Question

any help would be appreciated.. I have a test, and I am unsure of my probs below... ty.

Answer 1

where are the problems?

Answer 2

I have one...
Given that the polynomial f(x) has a zero at x=2 find the remaining zeros..
f(x)=x^3-10x^2+41x-50

Answer 3

You can factor (x-2) from f(x) if it has a zero at x=2.

Answer 4

What does that mean? It says my answer should be in a+bi form?

Answer 5

having a zero at x=2 means that (x-2) is one factor of f(x), you have to find the other factors.. using long division gives:
\[f(x)=(x-2)(x^2-8x+25)=0\] you already have the solution x=2
to find the other solutions solve the quadratic equation:
\[x^2-8x+25=0\]

Answer 6

using the quadratic formula gives us the two remaining solutions:
\[x=4-3i, x=4+3i\]

Answer 7

do you know how to use the quadratic formula?

Answer 8

Long division.. we are being taught syntheic.. is that ok?

Answer 9

there is only one solution to that division, and that step is really trivial it does not matter how you solve it.

Answer 10

Can you show me the way the quadratic formula is worked out on here??

Answer 11

The solution formula for \[ x^2 + px + q = 0\] is \[x = -\frac{p}{2} \pm \sqrt{\frac{p^2}{4} - q}\]

Answer 12

And thats the quadratic formula??

Answer 13

becasue I have never seen that before??

Answer 14

Some schools teach the form \[ax^2 + bx +c = 0\] where you would of course get a formula which is a bit different. But the above is the exact solution formula for such quadratic problems.

Answer 15

Thanks