Polynomials is not only can be operated by addition and subtraction but also multiplication and division.

Table of Contents

**Multiplying polynomials**

Multiplying polynomial is like simple algebra operation. It can be multiplied by a constant or another polynomial.

**1. Multiplying by a constant (a term)**

Multiplying polynomial with a constant or a term has same method. It is multiplying the constant or a term with each term in polynomial.

Let a as *a* constant and it multiply with polynomial bx^{n} + cx + d then

*a* (bx^{n} + cx + d) = *a*bx^{n} + *a*cx + *a*d

let *a*x^{m} as a term multiply with polynomial bx^{n} + cx + d then

*a*x^{m} (bx^{n} + cx + d) = *a*x^{m}.bx^{n} + *a*x^{m}.cx + *a*x^{m}.d

*= a*bx^{mn }+ acx^{m+1} + adx^{m}

**2. Multiplying by another polynomial**

Multiplying by another polynomial is also has same concept with multiplying by a term or a constant. But it be done term by term depend on the both polynomials.

Let polynomial *a*x^{m} + bx^{n} – c multiply with dx^{m} + e then

(*a*x^{m} + bx^{n} – c )(dx^{m} + e) = *a*x^{m}.dx^{m} + *a*x^{m}.e + bx^{n}.dx^{m} + bx^{n}.e – c. dx^{m} – c.e

= *a*dx^{2m} + *a*ex^{m} + bdx^{m+n} + bex^{n} – cdx^{m} – ce

**Dividing Polynomials**

There are two method that can be used.

**1. Horner’s Method**

Horner’s method usually used to simple divider. It is two term of divider or it has one in the power. For example the dividers are x+1, x-3, etc..

There are some steps using Horner’s method:

- Divider is equals with zero. Then change the form to be definition of the variable (example: x = …)
- Write down the polynomials (only the coefficients and the constant)
- Look at the pattern in example below.

If there is x^{4} – 2x^{3} – 5x + 1 divide by x-1

It can be written as (1)x^{4} – 2x^{3} – 5x + 1 divide by x-1 then

Divider is x -1 = 0 → x = 1

1 (coefficient of x^{4}) | -2(coefficient of x^{3}) | 0(coefficient of x^{2}) | -5(coefficient of x) | 1 (constant) | |

1 (divider) | (blank space) | 1 (divider x re-write = 1×1 = 1) | -1(divider x -1 | -1(divider x -1) | -6(divider x -6) |

1(just re-write) | -1(-2 + 1 = -1) | -1(0 – 1 = -1) | -6(-5 – 1 = -6) | -5(1 – 6 = -5) |

So, 1 -1 -1 -6 are the result.

It can be written using variable to be (1)x^{3} – (1)x^{2} – (1)x – 6 or x^{3} – x^{2} – x – 6.

And -5 in the last row-column is the remainder of division.

**2. Long Method**

Long method or classic method usually use to more complicated divider. If there are two polynomial that consist of more than two term, it will be easy using this method.

The main concept is dividing one by one and then subtract the polynomial and the result of multiplying between previous result and the divider. Look at the example to get the concept.

Example:

If there is 2x^{4} – 3x^{3} + x^{2} + x – 1 divide by x^{2} + x + 1 then

Result → 2x^{2} (from 2x^{4 }: x^{2}) – 5x (from 2^{nd} part-5x^{3} : x^{2}) + 4 (from 4x^{2} : x^{2}) | |

x^{2} + x + 1 | 2x^{4} – 3x^{3} + x^{2} + x – 1 |

(divider) | 2x^{4}(from result times 1^{st} term divider) + 2x^{3 }(from result times 2^{nd} term divider) + 2x^{2} (result times 3^{rd} divider) – |

-5x^{3} – x^{2} + x – 1 (from 2x^{4} – 3x^{3} + x^{2} + x – 1 – (2x^{4} + 2x^{3} + 2x^{2})) | |

-5x^{3} – 5x^{2} + 5x (same with previous ways with 2^{nd} term result) – | |

4x^{2} -4x – 1 (from -5x^{3} – x^{2} + x – 1 – (-5x^{3} – 5x^{2} + 5x)) | |

4x^{2} + 4x + 4 (same with previous ways with 3^{rd} term result) – | |

-8x – 5 (remainder) |

So, the result is 2x^{2} – 5x + 4 and the remainder is -8x – 5.

Notes:

- There is no certain situation / problem that make you must choose Horner’s or long method. It is all depend on you.
- The both methods need your carefulness.

**Examples**

- Determine the result of:

a. (2x^{3} – 4x – 1)(2x^{2})

b. (xz^{2} + yz – 1)(z^{3} – yz^{2} + z)

a. (2x^{3} – 4x – 1)(2x^{2}) = 2x^{3} (2x^{2}) – 4x(2x^{2}) – 1(2x^{2})

= 4x^{5} – 8x^{3} – 2x^{2}

b. (xz^{2} + yz – 1)(z^{3} – yz^{2} + z)

= (xz^{2})(z^{3}) – (xz^{2})(yz^{2}) + (xz^{2})(z) + (yz)(z^{3}) – (yz)(yz^{2}) + (yz)(z) – (1)(z^{3}) + (1)(yz^{2}) – (1)(z)

= xz^{5} – xyz^{4} + xz^{3} + yz^{4} – y^{2}z^{3} +yz^{2} – z^{3} + yz^{2} – z

= xz^{5} + (y-x)z^{4} + (x-y^{2}-1)z^{3} + 2yz^{2} -z

2. Determine the result and the remainder of:

a. x^{3} – 3x^{2} + 2x – 1 divide by x+2

b. 2b^{4} + b^{3} divide by b^{2}+b+1

It will be easier using Horner’s method

Because of x+2 = 0 then divider is x = -2

1 | -3 | 2 | -1 | |

-2 | -2 | 10 | -24 | |

1 | -5 | 12 | -25 |

So, the result is 1x^{2} – 5x + 12 or can be written as x^{2} – 5x + 12

And the remainder is 15.

b. It will be easier using long division method

2b^{2} – b – 1 | |

b^{2}+b+1 | 2b^{4} + b^{3} |

2b4 + 2b^{3} + 2b^{2} – | |

-b^{3} – 2b^{2} | |

-b^{3} – b^{2} – b – | |

-b^{2} + b | |

-b^{2} – b – 1 – | |

1 |

So, the result is 2b2 – b – 1 and the remainder is 1.

3. Determine the value of *a* if x + 3 is factor of x^{3} + x^{2} – *a*x – 9.

Using Horner’s method

Divider is x + 3 = 0 → x = -3

1 | 1 | –a | -9 | |

-3 | -3 | 6 | -18 + 3a_{} | |

1 | -2 | 6-a | 3a – 27 |

Because of x + 3 is factor of x^{3} + x^{2} – *a*x – 9 then the remainder must be zero.

3*a *– 27 = 0 → 3*a* = 27 → *a* = 9