The setup boxes in the synthetic division are (b)
<h3>How to determine the setup boxes?</h3>
The dividend is given as:
x^3 + 4x^2 + x - 6
The divisor is given as:
x - 2
Set the divisor to 0
x - 2 = 0
Solve for x
x = 2
Remove the variables in the dividend
1 + 4 + 1 - 6
Remove the arithmetic signs
1 4 1 - 6
So, the setup is:
2 | 1 4 1 - 6
Hence, the setup boxes are (b)
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Hhhhhhhhahahahahahahahahahahaahabahahah 4.7
2.16 x 10 to the power of -4
