# Factorising Polynomials with Complex Roots

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#1
I am quite comfortable with solving quadratics with complex roots by using the formula, and have experience of using the factor theorem from C2.

But how to I go about factorising quadratics and higher order polynomials that have complex roots? I am sure it is actually quite simple as I must be missing something.

I have an exercise of questions requiring factorising, the first two questions are as follows:

a) 9z^2 - 6z + 5
b) 4z^2 + 12z + 13

I have looked at the answers and tried to work backwards, but am still unsure. Is there a way to start these questions, perhaps by replacing z by a+bi?
0
5 years ago
#2
if the roots really are complex, then you use a form of the quadratic formula with just +sqrt(b^2-4ac) not +/-sqrt(b^2-4ac) then, you have to find out what the 2 square roots are of: check the discriminant first, though.

(real coefficients imply conjugate complex roots) - e.g. a cubic such as: has roots: 1
#3
Thank you very much for your advice. I can now factorise by calculating the roots and working backwards.
0
5 years ago
#4
(Original post by Jimm1y)
Thank you very much for your advice. I can now factorise by calculating the roots and working backwards. 0
#5
How would I go about solving equations that contain only complex roots, like the following:

z^4+z^3+5z^2+4z+4=0

I guess I could use the factor theorem by inputting values of i until I get zero, but this could take a long time.
0
5 years ago
#6
(Original post by Jimm1y)
How would I go about solving equations that contain only complex roots, like the following:

z^4+z^3+5z^2+4z+4=0

I guess I could use the factor theorem by inputting values of i until I get zero, but this could take a long time.
Are you sure you're meant to be solving that?

You won't be expected to solve general quartic equations unless there's a really obvious root, or unless the equation relates to something you've done in an earlier part of the question!
0
5 years ago
#7
(Original post by Jimm1y)
How would I go about solving equations that contain only complex roots, like the following:

z^4+z^3+5z^2+4z+4=0

I guess I could use the factor theorem by inputting values of i until I get zero, but this could take a long time.
I'd note that that quartic is the product of two quadratics, i.e. These you can both factorise individually easily enough. Generally quartics are going to be pretty hard to solve, unless you can see some trick like that. You could also have noted that were roots by a bit of trial and error and go from there. That's similar to how you would solve a cubic, except complex roots always come in pairs .
0
5 years ago
#8
Look for a root by just inspecting normally there will be 1 real root which you can typically find quite easy by trying values that are factors of the constant term (i.e. a) it is 5) etc. Then you can do long division until you get a quadratic which you can find roots using formula then just factorize easily.

I believe you can replace a+bi and then compare coefficients in some questions with complex coefficients.
0
5 years ago
#9
(Original post by Rinsed)
I'd note that that quartic is the product of two quadratics, i.e. These you can both factorise individually easily enough. Generally quartics are going to be pretty hard to solve, unless you can see some trick like that. You could also have noted that were roots by a bit of trial and error and go from there. That's similar to how you would solve a cubic, except complex roots always come in pairs .
How did you figure out that the quartic could be factorized as you suggested? Trial and error, or some other pattern?
0
5 years ago
#10
(Original post by Doctor_Einstein)
How did you figure out that the quartic could be factorized as you suggested? Trial and error, or some other pattern?
It's all a bit of filling in the gaps. Assume it can be the product of two quadratics, and assuming we're not dealing with fractional factors (which seems reasonable) then each quadratic has to start with a and the constant parts have to multiply to 4. Then it's just a question of what order of constants and z factors will multiply to give that quartic. Luckily this is a simple case so you don't have to trial and error for long. 4*1 = 4, 4+1 = 5 et cetera.

Essentially what we're doing is starting from generalised cubics and obtaining simultaneous equations by equating factors, so if you wanted you could write them all out and solve them, but that would take longer.
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