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 bxn + cx + d then
a (bxn + cx + d) = abxn + acx + ad
let axm as a term multiply with polynomial bxn + cx + d then
axm (bxn + cx + d) = axm.bxn + axm.cx + axm.d
= abxmn + acxm+1 + adxm
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 axm + bxn – c multiply with dxm + e then
(axm + bxn – c )(dxm + e) = axm.dxm + axm.e + bxn.dxm + bxn.e – c. dxm – c.e
= adx2m + aexm + bdxm+n + bexn – cdxm – 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 x4 – 2x3 – 5x + 1 divide by x-1
It can be written as (1)x4 – 2x3 – 5x + 1 divide by x-1 then
Divider is x -1 = 0 → x = 1
| 1 (coefficient of x4) | -2(coefficient of x3) | 0(coefficient of x2) | -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)x3 – (1)x2 – (1)x – 6 or x3 – x2 – 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 2x4 – 3x3 + x2 + x – 1 divide by x2 + x + 1 then
| Result → 2x2 (from 2x4 : x2) – 5x (from 2nd part-5x3 : x2) + 4 (from 4x2 : x2) | |
| x2 + x + 1 | 2x4 – 3x3 + x2 + x – 1 |
| (divider) | 2x4(from result times 1st term divider) + 2x3 (from result times 2nd term divider) + 2x2 (result times 3rd divider) – |
| -5x3 – x2 + x – 1 (from 2x4 – 3x3 + x2 + x – 1 – (2x4 + 2x3 + 2x2)) | |
| -5x3 – 5x2 + 5x (same with previous ways with 2nd term result) – | |
| 4x2 -4x – 1 (from -5x3 – x2 + x – 1 – (-5x3 – 5x2 + 5x)) | |
| 4x2 + 4x + 4 (same with previous ways with 3rd term result) – | |
| -8x – 5 (remainder) |
So, the result is 2x2 – 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. (2x3 – 4x – 1)(2x2)
b. (xz2 + yz – 1)(z3 – yz2 + z)
a. (2x3 – 4x – 1)(2x2) = 2x3 (2x2) – 4x(2x2) – 1(2x2)
= 4x5 – 8x3 – 2x2
b. (xz2 + yz – 1)(z3 – yz2 + z)
= (xz2)(z3) – (xz2)(yz2) + (xz2)(z) + (yz)(z3) – (yz)(yz2) + (yz)(z) – (1)(z3) + (1)(yz2) – (1)(z)
= xz5 – xyz4 + xz3 + yz4 – y2z3 +yz2 – z3 + yz2 – z
= xz5 + (y-x)z4 + (x-y2-1)z3 + 2yz2 -z
2. Determine the result and the remainder of:
a. x3 – 3x2 + 2x – 1 divide by x+2
b. 2b4 + b3 divide by b2+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 1x2 – 5x + 12 or can be written as x2 – 5x + 12
And the remainder is 15.
b. It will be easier using long division method
| 2b2 – b – 1 | |
| b2+b+1 | 2b4 + b3 |
| 2b4 + 2b3 + 2b2 – | |
| -b3 – 2b2 | |
| -b3 – b2 – b – | |
| -b2 + b | |
| -b2 – 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 x3 + x2 – ax – 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 x3 + x2 – ax – 9 then the remainder must be zero.
3a – 27 = 0 → 3a = 27 → a = 9