Question

If a= 3-√5 then the value of a^4-a^3-20a^2-16a + 24 is​

Answer 1

$\large\underline{\text{Step 1. Conjugate solution}}$

We know that a quadratic equation with rational coefficients has a pair of conjugate solution.

Let's derive the rational coefficient equation.

$\implies a-3=-\sqrt{5}$

$\implies(a-3)^{2}=(-\sqrt{5})^{2}$

$\implies a^{2}-6a+9=5$

$\red{\bigstar} a^{2}-6a+4=0$

$\large\underline{\text{Step 2. Factor theorem}}$

Meanwhile, by factor theorem

$a=2\implies a^{4}-a^{3}-20a^{2}-16a+24=0$

$a=3\implies a^{4}-a^{3}-20a^{2}-16a+24=0$

$\large\underline{\text{Step 3. Factorization}}$

Let's try the division of $(a^{4}-a^{3}-20a^{2}-16a+24)\div(a^{2}-5a+6)$. We get the quotient of $a^{2}-6a+4$ without a remainder. Hence, the quartic polynomial can be factorized into

$\red{\bigstar}a^{4}-a^{3}-20a^{2}-16a+24=(a^{2}-6a+4)(a-2)(a-3)$

However, we know that $\red{\bigstar}a^{2}-6a+4=0$, so the factor of the quartic polynomial is 0, and hence the required value is 0.

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

The required value is 0.

$\large\underline{\text{Learn more}}$

If a rational coefficient quadratic equation has the discriminant value $\red{\bigstar}D=b^{2}-4ac<0$, then the solutions are imaginary conjugates.

Answer 2

Step-by-step explanation:

given :

• If a= 3-√5 then the value of a^4-a^3-20a^2-16a + 24 is

to find :

• a^4-a^3-20a^2-16a + 24 iis

solution :

• a-3= -√5

• (a - 3)² = (-√√5)²

• a²6a +9=5

• a²6a+ 4 = 0

• a = 2a²-a³ - 20a² - 16a +24 = 0

• a = 3a²-a³ – 20a² - 16a +24 = 0

• hence, the answer is 0 value