Pre calculus question

Question

Use synthetic division to identify the integer bounds of the real zeros. Find the least upper bound and the greatest lower bound guarenteed by the upper and lower bounds of zeros theorem

m(x)=x^4-7x^3+11x^2+7x-12

Answer 1

P = {factors of 12} = {+-1, +-2, +-3, +-4, +-6, +-12}

Q = {factors of 1} = {+1}

P/Q = P = {+-1, +-2, +-3, +-4, +-6, +-12}

m(1) = 1 - 7 + 11 + 7 - 12 = -6+11+7-12 = 5+7-12 = 0, so 1 is a solution.

Synthetic division says:

1 | 1 -7 11 7 -12

1 -6 5 12

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1 -6 5 12 0

0 = (x-1)(x^3 - 6x^2 + 5x + 12)

The cubic polynomial has the same set for P/Q.

x=-1 is a solution since (-1)^3 - 6(-1)^2 + 5(-1) + 12 = -1 - 6 - 5 + 12 = -7 - 5 + 12 = -12+12 = 0

Synthetic division says:

-1 | 1 -6 5 12

-1 7 -12

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1 -7 12 0

The resulting quadratic has solutions:

x^2 - 7x + 12 = 0

(x -4)(x-3) = 0

x-4=0 or x-3 = 0

x=4 or x=3

the solution set is {-1,1,3,4}