check whether (x+1) is a factor of x³-x²-(2+√2)x+√2
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Answers
Answered by
21
No,(x+1) is not the factor of

, because:-
let p(x) be x+1
so x= -1
let g(x) be our polynomial
we use remainder theorem
g(x)=

g(-1)=



hence ,the remainder is not zero so p(x) is not the factor of g(x).
, because:-
let p(x) be x+1
so x= -1
let g(x) be our polynomial
we use remainder theorem
g(x)=
g(-1)=
hence ,the remainder is not zero so p(x) is not the factor of g(x).
Answered by
0
Step-by-step explanation:
x
3
−x
2
−(2+
2
)x+
2
, because:-
let p(x) be x+1
so x= -1
let g(x) be our polynomial
we use remainder theorem
g(x)=
{3}^{3} - {x}^{2} - (2 + \sqrt{2} )x + \sqrt{2}3
3
−x
2
−(2+
2
)x+
2
g(-1)=
- 1 - 1 + 2 + \sqrt{2} + \sqrt{2}−1−1+2+
2
+
2
- 2 + 2 + \sqrt{2}−2+2+
2
2 \sqrt{2}2
2
hence ,the remainder is not zero so p(x) is not the factor of g(x).
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