Question

obtain all zeros of x4 - x3 - 7x2 + x + 6 if 3 and 1 are zeroes

Answer 1

refer to the pic for answer.....zeroes are (-2), (-1),3&1

Answer 2

Answer:

-1,-2,3,1

Step-by-step explanation:

Dividend = $x^4 -x^3-7x^2+x +6$

Since we are given that two of its zeroes are 3 and 1

So, $(x-3)(x-1)$

 $x^2-x-3x+3$

 $x^2-4x+3$

Divisor =  $x^2-4x+3$

Since we know that :

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

$x^4 -x^3-7x^2+x +6=(x^2-4x+3 \times x^2+3x+2)+0$

Quotient =  $x^2+3x+2$

Factorize the quotient

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

$x^2+2x+x+2=0$

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

$(x+1)(x+2)=0$

$x=-1,-2$

Hence all zeros of  $x^4 -x^3-7x^2+x +6$ are -1,-2,3,1