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OpenStudy (lgbasallote):
are you sure the last nummber is 08?
OpenStudy (lgbasallote):
-8 *
OpenStudy (anonymous):
yes
OpenStudy (lgbasallote):
i doubt it's factorable then
OpenStudy (anonymous):
i dont know
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OpenStudy (lgbasallote):
are you sure @Wired ?
OpenStudy (anonymous):
4 times 2
OpenStudy (anonymous):
4+2=6 and 4x2=8
OpenStudy (lgbasallote):
if you have written this correctly @elainediaz then it is not factorable because the only factors of 8 that add up to -6 is -4 x -2
and that results to POSITIVE 8 not negative
OpenStudy (anonymous):
Bah, same damn typo twice lol.
\[(x+a)(x-b) = x^{2}+ax-bx-ab = x^{2}+(a-b)x-ab\]
Yeah, not factorable.
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OpenStudy (anonymous):
ok its not factorable
OpenStudy (anonymous):
lol
OpenStudy (anonymous):
i di positive
OpenStudy (anonymous):
This would be consistent with finding a complex root. Use the quadratic formula to find the root values of \[x ^{2} - 6x - 8; a = 1, b = -6, c = -8\]
OpenStudy (anonymous):
Quadratic formula
\[- b \pm \sqrt{b^2 -4ac}\div2a\]
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OpenStudy (anonymous):
@jenxcc question is to factorize and not to find the values of x..
OpenStudy (anonymous):
Factorizing this cannot be factored as a real root, true, but can be as a complex root.
So, \[(x-(3+\sqrt{17})(x-(3-\sqrt{17})\]