check whether the polynomial q(t)=t^3-4t^2-2t+6 is a multiple of 1-3/4t
Answers
Answer:
If (2t+1)is a factor then t=
2
−1
must be one of the zeroes of the given polynomial. Substituting in the given equation.
q((
2
−1
))=4(
2
−1
)
3
+4(
2
−1
)
2
−(
2
−1
)−1
=4(
8
−1
)+4(
4
1
)+(
2
1
)−1
=(
2
−1
)+1+(
2
1
)−1=0
Hence(2t+1) is a common factor of q(t)
Answer:
Step-by-step explanation:
If (2t+1)is a factor then t=
2
−1
must be one of the zeroes of the given polynomial. Substituting in the given equation.
q((
2
−1
))=4(
2
−1
)
3
+4(
2
−1
)
2
−(
2
−1
)−1
=4(
8
−1
)+4(
4
1
)+(
2
1
)−1
=(
2
−1
)+1+(
2
1
)−1=0
Hence(2t+1) is a common factor of q(t)If (2t+1)is a factor then t=
2
−1
must be one of the zeroes of the given polynomial. Substituting in the given equation.
q((
2
−1
))=4(
2
−1
)
3
+4(
2
−1
)
2
−(
2
−1
)−1
=4(
8
−1
)+4(
4
1
)+(
2
1
)−1
=(
2
−1
)+1+(
2
1
)−1=0
Hence(2t+1) is a common factor of q(t)If (2t+1)is a factor then t=
2
−1
must be one of the zeroes of the given polynomial. Substituting in the given equation.
q((
2
−1
))=4(
2
−1
)
3
+4(
2
−1
)
2
−(
2
−1
)−1
=4(
8
−1
)+4(
4
1
)+(
2
1
)−1
=(
2
−1
)+1+(
2
1
)−1=0
Hence(2t+1) is a common factor of q(t)If (2t+1)is a factor then t=
2
−1
must be one of the zeroes of the given polynomial. Substituting in the given equation.
q((
2
−1
))=4(
2
−1
)
3
+4(
2
−1
)
2
−(
2
−1
)−1
=4(
8
−1
)+4(
4
1
)+(
2
1
)−1
=(
2
−1
)+1+(
2
1
)−1=0
Hence(2t+1) is a common factor of q(t)If (2t+1)is a factor then t=
2
−1
must be one of the zeroes of the given polynomial. Substituting in the given equation.
q((
2
−1
))=4(
2
−1
)
3
+4(
2
−1
)
2
−(
2
−1
)−1
=4(
8
−1
)+4(
4
1
)+(
2
1
)−1
=(
2
−1
)+1+(
2
1
)−1=0
Hence(2t+1) is a common factor of q(t)If (2t+1)is a factor then t=
2
−1
must be one of the zeroes of the given polynomial. Substituting in the given equation.
q((
2
−1
))=4(
2
−1
)
3
+4(
2
−1
)
2
−(
2
−1
)−1
=4(
8
−1
)+4(
4
1
)+(
2
1
)−1
=(
2
−1
)+1+(
2
1
)−1=0
Hence(2t+1) is a common factor of q(t)If (2t+1)is a factor then t=
2
−1
must be one of the zeroes of the given polynomial. Substituting in the given equation.
q((
2
−1
))=4(
2
−1
)
3
+4(
2
−1
)
2
−(
2
−1
)−1
=4(
8
−1
)+4(
4
1
)+(
2
1
)−1
=(
2
−1
)+1+(
2
1
)−1=0
Hence(2t+1) is a common factor of q(t)If (2t+1)is a factor then t=
2
−1
must be one of the zeroes of the given polynomial. Substituting in the given equation.
q((
2
−1
))=4(
2
−1
)
3
+4(
2
−1
)
2
−(
2
−1
)−1
=4(
8
−1
)+4(
4
1
)+(
2
1
)−1
=(
2
−1
)+1+(
2
1
)−1=0
Hence(2t+1) is a common factor of q(t)If (2t+1)is a factor then t=
2
−1
must be one of the zeroes of the given polynomial. Substituting in the given equation.
q((
2
−1
))=4(
2
−1
)
3
+4(
2
−1
)
2
−(
2
−1
)−1
=4(
8
−1
)+4(
4
1
)+(
2
1
)−1
=(
2
−1
)+1+(
2
1
)−1=0
Hence(2t+1) is a common factor of q(t)