
How does ${x^5} = 32$, a 5th degree polynomial, have 5 zeroes?
Answer
537.9k+ views
Hint:In order to find the first root simply solve the question for x. Which will give us x=2 as our first real root. To find the other 4 roots, we will take a different way to solve the question. Since, x=2 was a root, the given equation must have a factor$(x - 2)$. Dividing this factor with the question we will get a quadratic equation which according to the rules must have 4 roots. Therefore, these 4 plus the 1 root found earlier makes a total 5 roots.
Complete step by step answer:
The equation ${x^5} = 32$ actually has 5 roots. In which the 1st root is 2 whereas, the other 4 roots are non-real complex numbers. And hence it has 5 roots. To know how, let’s go further and see how we concluded this. The initial equation we have is $(x - 2)$.Taking both the terms on one side, we will get:-
${x^5} - 32 = 0$
To solve this question, we will now have to find which number’s 5th power is equal to 32
With trial and error we can conclude
${2^5} = 32$
Therefore, for $x = 2$ the equation is satisfied, hence $x = 2$ is a solution for the given equation.
To find the other 4 non-real complex roots we will have to think a little bit differently.We know that $x = 2$ is a solution. This means that $x - 2$ must be a factor of the given equation. Therefore, we can write it as:-
${x^5} - 32 = \left( {x - 2} \right) \times X$
Now, to find the value of “X”, we will divide ${x^5} - 32$ by $\left( {x - 2} \right)$. Doing which we will get:-
${x^5} - 32 = \left( {x - 2} \right) \times \left( {{x^4} + 2{x^3} + 4{x^2} + 8x + 16} \right)$
Now, following the rule, we can now conclude that the second part of the equation has 4 roots.
So, 4 roots plus 1 constant root we found above makes a total of 5 roots.
Note:Division of the polynomial with the question must be done correctly. And the 4 roots we know of the quadratic equation are all non-real and complex numbers. Since, those numbers are not real numbers and can’t be used in the real world calculations, those roots are ignored and only the constant real root is considered.
Complete step by step answer:
The equation ${x^5} = 32$ actually has 5 roots. In which the 1st root is 2 whereas, the other 4 roots are non-real complex numbers. And hence it has 5 roots. To know how, let’s go further and see how we concluded this. The initial equation we have is $(x - 2)$.Taking both the terms on one side, we will get:-
${x^5} - 32 = 0$
To solve this question, we will now have to find which number’s 5th power is equal to 32
With trial and error we can conclude
${2^5} = 32$
Therefore, for $x = 2$ the equation is satisfied, hence $x = 2$ is a solution for the given equation.
To find the other 4 non-real complex roots we will have to think a little bit differently.We know that $x = 2$ is a solution. This means that $x - 2$ must be a factor of the given equation. Therefore, we can write it as:-
${x^5} - 32 = \left( {x - 2} \right) \times X$
Now, to find the value of “X”, we will divide ${x^5} - 32$ by $\left( {x - 2} \right)$. Doing which we will get:-
${x^5} - 32 = \left( {x - 2} \right) \times \left( {{x^4} + 2{x^3} + 4{x^2} + 8x + 16} \right)$
Now, following the rule, we can now conclude that the second part of the equation has 4 roots.
So, 4 roots plus 1 constant root we found above makes a total of 5 roots.
Note:Division of the polynomial with the question must be done correctly. And the 4 roots we know of the quadratic equation are all non-real and complex numbers. Since, those numbers are not real numbers and can’t be used in the real world calculations, those roots are ignored and only the constant real root is considered.
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