Hello everyone. In this video, we're going to tackle Fall 2000 question 8.
Let f of x be the following polynomial:
x cubed minus 3x squared plus a x plus b
in which the coefficients of a and b are real.
If minus 1 plus root 3i is a root of f of x,
determine the values of a and b and find all roots of f of x.
Okay, actually this has been my favorite problem on the exam thus far,
which is kind of cool.
So here we have a root and…
we have a complex root, and because the coefficients are real, we know that
by CJRT,
we have the complex conjugate of this must also be a root,
and so we know two factors.
I'm going to fix the bracket.
I've been good with my brackets today so let's fix that little bracket.
Okay.
So again, by CJRT, and I guess the Factor Theorem
so by CJRT and the Factor Theorem…
so let's put both of those in.
We know what two factors are.
So if we multiply these two factors together what do we get?
Well the middle term should be the...
the negation of the sum. The sum is minus 2, and the negation of that is positive 2,
and the constant term should be the product and the product of these two things
should just be the norm, which is 4, okay?
So by doing a little bit arithmetic, we know that
x squared plus 2x plus 4 must be a factor of f of x.
If it must be a factor, then if I do the long division,
my remainder must be 0.
So let's do the long division, x squared plus 2x plus 4 into this cubic polynomial.
Multiply that by x, I get x cubed plus 2x squared plus 4x,
simplify, I get minus 3
minus positive 2, so that’s negative 5.
a minus 4, that's a minus 4, then I bring the b down.
So the leading coefficient to cancel that out I need to multiply by negative 5, so I do that.
Now what do I get? I get a plus 6 x plus b plus 20.
This remainder suspiciously doesn't look like 0,
but it must be 0,
so therefore the a and b values must be chosen
so that this polynomial is 0.
Aye,
that's easy to do. So a x plus 6 times x plus b plus 20 equals 0,
let's compare coefficients. The coefficient of x is,
on both sides, should be 0 because this side it’s 0,
so that means that a plus 6 is 0, that is a equals minus 6,
and thus b plus 20 must be 0, that is b equals negative 20,
and that's it. Very quick question, very cool.
I like it, I think it's a neat little question.
Very quick to solve, and that's basically it.
So what was the key? The key point here was using CJRT.
So once you know one root, you know a lot of information about this polynomial.
Once you know one root you know the other root, and once you have two roots,
it shouldn't be too much work to find the third root.
In this case, I didn't actually find the third root I just divided and used a
nice little long division trick, so I used
DAP, I guess, Division Algorithm for Polynomials,
but yeah I think that's a…I think this is pretty cool problem.
This is, like I said, one of my favourite problems in this exam.
Nice, short, quick solution, but it might not be the first thing you come up with.
So thank you and good luck.