find all integers for "a" in which x

^{2}+ax-32 can be factored. Explain how you arrived at your conclusion.-
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-32 = -1*32 = -32*1 = -2*16 = -16*2 = -4*8 = -8*4

So, a = 31, -31, 14, -14, 4, -4

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Attn: a = sum of any pair of factors of -32.

Hi Ryan;

x^{2}+ax-32

I believe the question is all integers which can be inserted for a in the following FOIL...

(x+??)[x+(-?)] or [x+(-??)](x+?)

Let's first work with the number -32.

This can be...

(8)(-4)

(-8)(4)

(16)(-2)

(-16)(2)

(32)(-1)

(-32)(1)

To provide an example...

The first FOIL would be...

(x+8)(x-4)

FIRST...(x)(x)=x^{2}

OUTER...(x)(-4)=-4x

INNER...(8)(x)=8x

LAST...(8)(-4)=-32

x^{2}+8x-4x-32

x^{2}+4x-32

a=4

If I understand the question correctly, all integers are...

-31, -14, -4, 4, 14, 31

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