Question

How do you find the factor of ${x^4} + {x^3} + {x^2} + x + 1$ ?

Answer 1

Hint: Here we need to factorize the given polynomial. We can see that the given polynomial is of degree 4. We will write the given polynomial as the product of two quadratic equations with variable constants. Then we will find the product of these two quadratic equations and we will equate the coefficient of the product with the given polynomial. From there, we will get the value of the variables and hence the factors.

Complete step by step solution:
Here we need to factorize the given polynomial and the given polynomial is ${x^4} + {x^3} + {x^2} + x + 1$.
We can see that the given polynomial is of degree 4.
We will write the given polynomial as the product of two quadratic equations with variable constants.
$ \Rightarrow {x^4} + {x^3} + {x^2} + x + 1 = \left( {{x^2} + ax + 1} \right)\left( {{x^2} + bx + 1} \right)$
Now, we will find the product of the two quadratic equations on the left-hand side of the equation.
$ \Rightarrow {x^4} + {x^3} + {x^2} + x + 1 = {x^4} + \left( {a + b} \right){x^3} + \left( {2 + ab} \right){x^2} + \left( {a + b} \right)x + 1$
Now, we will compare the coefficients of the polynomials of both sides.
On comparing the coefficients, we get
$a + b = 1$ …………… $\left( 1 \right)$
$2 + ab = 1$ …………… $\left( 2 \right)$
Now, we will consider the equation $\left( 1 \right)$.
We will now subtract the term $b$ from both sides.
$a + b - b = 1 - b$
$ \Rightarrow a = 1 - b$ ………….. $\left( 3 \right)$
Now, we will substitute the value of the variable $a$ in equation $\left( 2 \right)$.
$2 + \left( {1 - b} \right)b = 1$
On multiplying the terms, we get
$ \Rightarrow 2 + b - {b^2} = 1$
On further simplifying the terms, we get
$ \Rightarrow {b^2} - b - 1 = 0$
We will directly apply the formula to find the roots.
We know if $a{x^2} + bx + c = 0$, then the roots of the quadratic equation will be equal to
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Using this formula of roots, we get
$b = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 1 \times - 1} }}{{2 \times 1}}$
On further simplification, we get
$ \Rightarrow b = \dfrac{{1 \pm \sqrt {1 + 4} }}{2} = \dfrac{{1 \pm \sqrt 5 }}{2}$
Rewriting the above equation, we get
$ \Rightarrow b = \dfrac{1}{2} \pm \dfrac{{\sqrt 5 }}{2}$
Now, we will substitute the value of $b$ in equation $\left( 1 \right)$.
$a + \dfrac{1}{2} \pm \dfrac{{\sqrt 5 }}{2} = 1$
On further simplifying the terms, we get
$\Rightarrow a = 1 - \dfrac{1}{2} \mp \dfrac{{\sqrt 5 }}{2} \\
   \Rightarrow a = \dfrac{1}{2} \pm \dfrac{{\sqrt 5 }}{2} \\ $
Hence, the factorization of the given polynomial is

${x^4} + {x^3} + {x^2} + x + 1 = \left( {{x^2} + \left( {\dfrac{1}{2} \pm \dfrac{{\sqrt 5 }}{2}} \right)x + 1} \right)\left( {{x^2} + \left( {\dfrac{1}{2} \pm \dfrac{{\sqrt 5 }}{2}} \right)x + 1} \right)$.

Note: Here we have used the formula to find the roots of the quadratic equation obtained after factoring the polynomial. Here roots of the quadratic equation are defined as the values which when put in the place of the variables will satisfy the equation. The number of roots of the polynomial is always equal to the degree of that polynomial. For example, for quadratic equations the highest degree is 2, so the solution is also 2.