Question

discussion module 7
factoring is important because it gives us a method for solving certain polynomial equations. for example, $x^{3}-2x^{2}-3x = 0$ factors to $x(x - 3)(x + 1)=0$. setting the x, the x - 3, and the x + 1 equal to zero, gives us the solutions x = 0, 3, -1.
for this discussion, lets go in reverse. choose three numbers to be solutions to an equation - ill choose 5, -2, and 1 as the solutions. that means an equations with those solutions should factor to $(x - 5)(x + 2)(x - 1)=0$. multiply the left side together to get $x^{3}-4x^{2}-7x + 10 = 0$. the equation $x^{3}-4x^{2}-7x + 10 = 0$ has the solutions x = 5, -2, and 1.
choose your own three numbers and find the equation that has those numbers as its solutions.

Explanation:

Step1: Choose three numbers

Let the three numbers be 2, - 3, 4.

Step2: Write the factored - form of the equation

If \(x = 2\), \(x=-3\), and \(x = 4\) are solutions, then the factored - form of the equation is \((x - 2)(x + 3)(x - 4)=0\).

Step3: Multiply the first two binomials

\((x - 2)(x + 3)=x^{2}+3x-2x - 6=x^{2}+x - 6\).

Step4: Multiply the result by the third binomial

\((x^{2}+x - 6)(x - 4)=x^{3}-4x^{2}+x^{2}-4x-6x + 24\).

Step5: Combine like terms

\(x^{3}-4x^{2}+x^{2}-4x-6x + 24=x^{3}-3x^{2}-10x + 24\).

Answer:

The equation is \(x^{3}-3x^{2}-10x + 24 = 0\).