Question 1173289
<pre>
I think you made two errors. Tou wrote this:
</pre>
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<pre>
I think you meant this:</pre>
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.<pre>
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.</pre>
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<pre>

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:

              <u>                           x +     (a-3)</u>
      x²+3x-28)x³+         ax²        -13x +         b
               <u>x³+         3x²        -28x</u>
                       (a-3)x²         15x           b
                       <u>(a-3)x²     (3a-9)x -   28(a-3)</u>
                              [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:

{{{15-(3a-9)=0}}}  and {{{b+28(a-3)=0}}}
{{{15-3a+9=0}}}
{{{24-3a=0}}}
{{{24=3a}}}
{{{8=a}}}

Substitute a=8 in {{{b+28(a-3)=0}}}

{{{b+28(8-3)=0}}}
{{{b+28(5)=0}}}
{{{b+140=0}}}
{{{b=-140}}}

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</pre>