I think you made two errors. Tou wrote this:
If (2+2) and (x-3) are factors of the polynomial f(x) =x^2+ax^2-7x+b where a
and b are constant determine the values of a and b and hence factorise f(x)
completely
I think you meant this:
If (x+2) and (x-3) are factors of the polynomial f(x) =x^3+ax^2-7x+b where a
and b are constant determine the values of a and b and hence factorise f(x)
completely.
Can you find your two errors?
But instead of doing that one for you I will do this one, which is exactly like yours step by step.
If (x+7) and (x-4) are factors of the polynomial f(x) =x^3+ax^2-13x+b where a
and b are constant determine the values of a and b and hence factorise f(x)
completely
Since they are both factors of f(x), their product is also a factor of f(x).
We multiply them together: (x+7)(x-4) = x²-4x+7x-28 = x²+3x-28.
Since that product is a factor we divide f(x) by it using long division. You
will have to spread the long division out like this since the coefficients
involve a and b:
x + (a-3)
x²+3x-28)x³+ ax² -13x + b
x³+ 3x² -28x
(a-3)x² 15x b
(a-3)x² (3a-9)x - 28(a-3)
[15-(3a-9)]x + b+28(a-3)
The quotient is x+(a-3).
The remainder must be identically 0 for all values of x, so:
and 
Substitute a=8 in
Therefore the quotient is x + (a-3) = x + (8-3) = x+5
So the complete factorisation is (x+7)(x-4)(x+5).
Now do your problem the same way.
Edwin
