SOLUTION: 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

Algebra.Com
Question 1173289: 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
Found 3 solutions by ikleyn, Edwin McCravy, mccravyedwin:
Answer by ikleyn(52805)   (Show Source): You can put this solution on YOUR website!
.
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
~~~~~~~~~~~~~~~~


Try to write your post correctly.


I see at least one error in your post.


But when I see one error, I may assume that there are 15 or 17 more other errors.



Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
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
Answer by mccravyedwin(408)   (Show Source): You can put this solution on YOUR website!

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