Remainder Theorum

[cisaclasslass]

I am learning about the Remainder Theorum which is: dIvide P(x) by x-r= P(r).
It does not make sense yet to me.
I just worked this problem:
f(x) is 2x^4-9x^3+6x-5 find F(5)
I got the solution of 150 by substituting 5 for (x), Is this correct? Or should I have divided the equation by x-5? What are I see practical uses for theorum?
Thank you for your help.

 

[lookagain]

I am learning about the Remainder Theorum which is: dIvide P(x) by x-r= P(r).

f(x) is 2x^4-9x^3+6x-5 find F(5)
I got the solution of 150 by substituting 5 for (x), Is this correct? Or should I have divided the equation by x-5?
What are I see practical uses for theorum?

\(\displaystyle f(5)\) does equal \(\displaystyle 150.\) Yes, you can substitute \(\displaystyle 5\) for \(\displaystyle x.\)

If you divide the polynomial (the dividend) by \(\displaystyle (x - 5),\) you should type this dividend as \(\displaystyle 2x^4 - 9x^3 + 0x^2 + 6x - 5\)
before working the division.

What to look for:

The remainder after doing this division is supposed to be \(\displaystyle 150,\) which is \(\displaystyle f(5).\)

So, as a practical use (in my opinion) that you asked about, if you were asked to find the remainder when
this particular \(\displaystyle f(x)\) is divided by \(\displaystyle (x - 5),\) then it would be \(\displaystyle f(5).\) You would then avoid writing out that
work of long division.