In today’s article, we will first factorize the expression x^{4}+x^{2}+1, and then we will solve the bi-quadratic equation x^{4}+x^{2}+1=0.

## How to Factorise x^{4}+x^{2}+1

**Answer: The factorisation of x ^{4}+x^{2}+1 is given by x^{4}+x^{2}+1 = (x^{2}-x+1)(x^{2}+x+1)**.

**Solution:**

At first, we will add and subtract x^{2} to the given expression.

x^{4}+x^{2}+1

= (x^{4}+x^{2}+1) + x^{2} – x^{2}

= (x^{4}+2x^{2}+1) – x^{2}

= {(x^{2})^{2} +2⋅x^{2}⋅1+1^{2}) – (x)^{2}

Applying the formula a^{2}+2ab+b^{2} = (a+b)^{2}, we get

= (x^{2}+1)^{2} – (x)^{2}

= (x^{2}+1-x)(x^{2}+1+x) as we know that a^{2}-b^{2} = (a-b)(a+b)

= (x^{2}-x+1)(x^{2}+x+1)

**Video Solution on How to Factorise x ^{4}+x^{2}+1:**

## Solve x^{4}+x^{2}+1=0

x^{4}+x^{2}+1= 0

⇒ (x^{2}-x+1)(x^{2}+x+1) = 0 by above.

Therefore, x^{2}-x+1 = 0 or x^{2}+x+1=0.

Apply the rule: ax^{2}+bx+c = 0 ⇒ x= $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

Now, x^{2}-x+1 = 0

⇒ x = $\dfrac{1 \pm \sqrt{1-4}}{2}$ = $\dfrac{1 \pm \sqrt{3}i}{2}$

And x^{2}+x+1 = 0

⇒ x = $\dfrac{-1 \pm \sqrt{1-4}}{2}$ = $\dfrac{-1 \pm \sqrt{3}i}{2}$

So the solutions of x^{4}+x^{2}+1= 0 are given by x= $\dfrac{1 + \sqrt{3}i}{2}$, $\dfrac{1 – \sqrt{3}i}{2}$, $\dfrac{-1 + \sqrt{3}i}{2}$ and $\dfrac{-1- \sqrt{3}i}{2}$.

So there are **four solutions** of the given equation x^{4}+x^{2}+1= 0.

**ALSO READ:**

## FAQs

**Q1: Factor x ^{4}+x^{2}+1.**

Answer: x^{4}+x^{2}+1 = (x^{2}-x+1)(x^{2}+x+1).

**Q2: How many solutions does the equation x ^{4}+x^{2}+1=0 has?**

Answer: The equation x^{4}+x^{2}+1= 0 has four solutions.