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Find the roots of the polynomial equation. x2 + 2x – 48 = 0 A. –6, 8 B. –4, 12 C. –12, –4 D. –8, 6
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@Mateaus
First factor then set them equal to 0
x2+(8-6)x-48 open bracket factor it
O_O no
\[x^2 + 2x -48 = 0\]
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im not good at math:(
I'm actually really bad at explaining factorials x_x
(x+8)(x-6)
then you set them to =0 x+8 = 0 x-6 = 0
solve for both x's
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help:(..
I am bad at explaining too, but here's my shot;\[(x+a)(x+b) = x^2+(a+b)x+ab,~~~\text{Right?}\] Since you are given \(x^2+2x-48\), you have you have \(a+b=2\) and \(ab=-48\) Can you figure out what \(a\) and \(b\) is?
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