Question

The polynomial ${x^6} + 18{x^3} + 125$, admits
A. No factor of degree between 2 and 5
B. ${x^2} - 3x + 5$ as a factor
C. $x + 1$ as a factor
D. $x - 125$ as a factor

Answer 1

Hint: We will check for each option, whether the condition holds or not. If the given expression is the factor of the polynomial ${x^6} + 18{x^3} + 125$, then, the factors of zeroes of the given expressions must also be the zeroes of the polynomial ${x^6} + 18{x^3} + 125$.

Complete step by step Answer:

We are given that the polynomial is ${x^6} + 18{x^3} + 125$
For option A, if there is no factor between 2 and 5, it means that the polynomial will not have any factor,
As the degree of the factor will always be less than the degree of the polynomial, then the degree of factors is less than 6.
If the degree of the factor of the polynomial not between 2 and 5, then there will be no factor of the polynomial, hence, option A is incorrect.
For option B, we will divide ${x^6} + 18{x^3} + 125$ by ${x^2} - 3x + 5$ and the remainder should be 0
We will substitute the value of ${x^2} = 3x - 5$ in the expression ${x^6} + 18{x^3} + 125$. If the result is 0, then ${x^2} - 3x + 5$ is factor of the given expression.
${x^6} + 18{x^3} + 125$ can be written as ${\left( {{x^2}} \right)^3} + 18x.{x^2} + 125$
Therefore, we get, ${\left( {3x - 5} \right)^3} + 18x\left( {3x - 5} \right) + 125$
Simplify the expression using the formula ${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3{a^2}b + 3a{b^2}$
Hence, we have
 $
  27{x^3} - 135{x^2} + 225x + 54{x^2} - 90x \\
   \Rightarrow 27{x^3} - 81{x^2} + 225x - 90x \\
   \Rightarrow 27x\left( {{x^2} - 3x + 5} \right) \\
$
But, we have ${x^2} = 3x - 5$
$27x\left( {3x - 5 - 3x + 5} \right) = 0$
Thus, ${x^2} - 3x + 5$ is a factor of ${x^6} + 18{x^3} + 125$
For option C, we will find the value of $x$ by equating the expression to 0 and if that value makes the polynomial 0, then it is the factor of the polynomial.
Hence, we have
$
  x + 1 = 0 \\
  x = - 1 \\
$
Substituting $x = - 1$ in the given polynomial gives,
${\left( { - 1} \right)^6} + 18{\left( { - 1} \right)^3} + 125 = 108$
Since, the value is not 0, $x + 1$ is not a factor of ${x^6} + 18{x^3} + 125$.
For option D, we will find the value of $x$ by equating the expression to 0 and if that value makes the polynomial 0, then it is the factor of the polynomial.
Hence, we have
$
  x - 125 = 0 \\
  x = 125 \\
$
Substituting $x = 125$ in the given polynomial gives,
${\left( {125} \right)^6} + 18{\left( { - 125} \right)^3} + 125 \ne 0$
Since, the value is not 0, $x - 125$ is not a factor of ${x^6} + 18{x^3} + 125$.
Hence, option B is correct.

Note: The zero of the factored polynomial will also be the zero of the polynomial. If the degree of the polynomial is \[n\], then it can have at most \[n\] zeroes. Also, if we multiply all the factors of the polynomial, it will give back the polynomial. If we know one factor of the polynomial, the other can be calculated by dividing the given polynomial by the known factor, the quotient will be the other factor.