Find a cubic polynomial if p(0)=0 and p(2)= -4 and p(x) > 0 only if x > 4, find the cubic polynomial.
p(x) = (x)* (x-a)(x-b)
if x = 2, p = -4
p(2) = -4 = (2)(2-a)(2-b)
-2 = 4 -2a - 2b + ab
-6 = -2a -2b + ab
p(4) = 0
0 = (4)(4-a)(4-b)
-16 = - 4a -4b + ab
10 = 2a +2b
5 = a + b
b = 5-a
p(x) = (x)* (x-a)(x-b)
p(x) = x * (x-a)(x+a-5)
if x = 2
-4 = 2*(2-a)(-3+a)
-2 = -6 +5a -a^2
a^2 - 5 a +4 = 0
(a-4)(a-1) = 0
a = 4 or a = 1
b = 1 or b = 4
p(x) = (x)* (x-a)(x-b)
using a = 4, b = 1
p = x(x-4)(x-1)
does that work - check it :)
To find the cubic polynomial, let's start by considering the general form of a cubic polynomial:
p(x) = ax^3 + bx^2 + cx + d
Given that p(0) = 0, we can substitute x = 0 into the equation:
p(0) = a(0)^3 + b(0)^2 + c(0) + d
0 = 0 + 0 + 0 + d
d = 0
Therefore, our polynomial now looks like this:
p(x) = ax^3 + bx^2 + cx
Now, let's substitute x = 2 using the second condition, p(2) = -4:
p(2) = a(2)^3 + b(2)^2 + c(2) = -4
8a + 4b + 2c = -4 (Equation 1)
Lastly, we need to consider the condition p(x) > 0 only if x > 4. This tells us that the graph of the polynomial should only lie above the x-axis for x values greater than 4. In other words, the polynomial should have a root at x = 4.
Since we know that p(0) = 0 and p(4) = 0, the polynomial should have factors (x - 0) and (x - 4). Expanding these factors gives us:
(x - 0)(x - 4) = x(x - 4) = x^2 - 4x
Now, let's substitute this into our polynomial:
p(x) = a(x^2 - 4x)
So far, we have:
p(x) = ax^3 + bx^2 + cx
p(x) = a(x^2 - 4x)
Now, let's substitute x = 2 using the second condition, p(2) = -4:
p(2) = a(2^2 - 4(2)) = -4
a(4 - 8) = -4
-4a = -4
a = 1
Therefore, the polynomial becomes:
p(x) = 1(x^2 - 4x)
p(x) = x^2 - 4x
Hence, the cubic polynomial that satisfies the conditions is p(x) = x^2 - 4x.
To find a cubic polynomial, we need to start by assuming the polynomial has the form: p(x) = ax^3 + bx^2 + cx + d.
Given that p(0) = 0, we have:
0 = a(0)^3 + b(0)^2 + c(0) + d
0 = d
So, we now know that the constant term d is equal to zero.
Next, we have p(2) = -4, which gives us:
-4 = a(2)^3 + b(2)^2 + c(2)
-4 = 8a + 4b + 2c
Since we know that d = 0, we can rewrite the equation as:
-4 = 8a + 4b + 2c + 0
-4 = 8a + 4b + 2c
Lastly, we have the condition that p(x) > 0 only if x > 4. To fulfill this condition, the cubic polynomial should have two real roots and one imaginary root. This guarantees that p(x) will be positive for x > 4 and negative for x < 4.
Since a cubic polynomial has three roots, we need to find three numbers that satisfy our conditions. Let's choose the roots: x = 0, x = 2, and x = 4.
For x = 0:
p(0) = 0a + 0b + 0c + 0 = 0
For x = 2:
p(2) = 8a + 4b + 2c = -4
For x = 4:
p(4) = 64a + 16b + 4c = k (where k is any constant)
Now that we have a system of three equations, we can solve it to find the coefficients a, b, and c.
From the first equation, we know that d = 0. Then, substitute this back into the second equation:
-4 = 8a + 4b + 2c
From the third equation, we can express k in terms of a, b, and c as:
k = 64a + 16b + 4c
Next, we can solve the system of equations using the above information. Substituting d = 0 into the second equation gives:
-4 = 8a + 4b + 2c
Rearranging the equation, we get:
2 = 4a + 2b + c
Now, multiply the equation by 2 to make the coefficients integers:
4 = 8a + 4b + 2c
Combining this equation with the first equation, we can eliminate the constant term, giving us a new system of equations:
4 = 8a + 4b + 2c
k = 64a + 16b + 4c
This system of equations can be solved using various methods, such as substitution or elimination. By solving this system, you will obtain the values of a, b, and c, which will allow you to determine the cubic polynomial that satisfies the given conditions.