by **Guest** » Wed Jan 22, 2014 5:12 pm

The original question was about "The Remainder Theorem Help"

The question was about evaluating polynomials, factors, quotients, remainders etc and the relationships between them.

A polynomial P(x) when divided by (x-3) had twice the remainder than when the same polynomial was divided by (x+1).

If P(x) was divisable by (x-3) and the remainder was Zero then (x-3) would be a factor of P(x) and the expression would

have a Root or Zero at (x+3).....but not in our case here.

Similar for the other divisor (x+1), it would be a factor if remainder was Zero.....but is not zero here

writing the other way around...

P(x) = (x-3)x(Q(x)) + R1 and P(x) = (x+1)x(Q(x)) + R2

(x-3) is 0 when (x=3) and (x+1) is 0 when (x=-1)

So P(x) = R1 when x=3 and P(x) = R2 when x=-1

and we are told R1 = 2 x R2

So at the values (x=3) and (x=-1) the values of the polynomial P(x) is equal to the remainder.

P(x) = X^3 - 2X^2 + aX + 5

At (x=3)....

(3x3x3) + (-2x3x3) + (3a) + 5

= 27 - 19 + 3a +5

= (3a + 14)

At (x=-1)....

(-1x-1x-1) + (-2x-1x-1) + (-a) + 5

= -1 - 2 - a + 5

= (2 - a)

Now we are told that x=3 remainder is twice the x=-1 remainder so.....

2(2 - a) = (3a + 14)

4 - 2a = 3a + 14

4 - 14 = 3a + 2a

- 10 = 5a

- 2 = a

So....The value of the constant "a" is -2

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.........................Simple.....................