How do you factor the trinomial #x^2+8x+24=0#?

1 Answer
Shwetank Mauria
Apr 26, 2016

#x^2+8x+24=(x+4+i2sqrt2)(x+4-i2sqrt2)#

Explanation:

As the determinant for equation #x^2+8x+24=0# is #8^2-4xx1xx24=64-96=-32#, the factors will be complex

Using quadratic formula, zeros of #x^2+8x+24# are #(-8+-sqrt(-32))/2# or #-4+-4/2sqrt(-2)# or #-4+-i2sqrt2#

Hence factors of #x^2+8x+24# are #(x+4+i2sqrt2)(x+4-i2sqrt2)#

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