Question

Obtain all the zeroes of the polynomial x4+x3-16x2-4x+48, if two of its zeroes are 2 and -4

Answer 1

THERE IS THE SOLUTION FOR YOUR QUESTION.




HOPE it helps

Answer 2

Answer:

-2,2,-3,-4

Step-by-step explanation:

Dividend = $x^4+x^3-16x^2-4x+48$

Since we are given that two of its zeroes are 2 and -4

So, $(x-2)(x+4)$

$x^2+4x-2x-8$

$x^2+2x-8$

Divisor =$x^2+2x-8$

Since we know that :

$Dividend =(Divisor \times Quotient)+Remainder$

$x^4+x^3-16x^2-4x+48=(x^2+2x-8 \times x^2-x-6)+0$

Quotient = $x^2-x-6$

Factorize the quotient

$x^2-x-6 =0$

$x^2+3x-2x-6 =0$

$x(x+3)-2(x+3) =0$

$(x-2)(x+3) =0$

$x=2,-3$

Hence all the zeros of the polynomial $x^4+x^3-16x^2-4x+48$ is -2,2,-3,-4