SOLUTION: factor this quadratic equation 30(p^2-1)=11p

Question 1163336: factor this quadratic equation
30(p^2-1)=11p
Found 4 solutions by Alan3354, Theo, ikleyn, greenestamps:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
factor this quadratic equation
30(p^2-1)=11p
---------------
Equations are solved, not factored.

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
your equation is 30 * (p^2 - 1) = 11 * p
simplify to get:
30 * p^2 - 30 = 11 * p
subtract 11 * p from both sides of the equation to get:
30 * p^2 - 11 * p - 30 = 0
factor this quadratic equation to get:
p = 1.2 or p = -.83333333......
when p = 1.2, the equation becomes:
30 * (1.2^2 - 1) = 11 * 1.2
this results in 13.2 = 13.2 which is true.
when p = -.83333333......, the equation becomes:
30 * ((-.83333333......)^2 - 1) = 11 * (-.83333333......)
this results in -9.1666666666...... = -9.1666666666...... which is true.
solution is confirmed to be good.
your values for p are:
p = 1.2 or p = -.83333333......

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

30*(p^2-1) = 11p This equation is EQUIVALENT to this = 0. Complete the square, step by step = = = = = = +/- = +/- p = +/- The roots are = = = and = = - = - . Hence, the polynomial can be factored in this way = = = . It is the factorization the problem wants from you.

Solved.

Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!

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.

The response from tutor @ikleyn shows you a path that always works, by completing the square.

There are numerous other methods for finding the answer, including some techniques for actually doing the factoring.

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

Plugging in a=30, b=-11, and c=-30...

or

With roots 6/5 and -5/6, the quadratic expression is

or

For actually performing the factorization, here is one popular technique:

(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)

And finally a good old-fashioned method for factoring the quadratic....

The factorization is going to be of the form

(ax+b)(cx-d)

The obvious conditions are
(1) the product of a and c is 30
(2) the product of b and d is 30

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

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.

It doesn't take too long now to list all the possible factorizations and to find the one that gives the correct middle term.

(30x+1)(x-30) no
(15x+2)(2x-15) no
(10x+3)(3x-10) no
(6x+5)(5x-6) YES

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