Then we add everything together, and we get our f of x is going to be equal to ax to the third let me scroll to the left a little bit ax to the third. x squared minus ax squared.

Break

Minus 2ax. ax times minus x is minus ax squared, right?

Break at Next

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What will this look like when we expand this into a trinomial?

I'll do it in very general terms.

Well this is the same thing as (ax b) times (ax b) which is the same thing as ax times ax. ax times ax is the same thing is a 2 x 2 plus ax times b which is abx plus b times ax which is another bax or abx plus b times b.

And now this a times this x.

And that's equal to x times x, x squared. x times that a, so that's plus ax.

Minus ax squared and then ax times x squared is ax to the third. ax to the third.

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And then I'm going to have to multiply that times ax plus 1, so I'm going to write the ax plus 1 down here just to save space. ax plus 1.

1 times x squared is x squared.

2x times b is plus 2bx. a times x is plus ax. a times b is plus ab.

And press WRlTE ENTER

But as you see, all we did is distribute it, this a times three.

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Now give me your initialization text.

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Let me write that because I went too far to the right.

So how do we do that?

Well, somebody slipped it in his drink!

Because if I introduced any products with any x terms, you're going to get one of these other terms over here.

But let's say that I have a quadratic equation that says ax squared plus bx plus c is equal to 0.

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Because if we have it in this form, or actually ax.

Well let's subtract c from both sides so we get ax squared plus the bx is equal to minus c.

I like just having one coefficient on my x squared term so let me divide everything by a.

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A root of a polynomial is the x coordinate of one of its x intercepts.

Well this is a plus here, so the parenthesis really don't matter.

However, I have also think that the amount of information you need in a week needs to be appropriate for a beginning class.

We can't enter a value in the Tolerance box since we only use that in Break Detect Mode

I just wanted to take a break because we have had almost one and a half hours plus, I think of video right now for today.

All right, I'll do it in blue. f of x is going to be equal to x plus 1 times x minus 2 times ax now minus a times r3.

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I'm trying to get into a format x minus a squared is equal to y minus b.

So the other binomial, just substituting that Ax squared plus Bx, is 2x squared plus Bx.

This will be x times x, which is x squared, plus x times b, which is bx, plus a times x, plus a times b plus ab.

That is working.

And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx.

Because if you were to multiply just the x terms here, when you actually do your distributive property and multiply out these binomials, you'll just get an x to the third, but we had an ax to the third.

I didn't derive every piece of math for you. But you've learned a number of things.

Because, if I just had let me write a different problem here if I just wrote a let me just make it clear, this is a separate problem if I just wrote a times x plus three, you know that that's just a x plus, you could say a three, or another way to say that would be three a, right?