Ramainder Theorem

[Ravingsofthesane]

Hello everyone .. got another issue that Ihave been trying to solve today

I appreciate anyone who can offer some advise on this.


Instruction :

Determine the remainder for the following divisions using the remainder theorem. If the divisor is a factor of the dividend, so state.

Problem : (3x^3 - 3x^2 - 5x + 16) / (x+3)

My understanding of how this works is the following

If F(X) / AX - B
then F(B/A)

so the result would be

3*2^3 - 3*2^2 - 5*2 + 16 = Remainder ( 18 in this case )

Is this correct ? or am i taking this the incorrect way?

thanks

 

[Matt]

Polynomial long division or synthetic division.

 

[soroban]

Hello, Ravings!

Your mind must be wandering . . .

Determine the remainder for the following divisions using the remainder theorem.
If the divisor is a factor of the dividend, so state.

Problem : \(\displaystyle \L\frac{3x^3 - 3x^2 - 5x + 16}{x+3}\)


My understanding of how this works is the following

If a polynomial \(\displaystyle F(x)\) is divisible by \(\displaystyle AX - B\), then: \(\displaystyle F\left(\frac{B}{A}\right)\,=\,0\)

so the result would be:
. . \(\displaystyle 3\cdot2^3\,-\,3\cdot2^2\,-\,5\cdot2\,+\,16\:=\:\text{Remainder}\;\;\text{(18 in this case)}\)
Why did you plug in x = 2 ?

There is a simpler version of the Remainder Theorem.

Given a polynomial \(\displaystyle f(x)\), the remainder after dividing by \(\displaystyle (x-a)\) is \(\displaystyle f(a).\)


You have: \(\displaystyle f(x)\:=\:3x^3\,-\,3x^2\,-\,5x\,+\,16\)

. .Hence: .\(\displaystyle f(-3)\:=\:3(-3)^3\,-\,3(-3)^2\,-\,5(-3)\,+\,16\:=\:-77\)

 

[Ravingsofthesane]

Hello, Ravings!

Your mind must be wandering . . .

Determine the remainder for the following divisions using the remainder theorem.
If the divisor is a factor of the dividend, so state.

Problem : \(\displaystyle \L\frac{3x^3 - 3x^2 - 5x + 16}{x+3}\)


My understanding of how this works is the following

If a polynomial \(\displaystyle F(x)\) is divisible by \(\displaystyle AX - B\), then: \(\displaystyle F\left(\frac{B}{A}\right)\,=\,0\)

so the result would be:
. . \(\displaystyle 3\cdot2^3\,-\,3\cdot2^2\,-\,5\cdot2\,+\,16\:=\:\text{Remainder}\;\;\text{(18 in this case)}\)
Why did you plug in x = 2 ?

There is a simpler version of the Remainder Theorem.

Given a polynomial \(\displaystyle f(x)\), the remainder after dividing by \(\displaystyle (x-a)\) is \(\displaystyle f(a).\)


You have: \(\displaystyle f(x)\:=\:3x^3\,-\,3x^2\,-\,5x\,+\,16\)

. .Hence: .\(\displaystyle f(-3)\:=\:3(-3)^3\,-\,3(-3)^2\,-\,5(-3)\,+\,16\:=\:-77\)

thanks for the help .. my mind has been wandering it has been a long day today