# Help , question in a

#### yangwah80@

I can't understand the last three line in a) ?

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#### romsek

Math Team
Do you understand what the binomial coefficients are?

The rest is just multiplication and a bit of algebra.

yangwah80@

#### yangwah80@

Do you understand what the binomial coefficients are?

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

#### romsek

Math Team
Hi , do you understand why number if real root can only be in odd numbers,while number of non-real root can only be in even nunbers?
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

#### yangwah80@

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
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 $x=\pm i$

#### algebratormath

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Math Team

#### skipjack

Forum Staff
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