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    How can synthetic division be used to factor a polynomial?

    Here is a reasonable Pre-Calculus example of to illustrate the concept.

    Let’s say you had:

    ##2x^4 – 3x^3 – 5x^2 + 3x + 8##

    Like Joan said, there is a trial and error aspect to this.

    Look at all the coefficients, and think about what a common factor might be. If you don’t get a zero remainder, then the factor doesn’t really work and you should try again. If the possible factors are all used up, maybe it isn’t factorable.

    Here, factors you might try include the ones that correspond to the fourth order coefficient (2) and the zeroth order coefficient (8).

    8 has factors of 1, 2, 4, and 8, and 2 has 1 and 2. So the possible factors might be said to be ##pmp/q##, where ##p## is the factors of the highest degree coefficient, and ##q## is the factors of the zeroth degree coefficient.

    You can have: ##pm[1, 2, 4, 8, 1/2]##

    So you can try all of these (##2/2##, ##4/2##, and ##8/2## are duplicates). Recall that if ##-a## is used as what is written in the synthetic division process on the left corner, it corresponds to ##x+a##. We will use ##-1## here. I tend to try ##1## and ##-1## first, and go up in value, and try the fractions last.

    ##-1 |## 2 -3 -5 3 8

    Drop down the 2, and multiply by the -1 to get -2.

    ##-1 |## 2 -3 -5 3 8 ………….-2……….. ………2

    Add -3 and -2, then multiply the resultant -5 by -1 again. ##-1 |## 2 -3 -5 3 8 ………….-2..5….. ………2..-5

    Repeat until you’re done.

    Add -3 and -2, then multiply the resultant -1 by -1 again. ##-1 |## 2…-3…-5…3…8 …………..-2…..1….0…-3 ………2…-5…0….3….5

    Your answer here happens to be this, where 2 corresponds to ##2x^3##. So, one way to express the result is:

    ##2x^3 – 5x^2 + 0x + 3 + 5/(x+1)##

    where the ##5/(x+1)## was written by saying that the last value below the horizontal bar (below -2, 1, 0, -3), being 5, is divided by the ##x pm a## equation such that ##x pm a = 0##. So, ##x+1## indicates that the factor we have just used is ##-1##.

    Naturally, if the remainder is 0, you do not have to divide by the ##x pm a## expression.

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