Verified answer

Use synthetic division

1 | .....6.....31.....–33.....–16.....12

..................6.......37........4...–12

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–6|....6......37........4.....–12.....| 0

..............–36.......–6......12

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........6........1.......–2.....| 0

We are left with 6x² + x – 2 = (3x + 2)(2x – 1)

The other roots are –2/3 and 1/2, B

If 1 and -6 are roots, then x-1 and x + 6 must be factors. Divide the original polynomial by each factor to find the remaining quadratic equation.

Divide 6x^4 + 31 x^2 - 33x^2 - 16x + 12 by x - 1 : 6x^3 + 37x^2 + 4x - 12

Divide 6x^3 + 37x^2 + 4x - 12 by x + 6: 6x^2 + x - 2

This factors into (3x + 2)(2x - 1)

So the remaining zeroes are -2/3 and 1/2, answer B

-1 | 6 31 -33 -16 12

-6 -26 59 -43

6 26 -59 43 -31

You must've copied the equation wrong.

if 1 and -6 are roots, then (x - 1)(x + 6) are factors

divide the given polynomial by the product of (x - 1)(x + 6)

(6x^4 + 31 x^3 − 33x^2 − 16x + 12) / (x - 1)(x + 6) = (6x^4 + 31 x^3 − 33x^2 − 16x + 12) /(x^2 +5x - 6)

x^2+5x-6) 6x^4+31x^3−33x^2−16x+12(6x^2

...............6x^4+30x^3-36x^2

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........................x^3 + 3x^2 - 16x (x

........................x^3 + 5x^2 - 6x

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.................................-2x^2 - 10x + 12(-2

.................................-2x^2 - 10x + 12

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............................................0

The quotient is 6x^2 + x - 2

=> 6x^2 + 4x - 3x - 2 = 0

=> 2x(3x + 2) - 1(3x + 2) = 0

=> (2x - 1)(3x + 2) = 0

x = 1/2 and - 2/3