You can
put this solution on YOUR website!Your answer is correct -- as far as it goes. If your assignment is to find just the real number roots, then you are done. But if you were asked to completely solve the equation, including finding any complex number roots, you are a long way from done.
The Fundamental Theorem of Algebra tells us a couple of important things in this context. First, any polynomial equation of degree n has exactly n roots. You have a 3rd degree equation, so there must be 3 roots. Furthermore, if
is a root of the polynomial equation, then
is a factor of the polynomial. That is to say that, given a polynomial function
, if
, then there exists a polynomial function
such that
.
Knowing that 5 is one of the roots of your equation, and we know this because
and
, makes the task of finding the other two roots much simpler. Solving the general cubic is a horror that I wouldn't wish on anyone, even if I particularly disliked them.
We can use the fact that
is a factor of
and the process of polynomial long division to find the other factor of
which will be a quadratic polynomial. But first we need to put
into standard form by putting in placeholders for the zero coefficient terms, thus:
goes into
times.
is the first term of the quotient.
times
is
. Subtract this from the first two terms of
to get
(remember: change the sign and add).
Bring down the
.
goes into
times.
is the second term of the quotient.
times
is
. Subtract this from
to get
.
Bring down the
.
goes into
times.
is the third term of the quotient.
times
is
. Subtract this from
to get
-- meaning no remainder.
The quotient polynomial is then
, and the zeros of this polynomial are the other two zeros of your original polynomial.
Using the quadratic formula:
or
Where
is the imaginary number defined by
In summary, your three roots are:
,
,
(