SOLUTION: to factor an equation of to forrm 24^2p-55p-24=0, find a pair of integers whose product is AC and addds up to B

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: to factor an equation of to forrm 24^2p-55p-24=0, find a pair of integers whose product is AC and addds up to B      Log On


   



Question 486733: to factor an equation of to forrm 24^2p-55p-24=0, find a pair of integers whose product is AC and addds up to B
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

factor an equation of to form 24p%5E2-55p-24=0,
24p%5E2-55p-24=0.......replace -55p with the 9p++-64p
24p%5E2%2B9p++-64p-24=0.....group
%2824p%5E2%2B9p%29++-%2864p%2B24%29=0
3p%288p%2B3p%29++-8%288p%2B3%29=0
%283p+-8%29%288p%2B3%29=0


find a pair of integers whose product is AC and adds up to B

"Each of the trinomials is the form ax%5E2+%2B+bx+%2B+c."
In this example, a+=+24, b+=+-55, and c+=+-24. We must find two integers whose product is ac, or 24%28-24%29%2C+or+%7B%7B%7B-576, and whose sum is a%2Bc=b.
If the problem is too difficult to yield readily to inspection, we can find the prime factors of |ac%7C, or 576 and divide them into two piles. The prime factors of 576 are
2*2*2*2*2*2*3*3=2^6 * 3^2
which can be found easily by factoring |a| and |c| separately and combining the result.

let’s try these:2^6 * 3^2
24%2B%28-24%29=-55
0=-55
This adds to 0, which has an absolute value greater than the required b=-55.

let’s try these:2^5(2 * 3^2)
32%2B%28-18%29=-55
69=-55

let’s try these:2^4(2^2 * 3^2)
16%2B%28-36%29=-55
-20=-55
let’s try these:2^3(2^3 * 3^2)
8%2B%28-72%29=-55
-64=-55
let’s try these:2^2*3(2^4 * 3)
12%2B%28-48%29=-55
-36=-55
let’s try these:2*3(2^5 * 3)
6%2B%28-96%29=-55
-90=-55


Therefore, no pair of integers can satisfy the necessary requirements.