Question 1190367
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Having trouble figuring out how to factor this problem.
Factor completely given that x-3 is a factor of x^3-5x^2-14x+60.
I tried grouping to factor and it isn't working.
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<pre>
Having given that (x-3) is a factor of x^3 - 5x^2 - 14x + 60  is a very 

significant piece of information.  Having it, you can divide the given polynomial

by linear factor (x-3) without a remainder and get the quotient, which is

a quadratic polynomial.  It is the standard way to proceed in this case.


    +----------------------------------------------------------------------------+
    |    But grouping-regrouping ALSO WORKS in this case, despite the statement  |
    |                                                                            |
    |    made by @greenestamps.  I will show you below HOW IT WORKS.             |
    +----------------------------------------------------------------------------+


So, I start with  this polynomial  (x^3 - 5x^2 - 14x + 60).



Since I am given that (x-3) is the factor, I will extract the leading part of the polynomial 

in the form (x^3 - 3x^2), leaving the rest for the other part 

    x^3 - 5x^2 - 14x + 60 = (x^3 - 3x^2) + (-2x^2 - 14x + 60) = x^2*(x-3) + (-2x^2 - 14x + 60).      (1)



Further, I will work with this remaining part, keeping the leading part of the polynomial as is.  So

    x^2*(x-3) + (-2x^2 - 14x + 60) = x^2*(x-3) - 2*(x^2 + 7x - 30).      (2)



Next, I can easily  factor the last trinomial, and the fact that (x-3) is the factor, WILL HELP ME AGAIN

    x^2*(x-3) - 2*(x^2 + 7x - 30) = x^2*(x-3) - 2*(x-3)*(x+10) = (x-3)*(x^2 - 2(x+10)) = (x-3)*(x^2 - 2x - 20).    (3)



Combining lines (1), (2) and (3), I get the final decomposition

    x^3 - 5x^2 - 14x + 60 = (x-3)*(x^2 - 2x - 20).



So, I did it using grouping-regrouping.



Surely, direct division (long division or synthetic division) is more straightforward way, 

but grouping-regrouping DOES WORK, TOO,  and the given info about a / (the) linear factor helps you to do it 

from the beginning to the end, through all intermediate steps. 
</pre>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Solved by grouping-regrouping, &nbsp;for your better understanding.