Question

solve for 'x'

${x}^{4} + {x}^{3} - 4 {x}^{2} + x + 1 = 0$


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Answer 1

Answer:

Step-by-step explanation:

Answer 2

${x}^{4} + {x}^{3} - 4 {x}^{2} + x + 1 = 0$

Put x = 1,

f(1) = 1⁴ + 1³ - 4(1²) + 1 + 1

= 1 + 1 - 4 + 1 + 1

= 0

Hence 1 is a zero of f(x).

Also (x-1) is a factor of f(x).

So other factor is,

____________________

x-1) x⁴ + x³ - 4x² + x + 1( x³ + 2x² - 2x - 1

x⁴ - x³

______________________

× 2x³ - 4x² +x + 1

2x³ - 2x²

_______________________

× - 2x² + x + 1

- 2x² +2x

_______________________

× - x +1

- x + 1

_______________________

× ×

So x³ + 2x² - 2x - 1 is also a factor of f(x).

Put x as 1,

=) f(1) = 1³ +2(1)² - 2(1) - 1

= 1 + 2 - 2 - 1

= 0

Hence x-1 is also a factor.

So third factor is

________________________

x-1) x³ + 2x² - 2x - 1 ( x² + 3x + 1

x³ - x²

___________________________

× 3x² - 2x - 1

3x² - 3x

___________________________

× x - 1

x - 1

____________________________

× ×

Hence f(x) = (x-1)(x-1)(x² + 3x + 1)

By factorizing x² + 3x + 1,

By quadratic eq formula,

x = (- 3 +/- √5) /2

& x = 1.