Question

Find all the zeroes of a polynomial 2x4+7x3-19x2-14x+30 if two of its zeroes are root 2 and -root 2

Answer 1

Hii friend,

✓2 and -✓2 are the two zeros of the polynomial 2X⁴+7X³-19X²-14X+30.

(X-✓2) and (X+✓2) are the two zeros of the polynomial.

(X-✓2) (X+✓2) => X²-(✓2)² = X²-2.


G(X) = X²-2

P(X) = 2X⁴+7X³-19X²-14X+30

On dividing P(X) by G(X) , we get

Quotient = 2X²+7X-15

and,

Remainder = 0.

Factories the Quotient then we will get the two other zeros of the P(X).

Quotient = 2X²+7X-15

=> 2X² +10X -3X -15

=> 2X(X+5) -3(X+5)

=> (X+5) (2X-3)

=> (X+5) = 0 and (2X-3)= 0

=> X = -5 and X = 3/2

Hence,

-✓2 , ✓2 , -5 and 3/2 are the four zeros of the polynomial 2X⁴+7X³-19X²-14X+30.

HOPE IT WILL HELP YOU..... :-)

Answer 2

Answer:

All zeroes of p ( x ) are 1 , - 1 / 2 , 2 + √ 3 and 2 - √ 3

Step-by-step explanation:

Given :

p ( x ) = 2 x⁴ - 9 x³ + 5 x² + 3 x - 1

Two zeroes of p ( x ) :

2 + √ 3 and 2 - √ 3

p ( x ) = ( x - 2 - √ 3 ) ( x - 2 + √ 3 ) × g ( x )

g ( x ) = p ( x ) ÷ ( x² - 4 x + 1 )

On dividing ( 2 x⁴ - 9 x³ + 5 x² + 3 x - 1 ) by ( x² - 4 x + 1 )

$\displaystyle{\begin{tabular}{c | c | c}(x^2-4x+1) & \overline{2x^4-9x^3+5x^2+3x-1}} & 2x^2-x-1\\ & 2x^4-8x^3+2x^2 \ \ \ \ \ \ \ \ \ \ \ \\ \\ & ( - ) ( + ) ( - ) \ \ \ \ \ \ \ \\ \\ & \overline{-x^3+3x^2+3x} \\ \\ & - x^3+4x^2-x\\ \\ & ( + ) ( - ) ( + ) \\ \\ & \overline{-x^2 + 4x-1} \\ \\ & -x^2 + 4x-1 \\ \\ & ( + ) ( - ) ( + ) \\ \\ & \overline{0}\endtabular$

We got ,

g ( x ) = 2 x² - x - 1

By splitting mid term

2 x² - 2 x + x - 1

2 x ( x - 1 ) + ( x - 1 )

( x - 1 ) ( 2 x + 1 )

Now zeroes of g ( x )

x = 1 or x = - 1 / 2 .