(6y^4 + 15y^3-28y-6

Tracey, it looks like you are being asked to factor this expression into a sum or product of simpler terms, using "synthetic division". Synthetic division is the process by which you go testing for possible factors (factors with integer coefficients).

But first here, you should notice something immediately ... there's NO y^2 term! That means, you *might* not be dealing with a normal (ay^2 + by + c) x (dy^2 + ey + f) (or some other kind of) product. Try grouping instead as (6y^4-28y) + (15y^3-6) and try to factor each of the parenthesized terms; maybe you'll find a *common factor* in both terms!

So: 2(3y^2-14y) +3(5y^2-2) .... nope, didn't get us anyplace! Too bad.

OK, next you want to try simple factors of the form (ay+b). You know the rules: a has to be a factor of the first coefficient (of 6), b has to be a factor of the last coefficient (of -6) That leaves a lot of possibilities! a= 1,2,3, or 6; b=1,2,3,6,or -1,-2,-3, or -6. This looks really time-consuming, but once you start actually doing the math, you will get better eventually at predicting what might work, based on the results you get during the trial divisions. If you can't "zero out" an intermediate term during your division, you know that your (a,b) choice is impossible, AND you may even be able to eliminate that entire a. Here, for example, trying (6y-1) as a divisor shows you that: (6,-1) isn't possible (you'd get a 16y^3 term during the division, and 6 doesn't divide into that), and that (6,-3), which would give an 18y^3 term at one point during the division (looks like OK for division by 6, right?) *can't* work because it would be divisible by 3, and the original expression as a whole *isn't* (the -28). That knocks out a=6 completely (nothing else for b works on the +15y^3 term except b=+3, and that's out as -3 was).

And so, you work your way through possible sets of combinations.

That's a lot of slogging! It's possible to work smarter than just slogging through, though.

Now, what you should do after you have a few possible factors that work all the way through to a non-zero remainder, is compare that with the graph you get when you graph your original function on a graphing calculator. You'll see a relationship between these remainders and the value of the function for the particular value of y which you get when you solve the equation for each trial factor (ay + b)=0. You can use that information!!!

In particular, you should be aware that your original expression, as an equation = 0, can have (at most) 4 places where it crosses the x-axis (because it has a y^4 term as the highest power); these places correspond to factors for your original expression!

Let me say that again: a root (axis crossing) of the function (your original expression) corresponds to a factor of your original expression.

So, if you had (let's say) a crossing at the point y=2/3, this would correspond to a factor (3y-2) in the original expression (because 3y-2=0 solves to y=2/3). So if you have two possible factors you tried such as (y-1) [remainder = -13] and (y+1) [remainder = +13] you know that there MUST be a value between y=1 and -1 where the original expression has a factor (where the original expression computes to 0). The function (your original expression) had to be zero somewhere in between where it was -13 and where it was +13! This is valuable information; it points you towards possible roots (factors!). Unfortunately, that place where the original expression is zero MAY not be a rational root, which is what you need for a factor with integer coefficients. But if you don't have a graphing calculator, you can do the same thing by hand; just poke a few values for y into the original expression, and calculate the values of the expression that result. Graph these up (as y on the x-axis, f(x)= the original expression value on the y-axis). Then use the result to try to guess factors, which you can try for synthetic division.

As you get better at this, you can hop back and forth between the graph for your function (however you manage to draw it) and your synthetic division, to arrive at your answer (factors) faster.

Happy dividing!

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