Help , question in a

romsek

Math Team
Sep 2015
2,961
1,674
USA
Do you understand what the binomial coefficients are?

The rest is just multiplication and a bit of algebra.
 
  • Like
Reactions: [email protected]
Feb 2020
5
0
Malaysia
Do you understand what the binomial coefficients are?

The rest is just multiplication and a bit of algebra.
Hi, do you understand why number of real root can only be in odd numbers, while number of non-real root can only be in even numbers?
 

romsek

Math Team
Sep 2015
2,961
1,674
USA
Hi, do you understand why number of real root can only be in odd numbers, while number of non-real root can only be in even numbers?
If the coefficients of the polynomial in question are all real. Then non-real roots must come in conjugate pairs. That's why their number must be even.

The number of real roots doesn't have to be odd though.

$f(x) = (x-1)(x-2)(x-3)(x-4)(x-i)(x+i)$

has 4 real roots and 1 pair of complex conjugate roots
 
Feb 2020
5
0
Malaysia
20200212_154721.jpg
If the coefficients of the polynomial in question are all real. Then non-real roots must come in conjugate pairs. That's why their number must be even.

The number of real roots doesn't have to be odd though.

$f(x) = (x-1)(x-2)(x-3)(x-4)(x-i)(x+i)$

has 4 real roots and 1 pair of complex conjugate roots
From the equation you gave, I get this graph.So the non-real root is ?
 

romsek

Math Team
Sep 2015
2,961
1,674
USA
View attachment 10914

From the equation you gave, I get this graph.So the non-real root is ?
You can't see the non-real roots from a 2D graph. You need to plot $|f(x)|$ against the complex plane.

You can see them from the form of the expression though.
It should be pretty clear the non-real roots of $f(x)$ are $x=\pm i$
 

skipjack

Forum Staff
Dec 2006
21,481
2,470
How do I get rid of the purchase offers?