Question 486733

factor an equation of to form {{{24p^2-55p-24=0}}}, 

{{{24p^2-55p-24=0}}}.......replace {{{-55p}}} with the  {{{9p  -64p}}}

{{{24p^2+9p  -64p-24=0}}}.....group

{{{(24p^2+9p)  -(64p+24)=0}}}

{{{3p(8p+3p)  -8(8p+3)=0}}}

{{{(3p -8)(8p+3)=0}}}



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


"Each of the trinomials is the form {{{ax^2 + bx + c.}}}"

In this example, {{{a = 24}}}, {{{b = -55}}}, and {{{c = -24}}}. We must find two integers whose product is {{{ac}}}, or {{{24(-24), or {{{-576}}}, and whose sum is {{{a+c=b}}}.

If the problem is too difficult to yield readily to inspection, we can find the prime factors of |{{{ac|}}}, 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+(-24)=-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+(-18)=-55}}}

 {{{69=-55}}}


let’s try these:2^4(2^2 * 3^2)

 {{{16+(-36)=-55}}}

 {{{-20=-55}}}

let’s try these:2^3(2^3 * 3^2)

 {{{8+(-72)=-55}}}

 {{{-64=-55}}}

let’s try these:2^2*3(2^4 * 3)

 {{{12+(-48)=-55}}}

 {{{-36=-55}}}

let’s try these:2*3(2^5 * 3)

 {{{6+(-96)=-55}}}

 {{{-90=-55}}}



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