Question

Find $(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)$ if the four solutions
of $x^{4}+4x^{3}+2x^{2}-4x+1=0$ are $x=\alpha,\beta,\gamma,\delta$.

Answer 1

$\large\underline{\text{Required answer}}$

$(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=4$

$\large\underline{\text{Step 1. Factor theorem}}$

By factor theorem, we know that,

$\red{\bigstar}(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)=x^{4}+4x^{3}+2x^{2}-4x+1$

$\large\underline{\text{Step 2. Substitution}}$

Then when $x=1$,

$\implies(1-\alpha)(1-\beta)(1-\gamma)(1-\delta)=1^{4}+4\cdot(1)^{3}+2\cdot(1)^{2}-4\cdot(1)+1$

Distributing $(-1)$ into each factor, we get,

$\implies(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1^{4}+4\cdot(1)^{3}+2\cdot(1)^{2}-4\cdot(1)+1$

$\implies(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1+4+2-4+1$

$\implies(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=4$

$\large\underline{\text{Conclusion}}$

The required value is 4.

$\purple{\bigstar}\large\underline{\text{Bonus question}}\purple{\bigstar}$

Find the value of $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}+\dfrac{1}{\delta}$ if $x=\alpha,\beta,\gamma,\delta$ are the four solutions to the equation $x^{4}-2x^{3}+5x^{2}+2x+9=0$.

$\purple{\bigstar}\large\underline{\text{Solution}}\purple{\bigstar}$

Let the given polynomial be $f(x)$, then since none of the solutions are zero, $f\left(\dfrac{1}{x}\right)$, which is a rational expression, has four reciprocal zeros.

Thus, $x^{4}\cdot f\left(\dfrac{1}{x^{4}}\right)$ is the minimal polynomial with the reciprocal zeros.

$\implies x^{4}\cdot f\left(\dfrac{1}{x^{4}}\right)=9x^{4}+2x^{3}+5x^{2}-2x+1$

Hence, by Vieta's formulas, the sum of the reciprocals is $-\dfrac{2}{9}$.

Answer 2

Step-by-step explanation:

given :

Find $(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)$ if the four solutions

of $x^{4}+4x^{3}+2x^{2}-4x+1=0$ are $x=\alpha,\beta,\gamma,\delta$.

to find :

alpha,\beta,\gamma,\delta[/tex].

solution :

• *(x-a) (x - 3)(x - y)(x − 8) = x¹ + 4x³ + 2x² - 4x + 1

• x= 1

• (1-a)(1 - 3)(1-7)(1 - 8) = 14 + 4 ∙ (1)³ + 2 ⋅ (1)² − 4. (1) + 1

• (a -1)

• (3 − 1) (y − 1)(8 − 1) = 14 + 4 ∙ (1)³ + 2 · (1)² − 4. (1) + 1

• (a-1)(3-1) (y − 1)(8 − 1) = 1 + - 4+2 4+1

• (a − 1)(31) (y − 1)(8 − 1) = 4

• polynomial be f(x),

• f(1/x)

• x= f (1/x) = 9x² + 2x³ + 5x².

• 2x + 1

• sum of reciprocal is -2/9