Question

NO LINKS!! Please help me with these questions:

1. How can synthetic division make the division more efficient, and when it is applicable?

2. When dividing a polynomial by a binomial, what does the remainder tell you?

3. How can polynomial equations be solved?

Answer 1

Problem 1

Synthetic division is more efficient because we focus on the coefficients only, and don't need to write the variables. This allows for a smaller footprint. Meaning that you don't need as much room on your paper. Plus it means there's less time devoted to it. After enough practice, synthetic division problems can be done very quickly.

While synthetic division is the better choice compared to polynomial long division, it's only applicable if you divide by a binomial and this binomial must be linear. We cannot divide by something like x^2+3 or x^4-x^2+5x-10. If the denominator is of this more complicated form, then we have no choice but to use polynomial long division instead.

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Problem 2

If the remainder is 0, then it tells us that the binomial is a factor of the numerator polynomial. The quotient polynomial then makes up the other factor. That quotient may or may not be able to be factored further (you may have to do another round of synthetic division).

An alternative is to use the remainder theorem. This is where we plug in the test root and see if we get 0 as a result. If so, then we get a remainder 0 and it concludes with the binomial being a factor.

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Problem 3

A polynomial is of the form

`a_nx^n+a_(n-1)x^(n-1)+a_(n-2)x^(n-2)+\cdots+a_(2)x^(2)+a_(1)x+a_(0)`

Focus on the last term
`a_0` and the leading coefficient
`a_n`

To find all possible rational roots, we divide factors of
`a_0` by factors of
`a_n`

Then we test each possible rational root with synthetic division. We're looking for cases where we get a remainder of 0. Refer to problem 2.

Use software like Desmos or GeoGebra to visually verify the answers. Recall that the x intercepts or roots represent the solutions to the equation.