# Can someone check to see if my first three answers are correct and help me with the last two, please? Using this function: f(x)=-x^5-2x^4+x+2 find the leading coefficient: 2 the degree of the...

Can someone check to see if my first three answers are correct and help me with the last two, please?

Using this function: f(x)=-x^5-2x^4+x+2

the degree of the polynomial function: 5

describe the end behavior of this function's graph: f(x)->+∞, as x->-∞

Use the rational zero theorem to list the possible rational zeros of f.:

Use synthetic division with the remainder and factor theorems to find all real roots of f.:

lemjay | Certified Educator

`f(x) = -x^5-2x^4+x + 2`

In this function, the highest exponent of x is 5. So, the leading terms is -x^5.

(a) Take note that leading coefficient is the coefficient of the polynomial's leading term.

Hence, the leading coefficient is -1.

(b) For the degree of the polynomial function, you are correct. Thus, the degree is 5.

(c) For the end behavior of the graph of the function, we have to consider the sign of the leading coefficient and also if the degree is odd or even. In our function, the leading coefficient is negative and the degree is odd. So the ends of the graph goes in opposite direction. The left end goes up, while the right end goes down. Thus, the end behavior of the graph is:

`f(x) -> +oo` , as `x ->-oo`

`f(x)->-oo` , as `x ->+oo`

(d) To determine the possible rational zeros of the function, enumerate the factors of the constant and the leading coefficient.

Factors of constant 2: -2, -1, 1, 2

Factors of leading coefficient -1: -1,1

And divide the each factors of the constant by the factors of the leading coefficient.

`( -2)/(-1)=2`      ,    `(-1)/(-1)=1`     ,    `1/(-1)=-1`   ,     `2/(-1)=-2`

`(-2)/1=-2`   ,    `(-1)/1=-1` ,        `1/1=1`      ,       `2/1=2`

Thus, the possible rational roots are -2, -1, 1 and 2.

(e) Applying synthetic division, we would have:

-2 |   -1   -2    0     0     1      2
|          2    0     0     0     -2

--------------------------------------

-1    0     0     0      1       0

The remainder is zero, thus, -2 is a root of the function.

-1 |   -1   -2    0     0     1      2
|          1    1     -1    1     -2

--------------------------------------

-1   -1    1    -1      2       0

The remainder is zero, so, -1 is a root of the function.

1  |   -1    -2     0      0       1      2
|          -1    -3     -3     -3     -2

--------------------------------------

-1    -3    -3     -3     -2       0

The remainder is zero, hence, 1 is a root of the function.

2 |   -1    -2      0        0        1         2
|          -2     -8     -16     -32      -62

--------------------------------------------

-1    -4     -8     -16      -31      -60

The remainder is 60, so 2 is not a root of the function.

Therefore, the real roots of the function are -2, -1 and 1.