Respuesta :
Answer: x = -5, -4, 4 and 5.
All four zeros are real solutions.
Step-by-step explanation:
Given the polynomial equation [tex]x^4-41x^2=-400[/tex].
Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get
[tex]x^4-41x^2+400=-400+400[/tex].
[tex]x^4-41x^2+400=0[/tex].
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,
[tex]x^4-25x^2-16x^2+400=0[/tex]
Making it into two groups, we get
[tex](x^4-25x^2)+(-16x^2+400)=0[/tex]
Factoring out GCF of each group, we get
[tex]x^2(x^2-25)-16(x^2-25)=0[/tex]
[tex](x^2-25)(x^2-16) =0[/tex]
Factoring out [tex](x^2-25)[/tex] and [tex](x^2-16)[/tex] separately by difference of the squares identity [tex]a^2-b^2=(a-b)(a+b)[/tex], we get
[tex](x^2-25) = x^2-5^2= (x-5)(x+5)[/tex] and
[tex]x^2-16 = x^2-4^2 =(x-4)(x+4)[/tex].
Therefore,
[tex](x-5)(x+5)(x-4)(x+4) =0[/tex]
Applying zero product rule,
x-5=0
x+5=0
x-4=0 and
x+4=0.
Therefore,
