given that root 3 is a zero of the polynomial x cube + x square - 3 x minus 3 find its other two zeros
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Answered by
108
Hello,
If p(a) =0 ; then by factor theorem we can say that (x-a) is a factor of p(x)
Similarly, above root 3 is given
So, (x-root3) should be a factor of p(x)
Now, please refer to the photo for the used process.
Now, explanation of part 1 from attachment :
In the second step of division we got a term as follows :
-3x
-(1+ root3) root3 x
So, down (-) changes to (+)
So, it becomes
-3x
+(1+root3)root3 x
In down number, (1 + root3) root3 x can be written as (root3 + 3)x
We multiplied the outside root3
Now the division becomes :
-3x
+(root3 + 3)x
Now , taking x common we get the answer as
(-3+root3 + 3)x= root3 x
after this the normal division process. Please refer the above photograph for the used process...
Hope this will be helping you ✌️
If p(a) =0 ; then by factor theorem we can say that (x-a) is a factor of p(x)
Similarly, above root 3 is given
So, (x-root3) should be a factor of p(x)
Now, please refer to the photo for the used process.
Now, explanation of part 1 from attachment :
In the second step of division we got a term as follows :
-3x
-(1+ root3) root3 x
So, down (-) changes to (+)
So, it becomes
-3x
+(1+root3)root3 x
In down number, (1 + root3) root3 x can be written as (root3 + 3)x
We multiplied the outside root3
Now the division becomes :
-3x
+(root3 + 3)x
Now , taking x common we get the answer as
(-3+root3 + 3)x= root3 x
after this the normal division process. Please refer the above photograph for the used process...
Hope this will be helping you ✌️
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Answered by
11
Answer:
-root 3 and - 1
Step-by-step explanation:
Hope that helps you
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