How do you find all the zeros of $x^{2} - 8 x - 48 = 0$ $x^2 - 8x - 48 = 0$ ?

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Answer 1 · 2016-04-29T13:06:51

The important thing is to be able to identify factors of $48$ $48$ that add / subtract to give $- 8$ $-8$ .

The important thing is to be able to identify factors of $48$ $48$ that add / subtract to give $- 8$ $-8$ .

$48$ $48$ could be $6 \cdot 8$ $6*8$ which doesn't work, or it could be $4 \cdot 12$ $4*12$ which does work - $4 - 12 = - 8$ $4-12 = -8$

Then $x^{2} - 8 x - 48 = (x - 12) (x + 4)$ $x^2 - 8x - 48 = (x-12)(x+4)$

How do you find all the zeros of x2−8x−48=0x^2 - 8x - 48 = 0?