Question

find all the zeros of the polynomoial 2x^4+7x^3-19x^2-14x+30 if two of its zeros are root 2 and -root 2

Answer 1

Heya !!!

( ✓2) and ( -✓2) are the two zeroes of the given polynomial.

( X - ✓2) ( X + ✓2) = ( X)² - (✓2)² = X²-2

G(X) = X²-2

P(X) = 2X⁴ + 7X³ - 19X² - 14X + 30

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

X²-2)2X⁴+7X³-19X²-14X - 15 (2X²+7X-15

*******2X⁴ ********-4X²

____________________
******0*****+7X³ -15X² -14X -15

************+7X³ ********-14X

___________________________

************0****-15X²***0****-15

*****************-15X² *********-15

__________________________

Remainder = 0

We get,

Remainder = 0

And,

Quotient = 2X² + 7X - 15

After Factorisation we will get two other zeroes of the given polynomial.

Now,

=> 2X² +7X - 15

=> 2X² + 10X -3X - 15

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

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

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

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

Hence,

-5 , 3/2 , ✓2 and -✓2 are four zeroes of the given polynomial.

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

Answer 2

Answer:

The other two zeros are 3 and -\dfrac{1}{2}−21

Step-by-step explanation:

Given: 2x^4-13x^3+19x^2+7x-32x4−13x3+19x2+7x−3

Two zeros are 2+\sqrt{3}2+3 and 2-\sqrt{3}2−3

The factor of the given polynomial,

(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))(x−(2+3))(x−(2−3))

x^2-4x+1x2−4x+1

Now divide the polynomial with their factor

\Rightarrow \dfrac{2x^4-13x^3+19x^2+7x-3}{x^2-4x+1}⇒x2−4x+12x4−13x3+19x2+7x−3

\Rightarrow 2x^2-5x-3⇒2x2−5x−3

Now factor the above quadratic

\Rightarrow 2x^2-6x+x-3⇒2x2−6x+x−3

\Rightarrow 2x(x-3)+1(x-3)⇒2x(x−3)+1(x−3)

\Rightarrow (x-3)(2x+1)⇒(x−3)(2x+1)

The zeros are,

x - 3 = 0  and  2x + 1 = 0

x=3,x=-\dfrac{1}{2}x=3,x=−21