Question

Obtain all the zeroes of the polynomial 3x4 – 12x3+ 5x2 + 16x - 12, if two of its zeroes are -2/√3 and 2/√3
.
( that x4and x3 are powers) ​

Answer 1

Step-by-step explanation:

If

$- \frac{2}{ \sqrt{3} } and \: \frac{2}{ \sqrt{3} } \\ \:$

are the zeros of polynomial,then

$\Big(x + \frac{2}{ \sqrt{3} } \Big)\Big(x - \frac{2}{ \sqrt{3} }\Big )$

are the factors of that polynomial,so

$apply \: identity \\ \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ \Big(x + \frac{2}{ \sqrt{3} } \Big)\Big(x - \frac{2}{ \sqrt{3} } \Big) = {x}^{2} - \frac{4}{3} \\ \\ or \\ \\ 3 {x}^{2} - 4 = 0 \\ \\ is \: polynomial \: that \: is \: factor \: of$

$3 {x}^{4} - 12 {x}^{3} +5x^2+16x-12$

Now divide this by the factor polynomial

$3 {x}^{2} - 4 \: )3 {x}^{4} - 12 {x}^{3} +5x^2+16x-12( {x}^{2} - 4x + 3\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3 {x}^{4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 4 {x}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( - ) \: \: \: \: \: \: \: \: \: \: \: \: \: ( + ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 12 {x}^{3 } + 9 {x}^{2} + 16x \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 12 {x}^{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 16x \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( + ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( - ) \\ \: \: \: \: \: \: \: \: \: \: \: \: - - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 9 {x}^{2} - 12 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 9 {x}^{2} - 12 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( - ) \: \: \: \: ( + ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0$

Now the quotient polynomial is

${x}^{2} - 4x + 3 \\ \\ find \: roots \: of \: it \\ \\ {x}^{2} - 4x + 3 = 0 \\ \\ {x}^{2} - 3x - x + 3 = 0 \\ \\ x(x - 3) - 1(x - 3) = 0 \\ \\ (x - 1)(x - 3) = 0 \\ \\ x = 1 \\ \\ x = 3$

So, all the roots of polynomial are

$\frac{ - 2}{ \sqrt{3} } , \: \frac{2}{ \sqrt{3} }, 1, \: 3$

Hope it helps you.

Answer 2

Answer:2/√3,-2/√3,3,1

Step-by-step explanation:

 The two zeroes of the given polynomial are 2/√3 and -2/√3.

 (√3x-2) and (√3x+2) are the factors of the given polynomial.

 => (√3x-2)(√3x+2) = (3x^2- 4)

Is also the factor of the given polynomial.

=> by the division method given below in the image <=

 After that factorise

 = 3x^4 - 12x^3 + 5x^2 + 16x - 12

 = (3x^2 - 4) (x^2 - 4x + 3)

 = (3x^2 - 4) [(x^2 - 3x)-(x + 3)]

 = (3x^2 - 4) [x (x - 3)-1 (x - 3)]

 = (3x^2 - 4) [(x - 3) (x - 1)]

 = (√3x-2)(√3x+2)(x - 3)(x - 1)

 Threrfore, the zeroes of the polynomial are 2/√3, -2/√3,3,1