How can you solve p2 - 12 =p ?
2 answers:
![- {p}^{2} - 12 = p \\ {p}^{2} - 12 - p = 0 \\ {p}^{2} - p - 12 = 0](https://tex.z-dn.net/?f=%20-%20%7Bp%7D%5E%7B2%7D%20%20-%2012%20%3D%20p%20%5C%5C%20%20%7Bp%7D%5E%7B2%7D%20%20-%2012%20-%20p%20%3D%200%20%5C%5C%20%20%20%7Bp%7D%5E%7B2%7D%20%20-%20p%20-%2012%20%3D%200)
Now, use the mid term splitting technique
![{p}^{2} - 4p + 3p - 12 \\ = p(p - 4) + 3(p - 4) \\ = (p + 3)(p - 4) \\ \\ \\ \\](https://tex.z-dn.net/?f=%20%7Bp%7D%5E%7B2%7D%20%20-%204p%20%2B%203p%20-%2012%20%5C%5C%20%3D%20%20p%28p%20-%204%29%20%2B%203%28p%20-%204%29%20%5C%5C%20%20%3D%20%28p%20%2B%203%29%28p%20-%204%29%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20)
From here
we'll do the same thing we do to find the zeroes,
p+3=0 or p -4 =0
p = -3 or p = 4
Therefore
p = -3 and 4
P= 4 and -3
Move all the terms to the left side and set equal to zero. Then set each factor to zero.
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