How do you solve (20x^3-8x^2+5x-5)/(5x-2) using synthetic division?
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I'm rusty on synthetic division, I forget what to do..
Whatever binomial you are dividing by, set equal to zero and solve for x to find the number to synthetically divide by. i.e. divisor = 5x-2. 5x-2=0 when x=2/5
Drop down the first term, multiply by the divisor (2/5) and add to the next term; then repeat the process.
|dw:1346809834228:dw| And so on.
Ah, I think I understand. If you were using a calculator would you just use the decimal of 2/5?
Yeah, that might be quicker: typing .4 instead of 2/5 (2 keystrokes instead of 3...) I think you'd be better off doing it by hand to sharpen your mental arithmetic skills. ;-)
None of the operations for this one even require working it out on paper. You should be able to do them all mentally without any trouble.
Alright thats good for me, I'm doing review questions and I havent done any of this in like 3 years
Let me know what you get and I'll check your answer.
So is it 8 0 5 0?
Those are the multiples that you then add to the subsequent terms.
the subsequent terms are 20-8 5 -5 right?
Right.
Ok, I'm getting kind of confused now hah. So all I have to do now is add the subsequent terms to the multiples? Do i add each subsequent term to each multiple?
Go one at a time in order. Start with the 20, drop it down and multiply by 2/5, that gives you 8. Add that 8 to the next term, -8 to make 0. 0 is the next digit in your answer, so multiply 0 by 2/5 to get . . . 0 and add that to the next term 5 to get 5 as your third digit. 2/5 of 5 is 2, so add 2 to -5 and get -3 as your remainder.
Now here is where you have to be careful about expressing your answer. This is synthetic division, so you have to put it in terms of the original divisor. This might look weird at first, but just take my word,.. er my numbers for it. \[q(x)=20x^2+5-\frac{3}{x-2/5}\] But your original divisor wasn't x-2/5 it was 5x-2, so you have to divide everything by 5 to get: \[q(x)=4x^2+1-\frac{3}{5x-2}\] ("q(x)" means quotient function.
I have to go now, so good luck!
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