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# Functions of polynomials with different degrees

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Question 1) Determine whether the function is a polynomial function. If it is, state the degree. If it is not, tell why not.
G(x)=7(x-5)^2(x^2+6)

a) Polynomial of degree 4
b) Polynomial of degree 7

Question 2) Form a polynomial whose real zeros and degree are given. Zeros: -1, 0, 5. Degree: 3
Write a polynomial with integer coefficients and a leading coefficient of 1.

https://brainmass.com/math/basic-algebra/polynomials-their-zeros-degrees-595128

## SOLUTION This solution is FREE courtesy of BrainMass!

1) Yes, 7(x - 5)^2(x^2 + 6) is a polynomial.

Note that "7" is a polynomial (a polynomial that's constant),

(x - 5)^2 is a quadratic polynomial (one that is of degree 2),

and x^2 + 6 is a quadratic polynomial (one that is of degree 2).

The product of finitely many polynomials (in the same variable, such as "x" in this case) is a polynomial in that same variable.

Also, the degree of a polynomial that is formed from the product of finitely many polynomials (all in the same variable) is the sum of the degrees of the individual polynomials in the product.

The degree of "7" is 0 (since it's a constant polynomial).

The degree of (x - 5)^2 is 2.

The degree of x^2 + 6 is 2.

Thus the degree of the polynomial which is formed by the product of 7, (x - 5)^2, and x^2 + 6 is 0 + 2 + 2 = 4.

Another way to look at this would be to use the fact that the leading term of a polynomial is the product of the leading terms of its factors.

The leading term of "7" is just 7.

The leading term of (x - 5)^2 is the "squared term," which is x^2.

The leading term of x^2 + 6 is x^2.

Thus the leading term of this polynomial is

7(x^2)(x^2) = 7 x^4, which is of degree 4 and has coefficient 7.

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2) A real number "a" is a zero of a polynomial if and only if (x - a) is a linear factor in that polynomial.

Here, we want a polynomial that has zeros at -1, 0, and 5.

Thus we want a polynomial that has the linear factors (x - (-1)), (x - 0), and (x - 5).

We can write (x - (-1)) as (x + 1), and we can write (x - 0) as just x.

Thus we want a polynomial that has (x + 1), x, and (x - 5) as factors.

Well, each of those factors is a polynomial of degree 1, so their product is a polynomial of degree 1 + 1 + 1 = 3.

Now their product is

(x + 1)x(x - 5) = (x + 1)[x(x - 5)]

= (x + 1)(x^2 - 5x)

= x(x^2 - 5x) + 1(x^2 - 5x)

= x^3 - 5x^2 + x^2 - 5x

Combining like terms (i.e., combining the two "squared terms"), we obtain the polynomial

x^3 + (-5 x^2 + x^2) - 5x

= x^3 + (-5 + 1)x^2 - 5x

= x^3 - 4x^2 - 5x.

This polynomial is of degree 3, and its leading coefficient is 1 (as desired).

Also, let's check to be sure that it's equal to 0 at x = -1, x = 0, and x = 5:

Substituting -1 for x, we find that

x^3 - 4x^2 - 5x = (-1)^3 - 4((-1)^2) - 5(-1) = -1 - 4(1) + 5 = -1 - 4 + 5 = -5 + 5 = 0

Substituting 0 for x, we have

x^3 - 4x^2 - 5x = (0^3) - 4(0^2) - 5(0) = 0 - 4(0) - 0 = 0 - 0 - 0 = 0

Substituting 5 for x, we see that

x^3 - 4x^2 - 5x = (5^3) - 4(5^2) - 5(5) = 125 - 4(25) - 25 = 125 - 100 - 25 = 25 - 25 = 0

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!