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when a polynomial f(x) is divided by (x-3) the remainder is -9 and when divided by (2x-1), the remainder is -6. Find the remainder when f(x) is divided by (x-3)(2x-1)

Hi Ndawana,

From the information provided about the remainders when f(x) is divided by each of x-3 and 2x-1, we know that f(3) = -9 and f(1/2) = -6.

Since the product (x-3)(2x-1) is quadratic then the remainder would in general be of the form ax+b. If the quotient is q(x), then we can write;

f(x) = (x-3)(2x-1)q(x) + (ax+b)

Now, we try to find f(3) and f(1/2) using this expression for f(x).

We have;

f(3) = 0 + (3a+b) = 3a + b

but it must be equal to -9 as we've been given.

Therefore,

3a + b = -9

In the same manner,

f(1/2) = 0 + (a/2+b) = a/2 + b

and

a/2 + b = -6

We now have a pair of simultaneous equations in a and b which can be solved to give;

a = -6/5 and b = -27/5

The remainder when f(x) is divided by (x-3)(2x-1) is therefore -6x/5 - 27/5.

Hope it is clear.

Regards

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