How to Factorise and Solve x^4+x^2+1=0

In today’s article, we will first factorize the expression x4+x2+1, and then we will solve the bi-quadratic equation x4+x2+1=0.

How to Factorise x4+x2+1

Answer: The factorisation of x4+x2+1 is given by x4+x2+1 = (x2-x+1)(x2+x+1).


At first, we will add and subtract x2 to the given expression.


= (x4+x2+1) + x2 – x2

= (x4+2x2+1) – x2

= {(x2)2 +2⋅x2⋅1+12) – (x)2

Applying the formula a2+2ab+b2 = (a+b)2, we get

= (x2+1)2 – (x)2

= (x2+1-x)(x2+1+x) as we know that a2-b2 = (a-b)(a+b)

= (x2-x+1)(x2+x+1)

Video Solution on How to Factorise x4+x2+1:

Solve x4+x2+1=0

x4+x2+1= 0

⇒ (x2-x+1)(x2+x+1) = 0 by above.

Therefore, x2-x+1 = 0 or x2+x+1=0.

Apply the rule: ax2+bx+c = 0 ⇒ x= $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

Now, x2-x+1 = 0

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

And x2+x+1 = 0

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

So the solutions of x4+x2+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 x4+x2+1= 0.


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Q1: Factor x4+x2+1.

Answer: x4+x2+1 = (x2-x+1)(x2+x+1).

Q2: How many solutions does the equation x4+x2+1=0 has?

Answer: The equation x4+x2+1= 0 has four solutions.

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