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Use synthetic division to show that \(x\) is a solution of the third-degree polynomial equation, and use the result to factor the polynomial completely. List all real solutions of the equation. $$x^{3}+2 x^{2}-3 x-6=0, \quad x=\sqrt{3}$$

Short Answer

Expert verified
By using synthetic division, the polynomial is factored into \( (x-\sqrt{3})(x^{2}+(2+\sqrt{3})x+(6+2\sqrt{3}))\). Then, by using the quadratic formula, the roots of the quadratic polynomial are calculated which would give us all the real solutions of the given polynomial equation.

Step by step solution

01

Synthetic Division

Perform synthetic division using \(x=\sqrt{3}\) on the given polynomial \(x^{3}+2 x^{2}-3 x-6=0\). Arrange the coefficients of the polynomial on a line (1,2,-3,-6) and write \(\sqrt{3}\) on the side, which is the given solution. Bring down the first coefficient (1), multiply it by \(\sqrt{3}\) and write this product under the next coefficient (2). Then add these numbers together and continue this pattern until it is done with all of the coefficients.
02

Write The Resulting Polynomial

The result of synthetic division gives us a quadratic polynomial. The numbers on the bottom line (1, \(2+\sqrt{3}\) and \(6+2\sqrt{3}\)) are the coefficients of this polynomial. The polynomial for these coefficients is \(x^{2}+(2+\sqrt{3})x+(6+2\sqrt{3})=0\).
03

Factor The Quadratic Polynomial

Find the roots of the quadratic polynomial. This can be done by applying the quadratic formula \(x = [-b±sqrt(b^{2}-4ac)]/(2a)\). Where `a`, `b` and `c` are the coefficients \(a=1\), \(b=2+\sqrt{3}\) and \(c=6+2\sqrt{3}\)
04

Identify All Real Solutions

Real solutions are ones that are not imaginary (do not contain the square root of a negative number). Calculate the discriminant \(b^{2}-4ac\) which would help to determine the type of solutions the quadratic equation has. If the discriminant is positive, there are two different real solutions, if it's zero, there's exactly one real solution and if it's negative, there are no real solutions.

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