Question

find all zeroes of polynomial 4 x square- 20 x cube + 23 x square + 5 x minus 6, if two of its zeros are 2 and 3

Answer 1

Hence..it's other two zeroes are -1/2 and 1/2

Answer 2

Answer:

$4x^4-20x^3+23x^2+5x-6=(x-2)(x-3)(x+\frac{1}{2})(x-\frac{1}{2})$

Step-by-step explanation:

Given : Expression $4x^4-20x^3+23x^2+5x-6$ if two of its zeros are 2 and 3.

To find : All zeroes of polynomial?

Solution :

Polynomial $4x^4-20x^3+23x^2+5x-6$

Two zeros, x=2 and x=3

i.e. (x-2)(x-3) factor of $4x^4-20x^3+23x^2+5x-6$

$x^2-5x+6$ factor of $4x^4-20x^3+23x^2+5x-6$

Now, we divide $4x^4-20x^3+23x^2+5x-6$ by $x^2-5x+6$

Divisor is $x^2-5x+6$

Dividend is $4x^4-20x^3+23x^2+5x-6$

As Dividend = Divisor × Quotient +remainder

We get, Quotient $4x^2-1$

Remainder = 0

Now, Factories further quotient

$4x^2-1=0$

$4x^2=1$

$x^2=\frac{1}{4}$

$x=\pm\frac{1}{2}$

So, The factors of the expression is

$4x^4-20x^3+23x^2+5x-6=(x-2)(x-3)(x+\frac{1}{2})(x-\frac{1}{2})$