How do I find the other roots of a polynomial if one root is given?

[skyza]

Here's the problem I have to do:

Let f(x)= x^3 -7x^2 +17x-15
One zero of f(x) is K=2-i find all others and write f(x) in a factored form.


I ended up getting:

f(x)= x^2 -x+2

Is this correct?


Thanks

 

[Curious3141]

Here's the problem I have to do:

Let f(x)= x^3 -7x^2 +17x-15
One zero of f(x) is K=2-i find all others and write f(x) in a factored form.


I ended up getting:

f(x)= x^2 -x+2

Is this correct?

Clearly not, because f(x) was given as a cubic, but you ended up with a quadratic. But the bigger problem is that the quadratic you wrote is not even a factor of f(x).

Say we set f(x) = 0 to get a polynomial equation in x with integer coefficients. Do you know any property that complex roots of such an equation must have? Hint: conjugate.

Once you use that hint, you can easily work out the correct quadratic factor of f(x), then perform long division.

If you want to use synthetic division using the complex root (2-i), it's possible, but fairly tedious (complex multiplication is quite a lot of work).

Of course, it's possible to work out the remaining real root (which happens to be an integer) by using the rational root theorem and a little trial and error, following which you can do the synthetic division easily, but that wouldn't be following the hint given in the question. They want you to work out the quadratic factor first, then divide to get the single linear factor.

 

[skyza]

This is how I've done it so far. I have to use synthetic division. I tried -1 & 1 and neither worked, so I tried 3

 

[Curious3141]

This is how I've done it so far. I have to use synthetic division. I tried -1 & 1 and neither worked, so I tried 3

In essence, you're using the Rational Root Theorem, as I mentioned. The way the question was phrased (giving you a complex factor) seems to be against this approach. Nevertheless, it'll give you the right answer.

BTW, you don't have to test each of those candidate roots by division! Just substitute them back into f(x) and see if the expression becomes zero. In this case, f(3) = 0, so (x-3) is a factor of f(x).

Now, looking at your synthetic division of f(x) over (x-3), the first step's fine. You ended up with (1, -4, 5, 0) in a row. With that, you're actually done. That row of numbers signifies a quadratic polynomial, with the first three numbers representing the respective coefficient from x² down to the constant term. The final zero represents the remainder after division - you expect this to be 0, because (x-3) is a factor of f(x). I don't know why you continued writing numbers below that row, it's not necessary, and that's why you got a wrong answer.

So the quadratic factor is x² - 4x + 5.

The final factorisation of f(x) = (x-3)(x² - 4x + 5).

Note that the quadratic factor x² - 4x + 5 has a negative discriminant (the b² - 4ac thing), so you can conclude that it has no real linear factors. If you want to try to factorise it, you'll end up with (x-z₁)(x-z₂), where z₁ and z₂ are complex conjugates. In fact, you were given the value of one of the complex numbers at the start of the question, and here it's (2-i). So x² - 4x + 5 = (x - (2-i))(x-(2+i)) = (x-2+i)(x-2-i).

 

[skyza]

In essence, you're using the Rational Root Theorem, as I mentioned. The way the question was phrased (giving you a complex factor) seems to be against this approach. Nevertheless, it'll give you the right answer.

BTW, you don't have to test each of those candidate roots by division! Just substitute them back into f(x) and see if the expression becomes zero. In this case, f(3) = 0, so (x-3) is a factor of f(x).

Now, looking at your synthetic division of f(x) over (x-3), the first step's fine. You ended up with (1, -4, 5, 0) in a row. With that, you're actually done. That row of numbers signifies a quadratic polynomial, with the first three numbers representing the respective coefficient from x² down to the constant term. The final zero represents the remainder after division - you expect this to be 0, because (x-3) is a factor of f(x). I don't know why you continued writing numbers below that row, it's not necessary, and that's why you got a wrong answer.

So the quadratic factor is x² - 4x + 5.

The final factorisation of f(x) = (x-3)(x² - 4x + 5).

Note that the quadratic factor x² - 4x + 5 has a negative discriminant (the b² - 4ac thing), so you can conclude that it has no real linear factors. If you want to try to factorise it, you'll end up with (x-z₁)(x-z₂), where z₁ and z₂ are complex conjugates. In fact, you were given the value of one of the complex numbers at the start of the question, and here it's (2-i). So x² - 4x + 5 = (x - (2-i))(x-(2+i)) = (x-2+i)(x-2-i).

I really appreciate the help. Some of my notes made me think to keep going, but I see how to do it now. I also figured out that I don't need to test each number.
Thank you.

 

[HallsofIvy]

Why are you trying 1 and -1? You are told that 2- i is a root. You could use synthetic division to divide by 2- i. Or, since this polynomial has only real coefficients, you can argue that 2+ i is also a root and so a quadratic factor is $(x- 2+ i)(x- 2- i)= ((x- 2)+ i)((x- 2)- i)= (x- 2) ^2- i^2= x^2- 4x+ 4- 1= x^2- 4x+ 3$ and divide by that to find the third linear factor. Of course, you cannot us "synthetic" division to do that.

 

[Curious3141]

Why are you trying 1 and -1? You are told that 2- i is a root. You could use synthetic division to divide by 2- i. Or, since this polynomial has only real coefficients, you can argue that 2+ i is also a root and so a quadratic factor is $(x- 2+ i)(x- 2- i)= ((x- 2)+ i)((x- 2)- i)= (x- 2) ^2- i^2= x^2- 4x+ 4- 1= x^2- 4x+ 3$ and divide by that to find the third linear factor. Of course, you cannot us "synthetic" division to do that.

That should read ... = (x-2)² - i² = x² - 4x + 4 + 1 = x² - 4x + 5.