` for the two other locations where the line crosses the `x`-axis?

If the curve crosses the `x`-axis at `x=3` then one of the factors must be `(x - 3)`. Carry out an algebraic division: note that when writing out the division, the `x^2` term is shown as a placeholder.

The division yields a quadratic of `x^2 + 3x -10`. Factorise this to obtain `(x+5)` and `(x - 2)`. the other points at which the curve crosses the `x`-axis are (-5, 0) and (2, 0).

`x^2` | `+ 3x` | `- 10` | ||

`(x - 3)` | `x^3` | `+ 0x^2` | `- 19x` | `+ 30` |

- | `x^3` | `- 3x^2` | ||

`3x^2` | `- 19x` | |||

- | `3x^2` | `- 9x` | ||

- | `10x` | `+ 30` | ||

`10x` | `- 30` |

Answer: `x = -5` and `x=2`

See also Long Division

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