Question 72237
When you are solving for x, you have the following form
{{{pq=0}}}Where p and q are factors of f(x). For instance p could equal (x-5) and q could equal (x+6). To solve for x, you set p and q equal to zero and isolate x. But this time you know what x is, you just need to find the factor that makes the polynomial. So if you have x=3 and x=-1, lets take x=3 first. What number adds to 3 to get 0? Well it would be -3. So in other words 3-3=0. If we replace the 1st 3 with x we get
{{{x-3=0}}}Theres one factor (if we plug in 3 for x, we get 0)
{{{x+1=0}}}There's the other factor, just do the same as before.
So now you multiply them together
{{{(x+1)(x-3)}}}FOIL
{{{x^2-3x+x-3}}}
{{{x^2-2x-3}}}Thats the polynomial with roots of x=-1 and x=3
Check:
{{{x^2-2x-3=0}}}Plug in x=-1
{{{(-1)^2-2(-1)-3=0}}}
{{{1+2-3=0}}}
{{{3-3=0}}}
{{{0=0}}}Works
{{{x^2-2x-3=0}}}Plug in x=3
{{{(3)^2-2(3)-3=0}}}
{{{9-6-3=0}}}
{{{3-3=0}}}
{{{0=0}}}Works
It turns out in a product of factors
{{{(x+a)(x+b)}}} The roots are x=-a and x=-b. Just a note for quick reference. So if you have any two roots, you can go backwards to get the equation.