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

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Answer 1 · 2016-04-26T16:32:11

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

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 $(x+4+i2sqrt2)(x+4-i2sqrt2)$

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