ctorize (x + 1) (x + 3) (x + 4) (x + 6) - 119 - HCF of the given polynomials by division met x3 + 5x2 - 6x -2; 3x3 - 5x2 + 6x - 4
Answers
Answer:
Apply remainder theorem
x+1=0
x=−1
Put the value of x=−1 in all equations.
(i) x
3
+x
2
+x+1=(−1)
3
+(−1)
2
+(−1)+1=−1+1−1+1=0
Then x+1 is the factor of equation
(ii) x
4
+x
3
+x
2
+x+1=(−1)
4
+(−1)
3
+(−1)
2
+(−1)+1=1−1+1−1+1=1
This is not zero.Then x+1 is not the factor of equation
(iii) x
4
+3x
3
+3x
2
+x+1=(−1)
4
+3(−1)
3
+3(−1)
2
+(−1)+1=1
This is not zero.Then x+1 is not the factor of equation
(iv)x
3
−x
2
−(2+
2
)x+
2
=(−1)
3
−(−1)
2
−(2+
2
)(−1)+
2
=−1−1+2−
2
+
2
=0
This is zero. Then x+1 is the factor of equation
Answer:
Apply remainder theorem
x+1=0
x=−1
Put the value of x=−1 in all equations.
(i) x
3
+x
2
+x+1=(−1)
3
+(−1)
2
+(−1)+1=−1+1−1+1=0
Then x+1 is the factor of equation
(ii) x
4
+x
3
+x
2
+x+1=(−1)
4
+(−1)
3
+(−1)
2
+(−1)+1=1−1+1−1+1=1
This is not zero.Then x+1 is not the factor of equation
(iii) x
4
+3x
3
+3x
2
+x+1=(−1)
4
+3(−1)
3
+3(−1)
2
+(−1)+1=1
This is not zero.Then x+1 is not the factor of equation
(iv)x
3
−x
2
−(2+
2
)x+
2
=(−1)
3
−(−1)
2
−(2+
2
)(−1)+
2
=−1−1+2−
2
+
2
=0
This is zero. Then x+1 is the factor of equation
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