what value of B makes x^2 + Bx - 24 factor-able?

Question

can you explain...

Answer 1

I assume you mean factorable as integers. Even just with integers, there are several answers. Start by figuring the factors of -24:
1, -24
2, -12
3, -8
4, -6
6, -4
8, -3
12, 2
24, 1

For any pair of factors, B will be the difference:

(x - 1)(x + 24) --> x² + 23x - 24
(x - 2)(x + 12) --> x² + 10x - 24
(x - 3)(x + 8) --> x² + 5x - 24
(x - 4)(x + 6) --> x² + 2x - 24
(x - 6)(x + 4) --> x² - 2x - 24
(x - 8)(x + 3) --> x² - 5x - 24
(x - 12)(x + 2) --> x² - 10x - 24
(x - 24)(x + 1) --> x² - 23x - 24

So the possible answers for B are:
{ -23, -10, -5, -2, 2, 5, 10, 23 }

Answer 2

Well to factor this, we need two equations

ab = -24

a+b = B

So B can be positive or negative {23, 10, 5, 2}

I just took all the possible ways you can multiply two numbers and get -24, then i took the two factors and added them together to get the values for B

Ways to get -24

-24*1
-12*2
-8*3
-6*4
-4*6
-3*8
-2*12
-1*24

-24+1 = -23
-12+2 = -10
-8+3 = -5
-6+4 = -2
-4+6 = 2
-3+8 = 5
-2+12 = 10
-1+24 = 23

So the integer values that will make the equation factorable are given above.

Answer 3

The difference of two factors of -24 is B.
B = ±2, ±5, ±10, ±23

Answer 4

Any real value of b makes
x^2+bx-24 factorable
Because b^2-4*1*(-24)=
b^2+96>0 (always positive)
So it has 2 roots r1,r2 hence
it can be written
(x-r1)(x-r2)

Answer 5

There'll be a bunch. For example, (x+6)(x-4)=x^2+2x-24, and (x+8)(x-3)=x^2+5x-24.

Answer 6

There's more than one value of B that will work. Do you need them all?

The general form of the factors is:
(x+α₁)(x+α₂) = x² + (α₁+α₂)x + α₁α₂

You want to find all values of α₁ and α₂ such that α₁α₂=-24, and determine the value of B = α₁+ α₂