Use long division to find the quotient and remainder. (6x^3...

Use long division to find the quotient and remainder.\\ \frac{6x^3 - 2x + 4}{2x + 1}\\ quotient $3x^2 - \frac{3x}{2} - \frac{1}{4}$\\ remainder $\frac{17}{4}$\\Write the answer in the form $p(x) = d(x) \cdot q(x) + R(x)$, as in equation (2) in the text.\ $6x^3 - 2x + 4 = (2x)(3x^2 - 1) + \frac{17}{4}$

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00:02 All right, we're using long division here.

00:05 And when you set them up, you're missing a few powers.

00:08 So you're going to write x squared 0x plus 1 to keep it in descending order.

00:17 Then you have 2x to the 4th.

00:19 There's no x to the 3rd.

00:20 You're right plus 0x to the 3rd, minus 5x squared, minus 2x plus 1.

00:33 First thing you do is divide your first terms.

00:36 So i just kind of here to the side and say if a 2x to the 4th, divided by x squared will give me 2x squared and always line them up with the same matching power now i know what i'm done 2x squared now you're going to multiply this 2x squared by everybody over here all three terms so you should get you get this way okay you should get 2x to the 4th plus this would be 0 x to the 3rd you know you don't have to write that x the 3rd i just put it there to show and then plus 2x squared.

01:21 And then we're going to subtract everything, so you're going to change the signs minus minus.

01:29 You should go away.

01:30 This is going to disappear to you.

01:32 So you have zero x to the third.

01:34 I'm just using putting the zero there for place value.

01:36 Then you have negative 7x squared and you're going to bring everybody else down.

01:40 All right...

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