Question

find the zeros of the polynomial x3-7x+6 plzzz slove any one urgent plzz guys
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Answer 1

HEY MATE HERE IS YOUR ANSWER ⭐

Step-by-step explanation:

P(x) = x^3 - 7x + 6

First we will factor 6 to find possible roots for the function f9x):

6 = 1, -1, 2, -1, 3, -3, 6, -6

Let us try and substitute x = 1:

==> P (1) = 1 - 7 + 6 = 0

Then x= 1 is one of the roots for P(x):

==> ( x -1) is a factor for P(x):

Then we could wrtie:

P (x) = (x-1) * R(x)

Now we will divide P(x) by (x-1) to determine the other factors:

==> P(x) = (x-1)(x^2 +x - 6)

= (x-1) (x+3)(x-2)

Then roots for P(x) are:

x = { 1, 2, -3}

HOPE THIS HELPS YOU

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Answer 2

x3-7x+6=0

then

x3-7x=-6

x(x2-7)=-6

x2-7=-6x

x2-7 +6x=0

x2-1x+6x-7=0

x(x-1)+7(x-1)=0

(x+7) (x-1)