Question

14. Given that
$x - \sqrt{5}$
is a factor of the cubic polynomial
${x}^{3} - 3 \sqrt{5} {x}^{2} + 13x - 3 \sqrt{5}$
find all the zeroes of the polynomial......pls answer fast....Pls answer in detail...... ​

Answer 1

Given that, x - √5 is a factor of cubic polynomial x³ - 3√5x² + 13x - 3√5

We have to find all the zeros of the polynomial.

x-√5 ) x³-3√5x²+13x-3√5 ( x²-2√5x+3

........+ x³-1√5x² (change the signs)

------------------------------------

..............-2√5x²+13x

..............-2√5x²+10x (change the signs)

------------------------------------------

..........................+3x-3√5

..........................+3x-3√5 (change the signs)

-------------------------------------------

...............................0

On dividing x³-3√5x²+13x-3√5 with x-√5, we get Quotient = x²-2√5x+3 and remained as 0.

Now, we can find the zeros of the polynomial by Splitting the middle term or by Quadratic formula.

By Quadratic formula

where, a = 1, b = -2√5 and c = 3

→ D = b² - 4ac

→ D = (-2√5)² - 4(1)(3)

→ D = 20 - 12

→ D = 8

Now,

→ x = (-b ± √D)/2a

→ x = (+2√5 ± √8)/2

→ x = (2√5 ± 2√2)/2

→ x = √5 ± √2

Also, x - √5 = 0, x = √5

So, zeros of the polynomial are √5 and √5 ± √2.

Answer 2

$\huge\boxed{Answer}$

(x - √5) is a factor of the cubic polynomial

(x³ - 3√5x² + 13x - 3√5)

So , When this cubic polynomial is divided

by (x - √5) , then the remainder will also

be factor of this polynomial .

x³ - 3√5x² + 13x - 3√5 = 0

(x - √5) (x² - 2√5x + 3) = 0

For quadratic polynomial,

x = 2√5 + √20-12 / 2

x = 2√5 + √8 / 2

x = 2√5 + 2√2 / 2

x = √5 + √2

So , all the three zeroes are √5 , √5 + √2

and √5 - √2 .