Find a polynomial f(x) of degree 4 that has the following zeros. 0,-1,6,-7
That would, of course, be
(x-0)(x+1)(x-6)(x+7)
To find a polynomial of degree 4 with the given zeros 0, -1, 6, and -7, we use the fact that the zeros of a polynomial are also the solutions to the equation f(x) = 0.
First, let's write the linear factors corresponding to the zeros of the polynomial:
x = 0
x + 1 = 0
x - 6 = 0
x + 7 = 0
Next, we multiply all these factors together to get the polynomial:
(x)(x + 1)(x - 6)(x + 7)
Expanding this expression gives:
f(x) = (x)(x + 1)(x - 6)(x + 7)
Multiplying this out further, we get:
f(x) = (x^2 + x)(x - 6)(x + 7)
Expanding again:
f(x) = (x^3 + x^2 - 6x^2 - 6x)(x + 7)
Simplifying:
f(x) = (x^3 - 5x^2 - 6x)(x + 7)
Finally:
f(x) = x^4 + 7x^3 - 5x^3 - 35x^2 - 6x^2 - 42x
Simplifying once more:
f(x) = x^4 + 2x^3 - 41x^2 - 42x
Therefore, the polynomial f(x) with the given zeros is:
f(x) = x^4 + 2x^3 - 41x^2 - 42x
To find a polynomial with the given zeros, you can use the zero-product property. According to this property, if a polynomial has a zero at a certain value, then the polynomial can be factored with (x - zero) as a factor.
Since the given zeros are 0, -1, 6, and -7, we can set up the following factors:
(x - 0)(x - (-1))(x - 6)(x - (-7))
Simplifying this expression, we obtain:
x(x + 1)(x - 6)(x + 7)
Now, we need to multiply these factors together to get the polynomial:
f(x) = x(x + 1)(x - 6)(x + 7)
Expanding this expression yields the desired polynomial:
f(x) = x(x^2 - 6x + x - 6)(x + 7)
f(x) = x(x^2 - 5x - 6)(x + 7)
f(x) = x^2(x^2 - 5x - 6)(x + 7)
f(x) = x^2(x - 6)(x + 7)
So, the polynomial of degree 4 with zeros 0, -1, 6, and -7 is f(x) = x^2(x - 6)(x + 7).