Factor this expression 8x^3+4x^2+6x+3
Are you familiar with synthetic division?
No sir.
Ok, give me a sec and I'll get you up and running...
x=-1/2
Maybe he is not familiar with the rational roots theorem?
maybe sooo
It's actually x = 1/2 not -1/2. Just saying.
suuuure
x - 1/2 = 0...x = 1/2. The rational roots theorem requires you to find the numbers that go into the last term, the constant, and the first term. Alll of them are as follows: \[\pm3, \pm1, \pm2, \pm4, \pm8,\pm \frac{ 3 }{ 2 },\pm \frac{ 3 }{ 4 }, \pm \frac{ 3 }{ 8 }\]\[\pm \frac{ 1 }{ 2 }, \pm \frac{ 1 }{ 4 },\pm \frac{ 1 }{ 8 }\]Considder the constant term the "c" and the first term, the 8, the d. All your possibilities for your divisor are either c, d, or c/d. Once you have them all you have to test them. Heres how it works:
In order for you to find a divisor, there has to be no remainder when you use synthetic division. Once you find a factor that goes in evenly, you have left over a polynomial to the second degree which you are most likely familiar with. Here's how the synthetic division goes. I will do one that doesn't work and then the one that does so you can see.
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