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Hello everyone. In this video, we're 
going to tackle Fall 2000 question 8.

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Let f of x be the following polynomial:

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x cubed minus 3x squared plus a x plus b

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in which the coefficients 
of a and b are real.

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If minus 1 plus root 3i is a root of f of x,

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determine the values of a and 
b and find all roots of f of x. 

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Okay, actually this has been my 
favorite problem on the exam thus far,

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which is kind of cool.

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So here we have a root and…

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we have a complex root, and because 
the coefficients are real, we know that

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by CJRT,

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we have the complex conjugate 
of this must also be a root,

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and so we know two factors.

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I'm going to fix the bracket.

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I've been good with my brackets 
today so let's fix that little bracket.

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Okay.

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So again, by CJRT, and I 
guess the Factor Theorem

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so by CJRT and the Factor Theorem…

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so let's put both of those in.

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We know what two factors are.

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So if we multiply these two 
factors together what do we get?

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Well the middle term should be the...

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the negation of the sum. The sum is minus 
2, and the negation of that is positive 2,

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and the constant term should be the 
product and the product of these two things

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should just be the norm, which is 4, okay?

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So by doing a little bit 
arithmetic, we know that

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x squared plus 2x plus 4 
must be a factor of f of x. 

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If it must be a factor, then 
if I do the long division,

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my remainder must be 0.

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So let's do the long division, x squared 
plus 2x plus 4 into this cubic polynomial.

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Multiply that by x, I get x cubed 
plus 2x squared plus 4x,

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simplify, I get minus 3

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minus positive 2, so that’s negative 5.

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a minus 4, that's a minus 4, 
then I bring the b down.

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So the leading coefficient to cancel that out
I need to multiply by negative 5, so I do that.

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Now what do I get? I get 
a plus 6 x plus b plus 20. 

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This remainder suspiciously 
doesn't look like 0,

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but it must be 0,

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so therefore the a and b
values must be chosen

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so that this polynomial is 0.

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Aye,

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that's easy to do. So a x plus 6 
times x plus b plus 20 equals 0,

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let's compare coefficients.
The coefficient of x is,

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on both sides, should be 0 
because this side it’s 0,

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so that means that a plus 6 is 
0, that is a equals minus 6,

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and thus b plus 20 must be 0, 
that is b equals negative 20,

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and that's it. Very quick
question, very cool.

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I like it, I think it's a
neat little question.

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Very quick to solve, and that's basically it.

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So what was the key? The key 
point here was using CJRT.

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So once you know one root, you know 
a lot of information about this polynomial.

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Once you know one root you know the 
other root, and once you have two roots,

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it shouldn't be too much 
work to find the third root.

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In this case, I didn't actually find the 
third root I just divided and used a

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nice little long division trick, so I used

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DAP, I guess, Division Algorithm for Polynomials,

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but yeah I think that's a…I think 
this is pretty cool problem.

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This is, like I said, one of my 
favourite problems in this exam.

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Nice, short, quick solution, but it might 
not be the first thing you come up with.

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So thank you and good luck.