What are the conventions for the division format to use the synthetic division algorithm? What can you do if the divisor is not in the required form?
See solution.
Practice makes perfect
Synthetic division is used to divide a polynomial by a binomial in a concise way. This requires us to follow certain conventions. In particular, the divisor must be of the form x-k. This reverses the sign of the k-value, which allows us to add instead of subtracting in the division algorithm. Let's see how to ensure this for different cases.
Dividing by a binomial of the form (x+a)
In these cases, we can rewrite the binomial as x-(- a) and use - a as our k-value.
Example [0.8em]
(2x^3-x^2-4) ÷ (x+3)
⇕
(2x^3-x^2-4) ÷ (x- ( - 3))
Therefore, to divide by x+3 we divide by x-(- 3) and use the k-value k= - 3.
Dividing by a binomial of the form bx-a
Adapting this binomial requires a bit of work. First, we can rewrite the binomial as b(x- ab). Then, to perform the division by b(x- ab) we can divide all the dividend coefficients by b first. Finally, we divide by x- ab using synthetic division with k= ab as our k-value.
Example
4x^3-2x^2-8 ÷ 2x-1
First, let's start by factoring out the coefficient of the x-term — in this case, 2. Then we can divide out this factor.
Now we we can rewrite our original division.
4x^3-2x^2-8 ÷ 2x-1
⇕
2x^3-x^2-4 ÷ x- 12
In this new format, we can use synthetic division by using k = 1 2 as the k-value.
Conclusion
The conventions for using the algorithm of synthetic division requires the divisor to be of the form x-k. Considering the cases mentioned above, as long as the binomial has the form ax+b we can rewrite our division so that the divisor take the form x-k. Then we proceed as usual by using the synthetic division algorithm.
