Question

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

Answer 1

Answer: x = -5, -4, 4 and 5.

All four zeros are real solutions.

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$.

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$.

$x^4-41x^2+400=0$.

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

$(x^2-25)(x^2-16) =0$

Factoring out  $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$, we get

$(x^2-25) = x^2-5^2= (x-5)(x+5)$ and

$x^2-16 = x^2-4^2 =(x-4)(x+4)$.

Therefore,

$(x-5)(x+5)(x-4)(x+4) =0$

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

x = -5, -4, 4 and 5.

All four zeros are real solutions.