Question

.Find the zeroes of the polynomial x

3

-6x

2

+3x+10, it is given that the zeroes are in

A.P.
plzz it's urgent answer it..​

Answer 1

Step-by-step explanation:

As it is given that zeroes are in A.P. let the zeroes be (A-B),A,(A+B). So we know that the sum of the roots in this polynomial = -(-6)= 6 = A+B+A+A-B =>3A=6=>A=2. Hence divide the polynomial

${x}^{3} - 6 {x}^{2} + 3x + 10$

by (x-2) then we will get the quotient which needs to be factorised.

(x-2)√ x^3-6x^2+3x+10 |(x^2-4x-5

-x^3+2x^2

______________

-4x^2+3x+10

4x^2 -8x

_____________

-5x+10

5x-10

__________

0

________

So the quotient thus obtained is

${x}^{2} - 4x - 5 = > {x}^{2} - 5x + x - 5 = >$

$x(x - 5) + 1(x - 5) = >$

$(x + 1)(x - 5)$

So the zeroes of the given polynomial is -1,2,5 respectively. Hope it helps you.