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Given that P(x) = x^4 + ax^3 - x^2 + bx - 12 has factors x - 2 and and x + 1, how do you solve the equation P(x) = 0?

Answer 1

This equation has #4# solutions: #x_1=-3#, #x_2=-2#, #x_3=-1# and #x_4=2#

According to Viete's Theorem if #P(x)# has a factor of #(x-a)# then #a# is a root of this polynomial, so in this case this polynomial has 2 roots: #x_3=-1# and #x_4=2#.

To find the other roots you have to divide #P(x)# by #(x-2)(x+1)#. The result will be: #x^2+5x+6#.

Then you can calculate the remaining roots.

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Answer 2

Eva Adamson

To solve the equation P(x) = 0, we can use the factor theorem. Since x - 2 and x + 1 are factors of P(x), we can set them equal to zero and solve for x.

Setting x - 2 = 0, we find x = 2. Setting x + 1 = 0, we find x = -1.

Therefore, the solutions to the equation P(x) = 0 are x = 2 and x = -1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.