sub x=1 into ax^3 - x^2 + 2x + b and set it equal to 10 to get
a - 1 + 2 + b = 10
a + b = 9 , this is your first equation
now x = 2 into the expression set equal to 51 to get your second equation.
Solve these two little equations for a and b
Can anyone start me off on these questions? Thanks!
When ax^3 - x^2 + 2x + b is divided by x-1 the remainder is 10. When it is divided by x-2 the remainder is 51. Find a and b.
The volume of a box V(x) = x^3 - 15x^2 + 66x - 80.
Determine expressions for the dimensions of the box in terms of x.
3 answers
Put
P(x) = ax^3 - x^2 + 2x + b
Then
P(x) = Q(x)(x-1) + 10
And
P(x) = R(x)(x-2) + 51
For certain (second degree) polynomials Q(x) and R(x).
This means that:
P(1) = 10
And
P(2) = 51
So we have:
a + b = 9
8 a + b = 51
Solution:
a = 6 and b = 3
In the second problem one has to make some extra assumptions about how the sides depend on x, so we'll assume that they are linear functions of x. We can then find these functions by factoring
V(x). This still does not uniquely fix the sides but we can try to write all the factors in the form x - p for integer p, so we'll assume that the sides are given by these factors.
All the zeroes of V(x) have to be factors of 80. you can see that x = 2 is a zero of V(x), and that another zero is x = 5. Now
(x-2)(x-5) has 10 as the constant term, while V(x) has -80 as the contant term, So, if V(x) is to factor like
V(x) = (x-p1)(x-p2)(x-p3)
with
p1 = 2 and p2 = 5,
then p3 must be 8. And you can check that x = 8 is indeed a zero of V(x). If you have three zeroes of the third degree polynomial, then that fixes the polynomial up to a constant factor. So, you know that:
V(x)= A (x-2)(x-5)(x-8)
And A must be equal to 1 because the coefficient of x^3 in V(x) is 1.
P(x) = ax^3 - x^2 + 2x + b
Then
P(x) = Q(x)(x-1) + 10
And
P(x) = R(x)(x-2) + 51
For certain (second degree) polynomials Q(x) and R(x).
This means that:
P(1) = 10
And
P(2) = 51
So we have:
a + b = 9
8 a + b = 51
Solution:
a = 6 and b = 3
In the second problem one has to make some extra assumptions about how the sides depend on x, so we'll assume that they are linear functions of x. We can then find these functions by factoring
V(x). This still does not uniquely fix the sides but we can try to write all the factors in the form x - p for integer p, so we'll assume that the sides are given by these factors.
All the zeroes of V(x) have to be factors of 80. you can see that x = 2 is a zero of V(x), and that another zero is x = 5. Now
(x-2)(x-5) has 10 as the constant term, while V(x) has -80 as the contant term, So, if V(x) is to factor like
V(x) = (x-p1)(x-p2)(x-p3)
with
p1 = 2 and p2 = 5,
then p3 must be 8. And you can check that x = 8 is indeed a zero of V(x). If you have three zeroes of the third degree polynomial, then that fixes the polynomial up to a constant factor. So, you know that:
V(x)= A (x-2)(x-5)(x-8)
And A must be equal to 1 because the coefficient of x^3 in V(x) is 1.
If you're looking at the answer provided for the first question and are lost at how they jump from having two equations to having the final answers it's because they subtract the two equations. a+b=9 and 8a+b=51, 51-9=47, 8a-a=7a, and b-b is nothing. The equation will look like this 47=7a. Solve for a, you get 6, and plug that back into any of the equations and find b = 3.
1 answer