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Division of Polynomials

Hi there,

I've been given a question which has me banging my head against the wall, the question is;

Find the equation in x3 which vanishes when x=-1 and x=2. has the value 8 when x=0 and leaves a remainder of 16/3 when divided by 3x+2.

From the question i can devise that d=8, i can also get 3 equations when i equate the coefficients by assuming that -2/3 is also a factor (this was given by another person on the course and i'm still unsure how he arrived at it) but my attempts at solving the equations has not yielded great results :frown:

Anyone care to point me in the right direction?

The 3 equations i get are;

1) -a+b-c+8=0
2) 8a+4b+2c+8=0
3) (-8a)/(27)+(4b)/(9)-(2c)/(3)+(8)/(3)=0
(edited 8 years ago)
Original post by Lakura225
Hi there,

I've been given a question which has me banging my head against the wall, the question is;

Find the equation in x3 which vanishes when x=-1 and x=2. has the value 8 when x=0 and leaves a remainder of 16/3 when divided by 3x+2.

From the question i can devise that d=8, i can also get 3 equations when i equate the coefficients by assuming that -2/3 is also a factor (this was given by another person on the course and i'm still unsure how he arrived at it) but my attempts at solving the equations has not yielded great results :frown:

Anyone care to point me in the right direction?

The 3 equations i get are;

1) -a+b-c+8=0
2) 8a+4b+2c+8=0
3) (-8a)/(27)+(4b)/(9)-(2c)/(3)+(8)/(3)=0

Hi.

If the equation vanishes when x=-1 and x=2, then both (x+1) and (x-2) are factors. So the equation is (x+1)(x-2)(px+q). This now leaves you with only two unknowns to find. Note that I do not need to start with (x+1)(x-2)(px+q)+r because if there are already two different roots, there has to be a third one, which may or may not be a repeated root.

You can substitute in x=0 as you did before, which will then allow you to find the value of q, as p will disappear.

You can substitute in x = -2/3, knowing that the answer will be 16/3 to find the value of p.

Having found both values, multiply out the brackets to get the equation, or you may even be able to leave it factorized.

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