Does a fraction fall under the definition any number?
C
c
Where is the variable located in a polynomial?
D
d
sqrt(a)=a^(0.5)
E
e
Start by multiplying the parentheses.
F
f
1/a^b=a^(- b)
G
g
Use what you learned in previous parts to write two expressions, where one is a polynomial and the other is not.
A
a
Polynomial
B
b
Polynomial
C
c
Not a polynomial, see solution.
D
d
Not a polynomial, see solution.
E
e
Polynomial
F
f
Not a polynomial, see solution.
G
g
Polynomial: g(x)= 7x^3+2x^2 Not a polynomial: h(x)= 2x^2+7x^(3.5)
Practice makes perfect
a
Let's rewrite some of the variables and the constant of the given equation, then check if it follows the description of a polynomial.
( any number)* x^(( whole number))
↓
f(x)= 8x^5+ 1x^2+ 6.5x^4+ 6x^0
As we can see, the equation is a polynomial since all terms follow the description of a polynomial expression.
b
Like in Part A, we will check if the equation fits the description of a polynomial.
( any number)* x^(whole number)
↓
y= 3/5x^6+ 19x^2
A fraction falls under the definition of any number. The equation is therefore a polynomial.
c
Examining the equation, we notice that it contains a term that does not follow the description of a polynomial.
( any number)* x^(whole number)
↓
y= 2^x+ 8x^0
The equation is not a polynomial, because x is in the exponent of one of the terms.
d
Before we start analyzing the equation, we can simplify it by combining like terms on the right-hand side.
f(x)=9+sqrt(x)-3 ⇔ f(x)=6+sqrt(x)
Notice that the square root of a number can be rewritten as the number raised to the power of 0.5. With this information we see that the function is not a polynomial.
( any number)* x^(whole number)
↓
f(x)= 6x^0+ 1x^(0.5)
Since 0.5 is not a whole number, the expression is not a polynomial.
e
Like in Part D, we will begin by simplifying the equation.
Now we can analyze the equation.
( any number)* x^(whole number)
↓
P(x)= 7x^3+ 49x^2- 35x^1- 525x^0
The equation is a polynomial.
f
Notice that an expression in the form 1a^b can be rewritten as a^(- b). With this information, we can rewrite the equation and determine if it is a polynomial.
( any number)* x^(whole number)
↓
f(x)= 1x^2+ 1(x^2+5)^(-1)
The equation is not a polynomial, because - 1 is not a whole number.
g
As long as each term we write follows the description from the exercise, we know that it will be a polynomial. Let's write two equations — one being a polynomial (I), and one that is not a polynomial (II).
