Question

How do you factor ${x^4} + 4$?

Answer 1

We have a biquadratic equation with the highest degree coefficients are zero. First, equate this coefficient with zero then just do some algebraic operations and note that the square root of minus one is iota ($i$) then just simplify the radical root of the number $4$.

Formula used:
$\sqrt[n]{{ab}} = \sqrt[n]{a} \cdot \sqrt[n]{b}$

Complete step by step solution:
Given polynomial: ${x^4} + 4$
We have to factorize this polynomial.
First, equate this polynomial with zero and make it an equation.
$\Rightarrow {x^4} + 4 = 0$
$\Rightarrow {x^4} = - 4$
Now this is of some square type
Let ${x^2} = u \Rightarrow {x^4} = {u^2}$
$\Rightarrow {u^2} = - 4$
Now, this negative number can be written as the product of two numbers one of them is positive as the main number is negative.
$\Rightarrow {u^2} = - 1 \times 4$
$\Rightarrow u = \sqrt { - 1 \times 4}$
And we know that, $\sqrt[n]{{ab}} = \sqrt[n]{a} \cdot \sqrt[n]{b}$
$\Rightarrow u = \sqrt { - 1} \times \sqrt 4$
And we know that $\sqrt { - 1}$ is known as the complex root and that is
$\sqrt { - 1} = i$, where $i$ is iota called complex root
$\Rightarrow u = i \times \sqrt 4$
Now we have to calculate the square root of $4$.
$4 = 2 \times 2$
$\Rightarrow \sqrt 4 = 2$
Since we assumed a new variable $u$ as ${x^2}$.
$\Rightarrow u = 2i, - 2i$
So, substitute back $u$ as ${x^2}$.
$\Rightarrow {x^2} = 2i, - 2i$
Now continuing with $2i$.
${x^2} = 2i$
On further simplifying, we get
$x = 1 + i$ and $x = - 1 - i$
And when we continue on $- 2i$.
$x = - 1 + i$ and $x = 1 - i$
Thus, on simplifying ${x^4} + 4 = 0$ we get
$x = 1 + i$, $x = - 1 - i$, $x = - 1 + i$ and $x = 1 - i$
Since these are the roots means these values for the $x$ equation will satisfy.
Now for factors,
$\left( {x - 1 - i} \right),\left( {x + 1 + i} \right),\left( {x + i - i} \right),\left( {x - 1 + i} \right)$.

Therefore, ${x^4} + 4 = \left( {x - 1 - i} \right)\left( {x + 1 + i} \right)\left( {x + i - i} \right)\left( {x - 1 + i} \right)$.

Note: We can also factorize given polynomials using algebraic identities.
Algebraic identity:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
So, first we have to add and subtract $4{x^2}$ to the given polynomial.
$\Rightarrow {x^4} + 4 + 4{x^2} - 4{x^2}$
It can be written as
$\Rightarrow {\left( {{x^2}} \right)^2} + {2^2} + 2\left( 2 \right)\left( {{x^2}} \right) - 4{x^2}$
Now, use algebraic identity (I), to factorize the polynomial.
$\Rightarrow {\left( {{x^2} + 2} \right)^2} - 4{x^2}$
It can be written as
$\Rightarrow {\left( {{x^2} + 2} \right)^2} - {\left( {2x} \right)^2}$
Now, use algebraic identity (II), to factorize the polynomial.
$\Rightarrow \left( {{x^2} + 2 - 2x} \right)\left( {{x^2} + 2 + 2x} \right)$
It can be written as
$\Rightarrow \left( {{x^2} + 2x + 2} \right)\left( {{x^2} - 2x + 2} \right)$
It can be further factorized using the Quadratic formula.
So, equate this polynomial with zero and make it an equation.
${x^2} + 2x + 2 = 0$ and ${x^2} - 2x + 2 = 0$
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$……(i)
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
First, we have to compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} + 2x + 2 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = 2$ and $c = 2$
Now, we have to substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( 2 \right)^2} - 4\left( 1 \right)\left( 2 \right)$
After simplifying the result, we get
$\Rightarrow D = 4 - 8$
$\Rightarrow D = - 4$
Which means the given equation has complex roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$\Rightarrow x = \dfrac{{ - 2 \pm 2i}}{{2 \times 1}}$
Divide both numerator and denominator by $2$, we get
$\Rightarrow x = - 1 \pm i$
Comparing ${x^2} - 2x + 2 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = - 2$ and $c = 2$
Now, we have to substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - 2} \right)^2} - 4\left( 1 \right)\left( 2 \right)$
After simplifying the result, we get
$\Rightarrow D = 4 - 8$
$\Rightarrow D = - 4$
Which means the given equation has complex roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$\Rightarrow x = \dfrac{{2 \pm 2i}}{{2 \times 1}}$
Divide both numerator and denominator by $2$, we get
$\Rightarrow x = 1 \pm i$
Final solution: Therefore, ${x^4} + 4 = \left( {x - 1 - i} \right)\left( {x + 1 + i} \right)\left( {x + i - i} \right)\left( {x - 1 + i} \right)$.