1) For finding the coefficients a and b, the given
polynomial has to be equal to the polynomial x^3 + 2x^2 - x - 2, for that we have to put
in a equality relation the corresponding
coefficients.
Let's see
how:
x^3 + (a-1)x^2 - x + b+1 = x^3 + 2x^2 - x -
2
The coefficient of x^2, from the left side polynomial, is
(a-1) and it has to be equal to the corresponding coefficient of x^2, from the right
side, which is
2.
a-1=2
a-1+1=2+1
a=3
The
same we'll proceed with the
terms:
b+1=-2
b+1-1=-2-1
b=-3
2)
For finding the coefficients a and b, we have to follow the constraint, which states
that the polynomial p(x) is divisible by x and x+2.
If p(x)
is divisible by x, that means that if p(x) will be divided by x, the reminder will be
0.
The constraint is mathematically
written:
p(0)=0
For
calculating p(0), we'll substitute x from polynomial by
0.
p(0) = 0^3 + (a-1)0^2 - 0 +
b+1
b+1=0
b=-1
If
p(x) is divisible by x+2, that means that if p(x) will be divided by x+2, the reminder
will be 0.
We'll write the division with
reminder:
p(x)=(x+2)*Q(x) +
R(x)
We notice that we substitute x with -2, we'll
have:
p(-2)=(-2+2)*Q(-2) + R(-2), where R(-2)=0, because
p(x) is divisible by
x+2.
p(-2)=0*Q(-2)+0
p(-2) =
(-2)^3 + (a-1)(-2)^2 - (-2) -1+1
p(-2) = -8 + 4a - 4 +
2
But
p(-2)=0
4a-10=0
4a=10
a=10/4
a=5/2
3)
If x+1 is a factor of p(x), that means that p(x) will be divided by x+1 and the reminder
will be 0.
We'll write the division with
reminder:
p(x)=(x+1)*Q(x) +
R(x)
We notice that we substitute x with -1, we'll
have:
p(-1)=(-1+1)*Q(-1) + R(-1), where R(-1)=0, because
p(x) is divisible by x+1.
p(-1) = (-1)^3 + (a-1)(-1)^2
- (-1) +b+1
p(-1) =
-1+a-1+1+b+1
But p(-1) = 0, so
a+b=0
If p(x) is divided by x+3 and the reminder is 30,
we'll write:
p(x)=(x+3)*Q(x) +
R(x)
We notice that we substitute x with -3, we'll
have:
p(-3)=(-3+3)*Q(-3) + R(-3), where
R(-3)=30.
p(-3) = (-3)^3 + (a-1)(-3)^2 - (-3)
+b+1
But
p(-3)=30
-27+9a-9+3+b+1=30, but
a=-b
-32+9a-a=30
8a=30+32
8a=62
a=62/8
a=31/4
and b=-31/4
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