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