Question

Obtain all zero of the polynomial f( x) = X ^4 - 3 x^3 -x^2+ 9 x - 6 if two of its zeros are -√3 and √3 ​

Answer 1

Answer:

√3, - √3 , 1 and 2

Step-by-step explanation:

Given :

√3, - √3 are the zeroes of x³ - x² + 9x - 6

Step 1 : Form a quadratic polynomial whose zero are √3 and - √3

If α, β are zeroes,

Quadratic polynomial = ( x - α )( x - β )

= ( x - √3 ){ x - ( - √3 ) }

= ( x - √3 )( x + √3 )

= x² - ( √3 )²

= x² - 3

Hence, x² - 3 is the is a factor of f( x )

Step 2 : So, Divide f( x ) by ( x² - 3 )

x² - 3 ) x^4 - 3x³ - x² + 9x - 6 ( x² - 3x + 2

*******+ x^4 + 0x³ - 3x²

*******(-)*****(-)*****(+)

_____________________

**************-3x³ + 2x² + 9x

**************-3x³ + 0x² + 9x

*************(+)***********(-)

_______________________

******************** 2x² - 6

******************** + 2x² - 6

*********************(-)****(+)

______________________________

***********************0

______________________________

So f( x ) = ( x² - 3 )( x² - 3x + 2 )

Hence the other zeroes of the polynomial would be the zeroes of x² - 3x + 2

Step 3 : Finding the zeroes of x² - 3x + 2

=> x² - 3x + 2 = 0

=> x² - 2x - x + 2 = 0

=> x( x - 2 ) - 1( x - 2 ) = 0

=> ( x - 1 )( x - 2 ) = 0

=> x - 1 = 0 or x - 2 = 0

=> x = 1 or x = 2

Therefore √3, - √3, 1 and 2 are all the zeroes of the given polynomial.

Answer 2

Answer:

• Polynomial : $\rm x^4-3x^3-x^2+9x-6$
• Zeroes : $-\:\sqrt{3}$ and $\sqrt{3}$

If $-\:\sqrt{3}$ and $\sqrt{3}$ are zeroes of the Polynomial then, Polynomial will be Divisible by $\rm[x-(-\:\sqrt{3})]$ and $\rm\rm[x-\sqrt{3}]$, and so to there Products.

⇢ $\rm[x-(-\:\sqrt{3})][x-\sqrt{3}]$

⇢ $\rm[x+\sqrt{3}][x-\sqrt{3}]$

⇢ $\rm x^2-3$

• (x² - 3) will completely Divide Polynomial. So by Division.

$\rule{150}{2}$

$\boxed{\begin{array}\quad\begin{tabular}{m{3.5em}ccccc}&&\bf x^2&\bf- 3x&\bf+ 2\\\cline{1-7}\multicolumn{2}{l}{\sf x{^\sf2} - \sf3\big)}&x^4&- 3x^3&- x^2&+ 9x&- 6\\&& - (x^4&&- 3x^2)&\\\cline{3-6}&&&- 3x^3&+ 2x^2&+ 9x\\&&&- (3x^3& &+ 9x)\\\cline{4-7}&&&&+ 2x^2&&- 6\\&&&&- (2x^2&&- 6)\\\cline{5-7}&&&&\times&&\times\\\end{tabular}\end{array}}$

• Remainder Obtained will be other two zeroes of the Polynomial.

$\underline{\bigstar\:\textbf{By Splitting Middle Term :}}$

$:\implies$ x² – 3x + 2 = 0

$:\implies$ x² – (2 + 1)x + 2 = 0

$:\implies$ x² – 2x – x + 2 = 0

$:\implies$ x(x – 2) – 1(x – 2) = 0

$:\implies$ (x – 2)(x – 1) = 0

$:\implies$ x = 2⠀or,⠀x = 1

⠀

$\therefore$ Hence, All four Zeroes of the Given Polynomial are - √3, √3, 2 and 1.