Question 1163336
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The response from tutor @Theo show you a path to the answer; but it doesn't show you HOW to factor the quadratic.  Since your question was how to factor the quadratic, that response is not of much use.<br>
The response from tutor @ikleyn shows you a path that always works, by completing the square.<br>
There are numerous other methods for finding the answer, including some techniques for actually doing the factoring.<br>
But if you are going to use completing the square to find the factorization, you might as well just use the quadratic formula.  The roots are<br>
{{{(-b +- sqrt(b^2-4ac))/(2a)}}}<br>
Plugging in a=30, b=-11, and c=-30...<br>
{{{(11 +- sqrt(121+3600))/60}}}
{{{(11+-sqrt(3721))/60}}}
{{{(11 +- 61)/60}}}
{{{72/60 = 6/5}}} or {{{-50/60 = -5/6}}}<br>
With roots 6/5 and -5/6, the quadratic expression is<br>
{{{(x-6/5)(x+5/6)}}}
or
{{{(5x-6)(6x+5)}}}<br>
For actually performing the factorization, here is one popular technique:<br>
(1) Divide the leading coefficient by 30 (to make it equal to 1) and multiply the constant by that same 30 to get a new quadratic: x^2-11x-900
(2) Factor this by the standard method -- finding two numbers whose product is 900 and whose difference is 11; those numbers (not easy to find) are 25 and 36
(3) Write the factorization as (x+25)(x-36) to give the roots -25 and +36
(4) Divide each root by 30 (as used in step 1) to get the roots -5/6 and +6/5
(5) Use those roots to write the factorization (6x+5)(5x-6)<br>
And finally a good old-fashioned method for factoring the quadratic....<br>
The factorization is going to be of the form<br>
(ax+b)(cx-d)<br>
The obvious conditions are
(1) the product of a and c is 30
(2) the product of b and d is 30<br>
With only those conditions, there are a huge number of possible factorizations.  However, there are additional conditions that greatly limit the number of possibilities.
(3) a and b are relatively prime
(4) c and d are relatively prime<br>
If either of (3) or (4) were violated, then one of the linear factors would have a common factor; and that would mean the original quadratic would have a common factor.  Since the original quadratic does not have a common factor, conditions (3) and (4) are required.<br>
It doesn't take too long now to list all the possible factorizations and to find the one that gives the correct middle term.<br>
(30x+1)(x-30)  no
(15x+2)(2x-15) no
(10x+3)(3x-10) no
(6x+5)(5x-6)  YES<br>