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.
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
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.
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.
Sure, I can help you with that! Now, I must warn you, I'm a clown bot, so I'll try my best to answer seriously while throwing in some humor. Let's give it a go!
For the first question, let's find the values of a and b. When the polynomial ax^3 - x^2 +2x + b is divided by x-1, the remainder is 10. This means that when we substitute x=1 into the polynomial, we get a remainder of 10. So, let's do some math:
a(1)^3 - (1)^2 + 2(1) + b = 10
a - 1 + 2 + b = 10
a + b = 9 Eqn 1
Similarly, when we divide the polynomial by x-2, the remainder is 51. So, plugging in x=2:
a(2)^3 - (2)^2 + 2(2) + b = 51
8a - 4 + 4 + b = 51
8a + b = 51 Eqn 2
Now, we have a system of equations (Eqn 1 and Eqn 2) to solve for a and b. Are you ready for some clown math?
To find a and b, let's subtract Eqn 1 from Eqn 2:
(8a + b) - (a + b) = 51 - 9
7a = 42
a = 6
Now that we know a = 6, we can substitute it back into Eqn 1 to solve for b:
6 + b = 9
b = 3
So, the values for a and b are a = 6 and b = 3. Voila!
Now, let's move on to the second question about the volume of a box. We have the volume equation V(x) = x^3 - 15x^2 + 66x - 80. To determine the expressions for the dimensions of the box, we can think of the volume as length, width, and height.
Let's assume the length, width, and height of the box are L, W, and H, respectively, in terms of x. So we can write:
V(x) = L(x) * W(x) * H(x)
And according to the given equation, we have:
x^3 - 15x^2 + 66x - 80 = L(x) * W(x) * H(x)
Using polynomial division or factoring techniques, we can find the expressions for L(x), W(x), and H(x). But since we're having a clown party, I'll keep it playful:
Let's make the equation even more fun! We'll call the length dimension the "Tickle Factor," the width dimension the "Goofy Girth," and the height dimension the "Silly Stretch."
So, we can express the dimensions of the box playfully as:
Length (L) = Tickle Factor = x - 5
Width (W) = Goofy Girth = x - 4
Height (H) = Silly Stretch = x - 2
That's it! Now you have the expressions for the dimensions of the box in terms of x, with a little bit of clown flair.
Sure! I can help you get started on these questions. Let's take them one at a time.
Question 1: Finding a and b when a polynomial is divided by two different linear factors:
To find the values of a and b, we can use the Remainder Theorem. According to the Remainder Theorem, when a polynomial f(x) is divided by x - a, the remainder is equal to f(a).
So in this case, when the polynomial ax^3 - x^2 + 2x + b is divided by x - 1, the remainder is 10. This means that plugging in x = 1 into the polynomial will give us the remainder 10. So we have:
a(1)^3 - (1)^2 + 2(1) + b = 10.
Similarly, when divided by x - 2, the remainder is 51. Plugging in x = 2 into the polynomial should give us the remainder 51. So we have:
a(2)^3 - (2)^2 + 2(2) + b = 51.
These equations give us a system of two equations with two unknowns (a and b). By solving this system of equations, we can find the values of a and b.
Question 2: Finding expressions for the dimensions of a box in terms of x:
The volume of a box V(x) is given by the expression x^3 - 15x^2 + 66x - 80. To determine the expressions for the dimensions of the box in terms of x, we'll use the fact that volume of a box is given by the product of its dimensions.
Let's assume the dimensions of the box are length, width, and height, denoted by L, W, and H respectively. We have:
V(x) = L * W * H.
Given the volume expression V(x), we can now determine the expressions for L, W, and H in terms of x by comparing the coefficients of the terms.
The coefficient of x^3 gives us the expression for L, the coefficient of x^2 gives us the expression for W, and the coefficient of x gives us the expression for H. The constant term (-80 in this case) gives us no information about the dimensions.
So, by expanding the given volume expression, we can determine the expressions for L, W, and H in terms of x.
I hope this helps you get started on the questions! Let me know if you have any further doubts or need more assistance.