Question 711966
In general, a polynomial that has r as a root will have a factor of (x-r). So the polynomial we are looking for will have factors of:
{{{P(x) = (x-(-1))(x-(-1))(x-1)}}}
which simplifies to:
{{{P(x) = (x+1)(x+1)(x-1)}}}<br>
Now all we have to is multiply this out. We can take advantage of the {{{(a+b)^2 = a^2+2ab+b^2}}} pattern to multiply the first two factors. Or we can take advantage of the {{{(a+b)(a-b) = a^2-b^2}}} pattern to multiply the last two factors. (Remember, with multiplication the order does not matter!) Since the second pattern has a simpler result I'm going to use it to multiply the last two factors:
{{{P(x) = (x+1)((x)^2-(1)^2)}}}
which simplifies to:
{{{P(x) = (x+1)(x^2-1)}}}
Now we multiply the remaining factors. They do not fit any pattern so we must resort to FOIL:
{{{P(x) = x*x^2-x*1+1*x^2-1*1}}}
which simplifies...
{{{P(x) = x^3-x+x^2-1}}}
Rearranging the terms so they are in standard form:
{{{P(x) = x^3+x^2-x-1}}}<br>
P.S. When I described the factored form of a polynomial, I left out a minor detail. The general factored form of a polynomial is:
P(x) = a * (x-r) ... (with as many (x-r) factors as there are roots).
The "a" can be any non-zero number and it is the part I left out. The answer we got while ignoring the "a" has, in effect, an "a" of 1. Fell free to re-do the solution and pick a different "a".