Simple Polynomial Question - FORGOT?!?!?! (1 Viewer)

[wb47]

P(x) = (x+1)(X-3)Q(x) + ax + b

if P(x) is divided by (x+1)(X-3), why is the answer ax + b?

I know i know this but I forgot the reasoning and it's really frustrating,

Thank You.

 

[math man]

by remainder thm, P(a) = R(x), where R(x) is your remainder, and the remainder is always at least one degree less than the divisor

Edit: above polynomial if of form P(x) = A(x)Q(x) + R(x), where A(x) is your divisor, Q(x) is your quotient and R(x) is your remainder

 

[SpiralFlex]

P(x) = (x+1)(X-3)Q(x) + ax + b

if P(x) is divided by (x+1)(X-3), why is the answer ax + b?

I know i know this but I forgot the reasoning and it's really frustrating,

Thank You.

You have just expressed your answer in division transformation form.



Dividend = divisor * quotient + remainder.

 

[wb47]

Thanks so much!

 

[mike12345678]

can someone help me please.i cant seem to do this ploynomial question.

i) reuce this to linear factors.
3x^3+3x^2-x-1

ii) hence solve the following equation 3tan^3@+3tan^2@-tan@-1=0
@=theta or alpha...

i cant seem to make my own thread thats why i posted here

 

[D94]

can someone help me please.i cant seem to do this ploynomial question.

i) reuce this to linear factors.
3x^3+3x^2-x-1

ii) hence solve the following equation 3tan^3@+3tan^2@-tan@-1=0
@=theta or alpha...

i cant seem to make my own thread thats why i posted here

You can factorize a out of the first two terms, so you're going to get:


Now factorize a out of the two expressions:

which equals (by difference of two squares in the second bracket)


For the second part, if you compare the two equations, we can let and solve each part by letting it equal zero.

 

[mike12345678]

yeaa thats rite thanx