Question 672928
(x - 1)(x - 4); f(x) = x<sup>4</sup> - 5x³ + 13x² - 45x + 36 
<pre>
We first find {{{f(x)/(x-1)}}} of {{{(x^4 - 5x^3 + 13x^2 - 45x + 36)/(x-1)}}}

to show that the remainder is 0.  That will show that (x - 1) 
is a factor of f(x).  I'll use synthetic division where we use +1
to divide by (x - 1):

 1 | 1 -5 13 -45  36 
   |<u>    1 -4   9 -36</u>
     1 -4  9 -36   0

So therefore (x - 1) is a factor, and we have now factored f(x) as

f(x) = (x - 1)(x³ - 4x² + 9x - 36)

Now we must also show that the factor in the second parentheses is
divisible by (x - 1)

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Now we must show that (x - 4) is also a fsctor

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We now find {{{(x^3 - 4x^2 + 9x - 36)/(x-4)}}} 

to show that the remainder is 0.  That will show that (x - 4) 
is a factor of x³ - 4x² + 9x - 36, and therefore a factor of
f(x).  I'll use synthetic division again this time where we 
use +4 to divide by (x - 4):

 4 | 1 -4  9 -36 
   |<u>    4  0  36</u>
     1  0  9   0

So therefore (x - 4) is a factor, and we have now factored f(x) as

We started with"

f(x) = x<sup>4</sup> - 5x³ + 13x² - 45x + 36

and factored it with synthetic division as this

f(x) = (x - 1)(x³ - 4x² + 9x - 36)

And further factored the right factor with synthetic
division as this:

f(x) = (x - 1)(x - 4)(x² + 0x + 9)

Eliminating the zero term, this is the factored form

f(x) = (x - 1)(x - 4)(x² + 9)

which proves that (x - 1)(x - 4) is a factor of f(x).

Edwin</pre>