How can what we know about the far right and left of the graph of an odd degree polynomial be used to show that any polynomial of odd
degree has at least one root?

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct

3 qrs + 3/2 xy hope this : /

step-by-step explanation:

“a term can be a signed number, a variable, or a constant multiplied by a + sign or j sign. in , the terms are 5x , 3y and 8. when a term is made up of a constant multiplied by a variable or variables, that constant is called a coefficient.” i hope this you.