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?

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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?

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Answer 1 · 2015-09-26T18:00:45

This equation has $4$ $4$ solutions: $x_{1} = - 3$ $x_1=-3$ , $x_{2} = - 2$ $x_2=-2$ , $x_{3} = - 1$ $x_3=-1$ and $x_{4} = 2$ $x_4=2$

Explanation:

According to Viete's Theorem if $P (x)$ $P(x)$ has a factor of $(x - a)$ $(x-a)$ then $a$ $a$ is a root of this polynomial, so in this case this polynomial has 2 roots: $x_{3} = - 1$ $x_3=-1$ and $x_{4} = 2$ $x_4=2$ .

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

Then you can calculate the remaining roots.

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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?

1 Answer

Explanation:

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