Question

give zeros of the polynomial p(x) =(x-1) (x-2) (x-3) (x-4)

Answer 1

To find zeroes , P(x) = 0

(x-1)(x-2)(x-3)(x-4) = 0

x-1 = 0 ; x = 1

x-2 = 0 ;  x = 2

x-3 = 0 ;  x = 3

x-4 = 0 ;  x = 4

hence the zeroes are :- 1,2,3,4

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Answer 2

The zeros of the polynomial p(x) = (x-1) (x-2) (x-3) (x-4) are 1 , 2 , 3 , 4

Given :

The polynomial p(x) = (x-1) (x-2) (x-3) (x-4)

To find :

The zeroes of the polynomial

Solution :

Step 1 of 2 :

Write down the given polynomial

The given polynomial is

p(x) = (x-1) (x-2) (x-3) (x-4)

Step 2 of 2 :

Find the zeroes of the polynomial

For zeroes of the polynomial we have

$\sf \: p(x) = 0$

$\sf \implies (x-1) (x-2) (x-3) (x-4) = 0$

$\sf \implies (x-1) = 0,(x-2) = 0, (x-3) = 0, (x-4) = 0$

x - 1 = 0 gives x = 1

x - 2 = 0 gives x = 2

x - 3 = 0 gives x = 3

x - 4 = 0 gives x = 4

Hence the required zeroes are 1 , 2 , 3 , 4

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