Question

How do you factor \[5{x^4} + 31{x^2} + 6\] ?

Answer 1

Hint: The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. The roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. If the polynomial of degree is ‘n’ then the number of roots or factors is ‘n’.

Complete step-by-step answer:
The degree of the equation \[5{x^4} + 31{x^2} + 6\] is 4, so the number of roots of the given equation is 4.
We can split the middle term ‘31’ as 30 and 1. (Because the multiplication of 5 and 6 is 30. If we multiply 30 and 1 we get 30 and if we add 30 and 1 we get 31)
We get,
 \[ = 5{x^4} + 30{x^2} + 1{x^2} + 6\]
Now taking \[5{x^2}\] in the first two terms of the equation and taking 1 common in the last two term we have,
 \[ = 5{x^2}({x^2} + 6) + 1({x^2} + 6)\]
Taking \[({x^2} + 6)\] as common we get,
 \[ = (5{x^2} + 1)({x^2} + 6)\] . These are the factors.
We can further simplify this by equating to zero.
 \[ \Rightarrow 5{x^2} + 1 = 0\] and \[{x^2} + 6 = 0\] .
 \[ \Rightarrow 5{x^2} = - 1\] and \[{x^2} = - 6\]
 \[ \Rightarrow {x^2} = \dfrac{{ - 1}}{5}\] and \[{x^2} = - 6\]
Taking square root on both sides we have,
 \[ \Rightarrow x = \pm \sqrt {\dfrac{{ - 1}}{5}} \] and \[x = \pm \sqrt { - 6} \]
We know that \[\sqrt { - 1} = i\] ,
 \[ \Rightarrow x = \pm i\sqrt {\dfrac{1}{5}} \] and \[x = \pm i\sqrt 6 \] .
Hence the factors are \[\left( {x + i\sqrt {\dfrac{1}{5}} } \right)\] , \[\left( {x - i\sqrt {\dfrac{1}{5}} } \right)\] , \[\left( {x + i\sqrt 6 } \right)\] and \[\left( {x - i\sqrt 6 } \right)\]
So, the correct answer is “ \[\left( {x + i\sqrt {\dfrac{1}{5}} } \right)\] , \[\left( {x - i\sqrt {\dfrac{1}{5}} } \right)\] , \[\left( {x + i\sqrt 6 } \right)\] and \[\left( {x - i\sqrt 6 } \right)\] ”.

Note: A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation $i^2 = −1$. Because no real number satisfies this equation, i is called an imaginary number