Which of the binomials below is a factor of this trinomial?
x^2+19x+84
a. x-7
b. x+84
c. x+7
d. x^2+12

You would use syntheitc division to find this out. If, after we do the division, we have a remainder of 0, then the binomial is a factor. If there is a remainder that is a real number (any number that's not a 0), then the binomial is not a factor. It's that simple. We will start at the top. Our first binomial is x - 7 = 0. That means that x = 7. We will put 7 outside the "box" and the coefficients of the terms inside the box, like this: 7 (1 19 84). Bring down the 1. Multiply that 1 by the 7 to get 7. Put that 7 up under the 19 and add to get 26. Multiply that 26 by the 7 to get 182. Put that 182 up under the 84 and add to get 266. Definitely NOT a 0, so x - is not a factor. Let's move to the next one. x + 84 = 0, so x = -84. Put -84 outside the box and the coefficients of the terms inside like this: -84 (1 19 84). Bring down the 1 and multiply it by -84 to get -84. Put that -84 up under the 19 and add to get -65. Multiply that -65 by the -84 to get 5460. Again, not a 0. Moving on...x + 7 = 0 so x = -7. Set up the same way. Bring down the 1 and multiply by the -7 to get -7. Put that -7 up under the 19 and add to get 12. Multiply 12 by the -7 to get -84. Put the -84 up under the 84 and add to get 0. That means that x + 7 is a factor of the polynomial. Just so you know, one of the rules of synthetic division is that you are only allowed to use binomials as your possible roots. x^2+12 is itself a polynomial so we can't use sythetic division to do this. Also, because both the polynomials are second degree polynomials, they won't divide evenly anyway. So your only factor is x + 7. There you go!

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