Answer to Question #103130 in Algebra for ea

Question

Find the value of a for which the polynomial x^3- a x^2- ax +1 has −1 as a root with multiplicity at least 2.

Answer 1

We know that:

If $x = a$ is a root of a polynomial then $(x - a)$ will be the factor of this polynomial.

If the multiplicity of this root $a$ is $n$ of the polynomial then $(x-a)^n$ will be the factor of the polynomial.

Actually, the multiplicity of the root is the power value of a factor ( Here, This root is involved in this factor) of the polynomial.

If the polynomial is the degree of $m$ then the highest power of the variable in the polynomial expression is $m$

And:

If the polynomial has a degree $m$ then this polynomial is created by using $m$ number of factors.

Now:

The polynomial is given by:

$x^3 - ax^2 -ax + 1$

Since:

The root of the given polynomial is $-1$

So:

$(x - (-1)) \, \, or \,\, (x+1)$ will be one factor of the given polynomial.

Again:

Since the root $-1$ has multiplicity at least $2$

So:

$(x + 1)^2$ will be the factors of the given polynomial.

Now:

We can see that our given polynomial has degree three so we have to required three factors but we have two factors which are $(x+1)^2$

For the third factors, we will assume it. As the last term of our polynomial is $1$ so our final factor must be $(x + 1)$ otherwise we can not get the last term $1$ of the given polynomial.

Hence:

We can write:

$x^3 - ax^2 - ax + 1 \,\,\, = (x+1)^3 \\ \hspace{1 cm} \hspace{1 cm} \hspace{1 cm} = x^3 + 3x^2 + 3x + 1$

Comparing both of side, we have:

$a = -3$

[ From the given information, it may also prove that the power of $( x +1)$ must be $3$ because the root $x = -1$ of the given polynomial has a multiplicity which is not exactly $2$ but at least $2$ ]