Question

if two zeros of a polynomial P of x is equal to 2 x to the power 4 minus 2 x cube + 5 x square + 3 x minus 1 are 2 + root 3 and 2 minus root 3. obtain all the zeros​

Answer 1

p(x) = 2x⁴ - 9x³ + 5x² + 3x - 1

Two of its factors are (2+√3) and (2-√3)

i.e. {x-(2+√3)}{x-(2-√3)}

=> x² - x(2+√3) - x(2-√3) + (2²-√3²)

=> x² - 2x - x√3 - 2x + x√3 + 4 - 3

=> x² - 4x + 1

Then, we have divided p(x) by (x² - 4x + 1) by long division method. (in attachment)

Now, we know that

p(x) = (x² - 4x + 1)(2x² - x - 1)

= (x² - 4x + 1)(2x² - 2x + x - 1)

= (x² - 4x + 1){2x(x-1) + 1(x-1)}

= (x² - 4x + 1)(x - 1)(2x + 1)

$\therefore$ the other two zeroes are 1 and - 1/2.

Answer 2

Answer: (√11 - 3)/2 and -(√11+3)/2

Step-by-step explanation:

We have,

p(x) = 2x⁴ -2x³ + 5x² + 3x - 1

Comparing  this equation with standard  form of 4th degree of equation,

ax⁴ + bx³ + cx² + d + E,

a = 2

b = -2

c = 5

d = 3

E = -1

Let the  other two zeroes of polynomial be α and β

Now,

Sum of  zeroes  = -b/a

α + β + (2+√3) + (2-√3)  = -(-2/2)

α + β + 4 = 1

α + β = 1 - 4

α + β = -3  ..............i)

Product  of  zeroes = E/a

α × β × (2+√3)(2-√3) = -1/2

α×β ×(4-3) = -1/2

αβ = -1/2

We know that,

(α - β)² = (α + β)² - 4αβ

(α - β)² = (-3)² - 4(-1/2)

(α - β)² = 9 + 2

(α - β)² = 11

α - β = √11 .......ii)

Adding equation i) and ii)

2α = √11 - 3

α = (√11 - 3)/2

Subtracting i) from ii)

-2β = √11 + 3

β = -(√11+3)/2

Hence other zeroes of  polynomial p(x) are (√11 - 3)/2 and -(√11+3)/2

Note:- In a polynomial ax⁴ + bx³ + cx² + d + E, if the zeroes are $\alpha,\;\beta,\;\gamma,\;\delta$

$\alpha+\beta+\gamma+\delta=\frac{-b}{a}\\\;\\\alpha\beta+\beta\gamma+\gamma\delta+\delta\alpha+\beta\alpha+\gamma\alpha=\frac{c}{a}\\\;\\\alpha\beta\gamma+\beta\gamma\delta+\delta\gamma\alpha+\delta\beta\gamma=\frac{-d}{a}\\\;\\\alpha\beta\gamma\delta=\frac{E}{a}$