Find a polynomial of the specified degree that satisfies the given conditions.
Degree 4; zeros −3, 0, 1, 4; coefficient of x3 is 4
zeros −3, 0, 1, 4 -->x(x+3)(x-1)(x-4)
expanding this gives us
x^4 - 2x^3 - 11x^2 + 12x
but we are told that the x^3 terms is 4x^3
and we have it as -2x^3
so there had to be a -2 in front of the whole mess
polynomial is -2x(x-1)(x-4)(x+3)
or
-2x^4 + 4x^3 + 22x^2 - 24x
Why did the polynomial go to the circus? Because it wanted to find its zeros and coefficients!
But let's get serious now. To find a polynomial of degree 4 with zeros −3, 0, 1, and 4, we know that the factors of the polynomial are (x + 3), x, (x - 1), and (x - 4). Since the coefficient of x^3 is 4, we need an extra factor of (x) in our polynomial.
Multiplying these factors together, we get:
f(x) = 4x(x + 3)(x)(x - 1)(x - 4)
Simplifying this expression gives us the polynomial you're looking for.
To find a polynomial with the specified conditions, we can start by determining the factors of the polynomial using the zeros given.
Since the zeros are -3, 0, 1, and 4, the factors of the polynomial can be written in factored form as follows:
(x + 3)(x - 0)(x - 1)(x - 4)
Next, we can expand this factored form to obtain the polynomial. Multiplying the factors together, we have:
(x + 3)(x - 0)(x - 1)(x - 4) = (x + 3)(x)(x - 1)(x - 4)
To find the coefficient of x^3, we need to look at the terms that contain x^3. Since the x^3 term can be formed by multiplying x from the first factor and x - 4 from the fourth factor, the coefficient of the x^3 term will be the product of the coefficients of these terms.
So, the coefficient of x^3 will be:
(1)(-4) = -4
However, in the given conditions, we are looking for a coefficient of x^3 that is 4. To achieve this, we can simply introduce a factor of -1 to the polynomial to reverse the sign. This will not change the positions of the zeros.
Hence, the polynomial that satisfies the given conditions is:
-4(x + 3)(x)(x - 1)(x - 4)
If you expand it further, you will get the complete polynomial.
To find a polynomial that satisfies the given conditions, we can start by writing the general form of a polynomial of degree 4:
P(x) = a(x - r1)(x - r2)(x - r3)(x - r4)
where a is the leading coefficient and r1, r2, r3, and r4 are the zeros or roots of the polynomial.
Given that the zeros are -3, 0, 1, and 4, we can write:
P(x) = a(x + 3)(x - 0)(x - 1)(x - 4)
Now, we know that the coefficient of x^3 is 4. We can expand the polynomial and equate it to the general form, and then solve for the leading coefficient a.
Expanding P(x):
P(x) = a(x + 3)(x)(x - 1)(x - 4)
= a(x^2 + 3x)(x - 1)(x - 4)
= a(x^3 - x^2 - x^2 + 3x)(x - 4)
= a(x^3 - 2x^2 + 3x)(x - 4)
= a(x^4 - 2x^3 + 3x^2 - 4x^3 + 8x^2 - 12x)(x - 4)
= a(x^4 - 6x^3 + 11x^2 - 12x)(x - 4)
= a(x^5 - 4x^4 - 6x^4 + 24x^3 + 11x^3 - 44x^2 - 12x^2 + 48x)(x - 4)
Now, we compare the expanded polynomial with the general form:
P(x) = a(x^5 - 10x^4 + 35x^3 - 56x^2 + 48x)(x - 4)
Comparing the coefficients of x^3:
4 = 35a
Solving for a:
35a = 4
a = 4/35
Therefore, the polynomial that satisfies the given conditions is:
P(x) = (4/35)(x^5 - 10x^4 + 35x^3 - 56x^2 + 48x)(x - 4)