divide: p(x) = x⁵ + x³ + x² + 6x - 7 g(x) = x³ + x² + 1
Answers
Answer:
If (x+1) is a factor of given polynomial say p(X) , then at x= -1, p(x) will become zero, otherwise it is not a factor of given polynomial.
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Solution:
(i) x³+x²+x+1
Let p(x)= x³+x²+x+1
The zero of x+1 is -1.
On putting x= -1
p(−1)=(−1)³+(−1)²+(−1)+1
=−1+1−1+1=0
Hence, by factor theorem, x+1 is a factor of x³+x²+x+1
(ii) x4 + x3 + x2 + x + 1
Let p(x)= x⁴+x³+x²+x+1
The zero of x+1 is -1.
On putting x= -1
p(−1)=(−1)⁴+(−1)³+(−1)²+(−1)+1
=1−1+1−1+1=1≠0
Hence, by factor theorem, x+1 is not a factor
of x⁴+x³+x²+x+1
(iii) x4 + 3x3 + 3x2 + x + 1
Let p(x)= x⁴+3x³+3x²+x+1
The zero of x+1 is -1.
On putting x= -1
p(−1)=(−1)⁴+3(−1)³+3(−1)²+(−1)+1
=1−3+3−1+1
=1≠0
Hence,by factor theorem, x+1 is not a factor of x⁴+3x³+3x²+x+1
(iv) x³–x²–(2+√2)x+√2
Let p(x)= x³–x²–(2+√2)x+√2
The zero of x+1 is -1.
On Putting x= -1
p(−1)=(−1)³–(−1)²–(2+√2)(−1)+√2
Hence, by factor theorem, x+1 is not a factor of x³–x²–(2+√2)x+√2.
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