Exam Two Review

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This is a (lengthy) practice exam for Exam 2 on Polynomials.

NOTE: There may be random lines that look like \texttt{(SOME TEXT)} and maybe even some { } within that text. You can ignore everything that isn’t the text itself; ie the texttt bit and the braces, they are an artifact of how some randomization is being done that I haven’t had time to fix yet. If you literally delete the \texttt and the braces from what you see, and just keep the text itself, it will read exactly as intended. So just ignore all those texttt and braces whenever you see them.

Also note: Some of the below has randomized elements, some do not. Currently Xronos does not support randomized graphing (although we’re working on it!) so a lot of the graphing problems won’t randomize, but a surprising amount of the other problems will change values or text if you hit the green “another” button in the top right corner to get another version of this practice exam. This also means it may take some time for the entire test to render because we have to rely on public servers to do the randomization for us currently, so please be patient. If it takes more than 3-5 minutes to fill out the random values (meaning: if there are still spinning wheels of death going after 3-5 minutes) there is a problem. Try hitting the “another” button to see if it resolves itself. If it keeps doing this, please contact your instructor and let him know!

Simplify the following: $\dfrac {\sage {p1CNEc5}}{\sage {p1CNEc6}}$ .

$\sage {p1CNEans4}$ $\sage {p1CNEans3}$ $\sage {p1CNEans2}$ $\sage {p1CNEans1}$ $\sage {p1CNEans5}$

Simplify the following: $\sage {p2CNEc5} + \sage {p2CNEc6}$ .

$\sage {p2CNEans1a} + (\sage {p2CNEans1b}) \cdot i$ $\sage {p2CNEans2a} + (\sage {p2CNEans2b}) \cdot i$ $\sage {p2CNEans3a} + (\sage {p2CNEans3b}) \cdot i$ $\sage {p2CNEans5a} + (\sage {p2CNEans5b}) \cdot i + (\sage {p2CNEans5c})\cdot i$ $\sage {p2CNEans4a} + (\sage {p2CNEans4b}) \cdot i$

Consider the polynomial: $p(x) = (\sage {p1CNMf1}) + (\sage {p1CNMf2}) + (\sage {p1CNMf3}) + (\sage {p1CNMf4}) + (\sage {p1CNMf5})$ . If we know one of the roots is a non-real complex number, what is the best we can say about the number of real roots that $p(x)$ has?

$p(x)$ has at most $\sage {p1CNMpwr1t-1}$ real roots. $p(x)$ has at most $\sage {p1CNMpwr1t-2}$ real roots. $p(x)$ has at most $\sage {p1CNMpwr1t}$ real roots.Nothing; there is not enough information. $p(x)$ has at least $\sage {p1CNMpwr1t-1}$ real roots.

$p(x)$ has at most $1$ non-real roots. $p(x)$ has at most $2$ non-real roots. $p(x)$ has at least $1$ non-real roots. $p(x)$ has at least $2$ non-real roots.Nothing; there is not enough information.

Simplify the following expression fully: $\frac {\sage {p1CNHc7}}{(\sage {p1CNHc8})(\sage {p1CNHc9})}$

$\sage {p1CNHans2a} + \left (\sage {p1CNHans2b}\right )\cdot i$ $\sage {p1CNHans2b} + \left (\sage {p1CNHans2a}\right )\cdot i$ $\sage {p1CNHans1a} + \left (\sage {-p1CNHans1b}\right )\cdot i$ $\sage {p1CNHans1b} + \left (\sage {p1CNHans1a}\right )\cdot i$ $\sage {p1CNHans1a} + \left (\sage {p1CNHans1b}\right )\cdot i$

Simplify the following expression fully: $\frac {\sage {p2CNHc7}}{(\sage {p2CNHc8})\overline {(\sage {p2CNHc9})}}$

$\sage {p2CNHans1a} + \left (\sage {-p2CNHans1b}\right )\cdot i$ $\sage {p2CNHans2a} + \left (\sage {p2CNHans2b}\right )\cdot i$ $\sage {p2CNHans1a} + \left (\sage {p2CNHans1b}\right )\cdot i$ $\sage {p2CNHans2b} + \left (\sage {p2CNHans2a}\right )\cdot i$ $\sage {p2CNHans1b} + \left (\sage {p2CNHans1a}\right )\cdot i$

Consider the polynomial $p(x) = \sage {p1PDHf4}$ . What is the sum of the zeros of $p(x)$ ? Hint: $\sage {p1PDHcA}$ is a zero of $p(x)$ .

$\sage {p1PDHans2}$ $\sage {p1PDHans5}$ $\sage {p1PDHans3}$ $\sage {p1PDHans1}$ $\sage {p1PDHans4}$

Consider the polynomial $p(x) = \sage {p2PDHf4}$ . What is the sum of the zeros of $p(x)$ ? Hint: $\sage {x - p1PDHcA}$ is a factor of $p(x)$ .

$\sage {p2PDHans4}$ $\sage {p2PDHans2}$ $\sage {p2PDHans5}$ $\sage {p2PDHans1}$ $\sage {p2PDHans3}$

When using polynomial long division, if the remainder is zero, what does this mean?

The divisor is a factor of the polynomial.The divisor is not a factor of the polynomial.The polynomial is irreducible.There is not enough information.That the long division worked.

What degree factor can be used as a divisor with polynomial long division?

Only degree 1 or 2 polynomials can be used.Only linear polynomials (degree 1) can be used.Any degree polynomial can be used with long division.Any degree can be used, but it must have leading coefficient of 1.Only degree 1 polynomials with leading coefficient of 1 can be used.

What is the remainder when $\sage {p1PDMf7}$ is divided by $\sage {p1PDMf6}$ ?

$\sage {p1PDMf5}$ $\sage {p1PDMf2}$ $\sage {-p1PDMf5}$ $\sage {p1PDMf3}$ 0

What method is best for factoring out a complex-valued non-real zero of a polynomial?

Synthetic division, since a single zero means it’s a linear factor.Factor by Grouping, since we know the conjugate is also a zero.Polynomial long division, using the conjugate of the provided zero.It could be any of our methods; which is best depends on the specific polynomial.Quadratic Formula, since the provided zero is non-real.

Consider the polynomial $p(x) = \sage {p1RRTMf1(x)}$ . According to the rational root test, which of the following is NOT a possible zero of $p(x)$ ?

$x = \sage {p1RRTMans1}$ $x = \sage {p1RRTMans2}$ $x = \sage {p1RRTMans3}$ $x = \sage {p1RRTMans4}$ $x = \sage {p1RRTMans5}$

Consider the polynomial $p(x) = \sage {p1RRTMf1(x)}$ . Which of the following can not be a root of the polynomial?

$\sage {p2RRTMans1}$ $\sage {p2RRTMans5}$ $\sage {p2RRTMans3}$ $\sage {p2RRTMans4}$ $\sage {p2RRTMans2}$

Consider the polynomial $p(x) = \sage {p1RRTHf1}$ . Which of the following are true?

I:: $p(x)$ is degree 4.
II:: $p(x)$ has 4 real zeros.
III:: The sum of the zeros of $p(x)$ is $\sage {p1RRTHc3}$ .
IV:: The product of the zeros of $p(x)$ is 0.

All of these are correct.Only I and IV are correctOnly I, III, and IV are correct.Only I is correctOnly I, II and IV are correct

Can the Rational Root theorem be used to find zeros of a polynomial that has (some) complex-valued zeros?

Yes, but only if it has roots that are rational in addition to any complex-valued roots.Yes, the rational root theorem works on all polynomials.No, rational root theorem only works on polynomials with all rational roots.No, rational root theorem only works on polynomials with all real roots.None of the other responses are correct.

If you have tried every number listed by rational root theorem and found no roots (and made no mistakes), what does that tell you about the polynomial?

The polynomial will not be factorable using real numbers.The polynomial may be factorable and it may have rational zeros.Nothing; there is not enough information.The polynomial may be factorable, but it will have irrational zeros.All the zeros of the polynomial are complex-valued.

What is the minimal degree polynomial one could use the rational root theorem on?

$0$ $3$ There is no minimum. $2$ $1$

Which of the following is a zero of the polynomial $p(x) = \sage {p1SDHf5}$ ?

$\sage {p1SDHans3}$ $\sage {p1SDHans4}$ $\sage {p1SDHans5}$ $\sage {p1SDHans1}$ $\sage {p1SDHans2}$

Which of the following is a factor of the polynomial $p(x) = \sage {p2SDHf5}$ ?

$\sage {x - p2SDHans2}$ $\sage {x - p2SDHans3}$ $\sage {x - p2SDHans5}$ $\sage {x - p2SDHans4}$ $\sage {x - p2SDHans1}$

Which of the following is a zero of the polynomial $p(x) = \sage {p1SDMf5}$ ?

$\sage {p1SDMans5}$ $\sage {p1SDMans2}$ $\sage {p1SDMans1}$ $\sage {p1SDMans4}$ $\sage {p1SDMans3}$

Which of the following is a factor of the polynomial $p(x) = \sage {p2SDMf5}$ ?

$\sage {x - p2SDMans5}$ $\sage {x - p2SDMans1}$ $\sage {x - p2SDMans3}$ $\sage {x - p2SDMans2}$ $\sage {x - p2SDMans4}$

Can you use Synthetic Division to divide the polynomial $p(x) = \sage {p1SDEf1}$ by $d(x) = \sage {p1SDEdiv}$ ?

$\sage {p1SDEans3a} \sage {p1SDEdiv} \sage {p1SDEans3}$ $\sage {p1SDEans5a} \sage {p1SDEdiv} \sage {p1SDEans1}$ $\sage {p1SDEans1a} \sage {p1SDEdiv} \sage {p1SDEans1}$ $\sage {p1SDEans2a} \sage {p1SDEdiv} \sage {p1SDEans2}$ $\sage {p1SDEans4a} \sage {p1SDEdiv} \sage {p1SDEans4}$

Which of the following is a proper factorization of the polynomial $p(x) = \sage {p1CMEf3}$ (using real coefficients):

This is not factorable with real coefficients. $(\sage {p1CMEf1F})(\sage {p1CMEf2})$ $(\sage {p1CMEf1})(\sage {p1CMEf2F})$ $(\sage {p1CMEf1F})(\sage {p1CMEf2F})$ $(\sage {p1CMEf1})(\sage {p1CMEf2})$

When factoring using the coefficient method the original polynomial must...

have at least three terms.be a quadratic form with leading coefficient of 1.be a degree two polynomial (with any leading coefficient).have a non-zero constant term.have a non-zero coefficient for the degree one term.

Which of the following is a proper factorization (using real coefficients) of the polynomial $p(x) = \sage {p1ACMMf3}$

$(\sage {p1ACMMf1F})(\sage {p1ACMMf2F})$ $(\sage {p1ACMMf1F})(\sage {p1ACMMf2})$ $(\sage {p1ACMMf1})(\sage {p1ACMMf2F})$ $(\sage {p1ACMMf1})(\sage {p1ACMMf2})$ This is not factorable with real coefficients.

When using the AC-method, we are really...

using the original polynomial to factor an associate polynomial via the coefficient method.using an associated polynomial to figure out how to split the middle term to factor by grouping.using the original polynomial and guessing at values to figure out how to group the terms.using an associated polynomial which allows us to use coefficient method on the original polynomial.using an associate polynomial to fact via the quadratic formula.

Which of the following is a proper factorization (using real coefficients) of the polynomial $p(x) = \sage {p1ACMHf3}$

$(\sage {p1ACMHf1})(\sage {p1ACMHf2F})$ $(\sage {p1ACMHf1F})(\sage {p1ACMHf2F})$ This is not factorable with real coefficients. $(\sage {p1ACMHf1})(\sage {p1ACMHf2})$ $(\sage {p1ACMHf1F})(\sage {p1ACMHf2})$

Which of the following is a proper factorization (using real coefficients) of the polynomial $p(x) = \sage {p2ACMHf5}$

$(\sage {p2ACMHf1F})(\sage {p2ACMHf2F})(\sage {p2ACMHf3F})(\sage {p2ACMHf4F})$ This is not factorable with real coefficients. $(\sage {p2ACMHf1F})^2(\sage {p2ACMHf3F})^2$ $(\sage {p2ACMHf1})^2(\sage {p2ACMHf3})^2$ $(\sage {p2ACMHf1})(\sage {p2ACMHf2})(\sage {p2ACMHf3})(\sage {p2ACMHf4})$

Which of the following is a proper factorization (using real coefficients) of the polynomial $p(x) = \sage {p1ACMEf3}$

$(\sage {p1ACMEf1})(\sage {p1ACMEf2F})$ $(\sage {p1ACMEf1})(\sage {p1ACMEf2})$ $(\sage {p1ACMEf1F})(\sage {p1ACMEf2F})$ This is not factorable with real coefficients. $(\sage {p1ACMEf1F})(\sage {p1ACMEf2})$

The AC method can be used on...

any polynomial that can be appropriately grouped for factoring.any polynomial with exactly three terms.any polynomial with three or fewer terms.any polynomial with any number of terms.any quadratic form

Which of the following is a proper factorization of the polynomial $p(x) = \sage {p1CMHf3}$ (using real coefficients):

$(\sage {p1CMHf1})(\sage {p1CMHf2})$ $(\sage {p1CMHf1F})(\sage {p1CMHf2F})$ $(\sage {p1CMHf1F})(\sage {p1CMHf2})$ $(\sage {p1CMHf1})(\sage {p1CMHf2F})$ This is not factorable with real coefficients.

Consider the polynomial $p(x) = \sage {p2CMHf1}$ . What is the sum of the zeros of $p(x)$ after is has been shifted to the right by $\sage {p2CMHc3}$ ?

$\sage {p2CMHans1}$ The answer cannot be determined. $\sage {p2CMHans2}$ $\sage {p2CMHans3}$ $\sage {p2CMHans4}$

What number would you add and subtract to the polynomial $p(x) = \sage {p1CtSEf1}$ in order to complete the square?

$\sage {p1CtSEans1}$ $\sage {p1CtSEans4}$ $\sage {p1CtSEans3}$ $\sage {p1CtSEans2}$ There is not enough information.

When using the technique of Completing the Square, you will end up with...

a perfect square, always.a binomial to a power with (possibly many more) terms added or subtracted.some perfect power (not necessarily square) plus or minus a constant.a perfect square, but you may have to add or subtract a constant to it.None of these options are correct.

Which of the following is a proper factorization of the polynomial $p(x) = \sage {p1CMMf3}$ (using real coefficients):

$(\sage {p1CMMf1})(\sage {p1CMMf2})$ $(\sage {p1CMMf1})(\sage {p1CMMf2F})$ $(\sage {p1CMMf1F})(\sage {p1CMMf2F})$ This is not factorable with real coefficients. $(\sage {p1CMMf1F})(\sage {p1CMMf2})$

Consider the polynomial $f(x) = \sage {p2CMMf1}$ . Which of the following statements are true?

I:: $f(x)$ has 2 roots but no real zeros.
II:: $f(x)$ has 2 roots and 2 real zeros.
III:: $f(x)$ is degree 2 which is why it must have 2 real zeros.
IV:: $f(x)$ is degree 2 but has no real zeros.

Only II and III are correctOnly I is correctOnly II is correct.None of these are correct.Only I and IV are correct

Complete the square on the expression $\sage {p1CtSHf1}$ to get it into the form $A(x+B)^2 + C$ . Which of the following is correct:

$A \in [\sage {p1CtSHansAf1},\sage {p1CtSHansAe1}]$ , $B\in [\sage {p1CtSHansBf2},\sage {p1CtSHansBe2}]$ , and $C\in [\sage {p1CtSHansCf1},\sage {p1CtSHansCe1}]$ $A \in [\sage {p1CtSHansAf1},\sage {p1CtSHansAe1}]$ , $B\in [\sage {p1CtSHansBf1},\sage {p1CtSHansBe1}]$ , and $C\in [\sage {p1CtSHansCf1},\sage {p1CtSHansCe1}]$ $A \in [\sage {p1CtSHansAf2},\sage {p1CtSHansAe2}]$ , $B\in [\sage {p1CtSHansBf2},\sage {p1CtSHansBe2}]$ , and $C\in [\sage {p1CtSHansCf2},\sage {p1CtSHansCe2}]$ $A \in [\sage {p1CtSHansAf1},\sage {p1CtSHansAe1}]$ , $B\in [\sage {p1CtSHansBf1},\sage {p1CtSHansBe1}]$ , and $C\in [\sage {p1CtSHansCf2},\sage {p1CtSHansCe2}]$ $A \in [\sage {p1CtSHansAf2},\sage {p1CtSHansAe2}]$ , $B\in [\sage {p1CtSHansBf1},\sage {p1CtSHansBe1}]$ , and $C\in [\sage {p1CtSHansCf1},\sage {p1CtSHansCe1}]$

Consider the polynomial $p(x) = \sage {p1CtSMf1} + C$ . What would the value of $C$ need to be in order for $p(x)$ to be a perfect square? (Hint: Try to complete the square and see what number you must add and subtract)

$\sage {p1CtSMans1}$ There is not enough information. $\sage {p1CtSMans4}$ $\sage {p1CtSMans2}$ $\sage {p1CtSMans3}$

Complete the square on the following quadratic form. (Note: You do not need to fully factor, just complete the square.) $\sage {p2CtSMf1}$

$\left (\sage {p2CtSMans1a}\right )^2 + \left (\sage {p2CtSMans1b}\right )$ $\left (\sage {p2CtSMans2a}\right )^{\sage {p2CtSMpwr1}} + \left (\sage {p2CtSMans1b}\right )$ $\left (\sage {p2CtSMans1a}\right )^2 + \left (\sage {p2CtSMans3b}\right )$ $\left (\sage {p2CtSMans1a}\right )^{\sage {p2CtSMpwr1}} + \left (\sage {p2CtSMans2b}\right )$ $\left (\sage {p2CtSMans2a}\right )^2 + \left (\sage {p2CtSMans1b}\right )$

Which of the following is a proper factorization of the polynomial $p(x) = \sage {p1GMEf3}$ (using real coefficients):

This is not factorable with real coefficients. $(\sage {p1GMEf1F})(\sage {x-p1GMEc2})(\sage {x+p1GMEc2})$ $(\sage {p1GMEf1})(\sage {p1GMEf2})$ $(\sage {p1GMEf1F})(\sage {p1GMEf2})$ $(\sage {p1GMEf1})(\sage {p1GMEf2F})$

When factoring by grouping, you need to ensure that...

The original polynomial is at least degree 3.Each grouping has a linear polynomial remainder after factoring out the GCD.There are exactly 4 terms in the original polynomial.The leading coefficient is one.Each group has the same remainder after factoring out the GCD

Which of the following is a proper full factorization (using real coefficients) of the polynomial: $p(x) = \sage {p1GMHf3}$

$(\sage {p1GMHf1F})(\sage {p1GMHf2})$ $(\sage {p1GMHf1})(\sage {p1GMHc3*x-p1GMHc2})(\sage {p1GMHc3*x+p1GMHc2})$ $(\sage {p1GMHf1})(\sage {p1GMHans1*x-p1GMHans2})(\sage {p1GMHans1*x+p1GMHans2})$ This is not factorable with real coefficients. $(\sage {p1GMHf1})(\sage {p1GMHf2})$

Which of the following is a proper full factorization (using real coefficients) of the polynomial: $p(x) = \sage {p2GMHf3}$ Hint: Use factor by grouping.

$(\sage {p2GMHf1})^2(\sage {x-p2GMHc2})(\sage {p2GMHans1})$ $(\sage {p2GMHf1}F)(\sage {x-p2GMHc2})(\sage {p2GMHans1})$ $(\sage {p2GMHf1})^2(\sage {p2GMHf2})$ This is not factorable with real coefficients. $(\sage {p2GMHf1F})(\sage {p2GMHf2})$

Which of the following is a proper full factorization of the polynomial $p(x) = \sage {p1GMMf3}$ (using real coefficients):

$(\sage {p1GMMf1})(\sage {p1GMMf2F})$ This is not factorable with real coefficients. $(\sage {p1GMMf1})(\sage {p1GMMf2})$ $(\sage {p1GMMf1})(\sage {x-p1GMMc2})(\sage {x+p1GMMc2})$ $(\sage {p1GMMf1F})(\sage {p1GMMf2})$

Consider the polynomial $p(x) = \sage {p2GMMf1}$ . What is the sum of the zeros of $p(x)$ ?

$\sage {p2GMMans5}$ $\sage {p2GMMans2}$ $\sage {p2GMMans4}$ $\sage {p2GMMans1}$ $\sage {p2GMMans3}$

Which of the following is a proper factorization of the polynomial $p(x) = \sage {p1SFEf1}$ (using real coefficients):

This is not factorable with real coefficients. $(\sage {p1SFEans1F})(\sage {p1SFEans2F})$ $(\sage {p1SFEans1F})(\sage {p1SFEans2})$ $(\sage {p1SFEans1})(\sage {p1SFEans2})$ $(\sage {p1SFEans1FF})(\sage {p1SFEans2})$

Which of the following is not a special form for factoring polynomials with real coefficients?

These are all special forms.Sum of CubesSum of SquaresQuadratic FormDifference of Squares

Which of the following is a proper factorization of the polynomial $p(x) = \sage {p1SFHf1}$ (using real coefficients):

$\sage {p1SFHans2}$ $\sage {p1SFHans4}$ This is not factorable with real coefficients. $\sage {p1SFHans1}$ $\sage {p1SFHans3}$

Factor the polynomial $p(x) = \sage {p2SFHf3}$ using real coefficients.

$(\sage {p2SFHf1})(\sage {p2SFHf2})$ $p(x)$ cannot be factored further. $(\sage {p2SFHans1a})(\sage {p2SFHans1b})(\sage {p2SFHans1c})$ $\sage {p2SFHans2a}\sage {p2SFHans2b}(\sage {p2SFHans3a})$ $(\sage {p2SFHans1a})(\sage {p2SFHans1b})\sage {p2SFHans2a}\sage {p2SFHans2b}$

Which of the following is a proper factorization of the polynomial $p(x) = \sage {p1SFMf1}$ (using real coefficients):

$\sage {p1SFMans4}$ $\sage {p1SFMans2}$ $\sage {p1SFMans3}$ This is not factorable with real coefficients. $\sage {p1SFMans1}$

Factor the polynomial $p(x) = \sage {p2SFMf3}$ using real coefficients.

$(\sage {p2SFMans1a})(\sage {p2SFMans1b})(\sage {p2SFMans1c})$ $(\sage {p2SFMans1a})(\sage {p2SFMans1b})(\sage {p2SFMans2a})(\sage {p2SFMans2b})$ $(\sage {p2SFMans2a})(\sage {p2SFMans2b})(\sage {p2SFMans3a})$ $(\sage {p2SFMf1})(\sage {p2SFMf2})$ $p(x)$ cannot be factored further.

Given a polynomial of degree $\sage {p1EEpwr1}$ what can be said about the absolute extrema it has?

It has no absolute extrema.It must have an absolute minimum.There is not enough information to tell.It must have an absolute maximum.It must have at least one absolute extrema; but we don’t know if it is a max or min.

Given a polynomial of degree $\sage {p1EEpwr1+1}$ what can be said about the absolute extrema it has?

There is not enough information to tell.It must have at least one absolute extrema; but we don’t know if it is a max or min.It has no absolute extrema.It must have an absolute maximum.It must have an absolute minimum.

Given a polynomial with leading term $\sage {p1EHf1}$ what can be said about the number of absolute and relative extrema it has?

It has 1 absolute extrema and an even number of relative extrema.It has no absolute extrema and an odd number of relative extrema.It has 1 absolute extrema, but there is not enough information for relative extrema.It has 1 absolute extrema and an odd number of relative extrema.It has no absolute extrema and an even number of relative extrema.

Given a polynomial with leading term $\sage {p1EHf2}$ what can be said about the number of absolute and relative extrema it has?

It has no absolute extrema and an even (possibly 0) number of relative extrema.It has no absolute extrema, but there is not enough information for relative extrema.It has 1 absolute extrema and an even (possibly 0) number of relative extrema.It has 1 absolute extrema and an odd number of relative extrema.It has no absolute extrema and an odd number of relative extrema.

Consider the monomial $p(x) = \sage {p1CEf1}$ . Does $p(x)$ have any absolute extrema, if so which type?

No, $p(x)$ does not have any absolute extrema, but it does have relative extrema.Yes, $p(x)$ has both an absolute maximum and minimum. $\sage {p1CEans1a} p(x) \sage {p1CEans1b}$ $\sage {p1CEans3a} p(x) \sage {p1CEans3b}$ $\sage {p1CEans2a} p(x) \sage {p1CEans2b}$

Consider the monomial $p(x) = \sage {p2CEf1}$ . Is $p(x)$ increasing, or decreasing to the right of the origin?

$p(x)$ is $\sage {p2CEansA}$ $p(x)$ is neither increasing nor decreasing over that interval.There is not enough information to tell. $p(x)$ is $\sage {p2CEansB}$ $p(x)$ is increasing over some of the interval and decreasing over other parts.

Given a polynomial with leading term $\sage {p1EMf1}$ what can be said about the absolute extrema it has?

It must have an absolute $\sage {p1EMans1}$ .It has no absolute extrema.It must have at least one absolute extrema; but we don’t know if it is a max or min.It must have an absolute $\sage {p1EMans2}$ .There is not enough information to tell.

Given a polynomial with leading term $\sage {p1EMf2}$ what can be said about the absolute extrema it has?

It must have an absolute minimum.It must have at least one absolute extrema; but we don’t know if it is a max or min.There is not enough information to tell.It has no absolute extrema.It must have an absolute maximum.

Consider the following graph:

Based on the number of extrema, what is the smallest degree possible for the leading term of $f$ ?

6534

Consider the following graph:

Within which of the following segments is the function increasing and concave down?

Between a and bBetween d and eBetween b and cBetween c and dTo the right of e

What is the leading coefficient of $(\sage {p1idFTAEf1}) + (\sage {p1idFTAEf2}) + (\sage {p1idFTAEf3}) + (\sage {p1idFTAEf4}) + (\sage {p1idFTAEf5})$ ?

$\sage {p1idFTAEc3t}$ $\sage {p1idFTAEc5t}$ $\sage {p1idFTAEc2t}$ $\sage {p1idFTAEc4t}$ $\sage {p1idFTAEc1t}$

Which of the following equations are polynomials?

(a): $f(x) = x^2 + 1$
(b): $g(x) = \frac {1}{3}x^3 + 7$
(c): $h(x) = 7$
(d): $k(x) =x^3 + 3x^{\frac {1}{3}} + 1$

I, II, and III Only.IV Only. All of them are Polynomials.I Only.I and II Only.

Which of the following is the graph of $f(x) = (-x+2)^3 - 1$ ?

Consider the polynomial $p(x) = \sage {p1idFTAMf4}$ . How many zeros does this polynomial have counting multiplicity?

$\sage {p1idFTAMans3}$ $\sage {p1idFTAMans4}$ $\sage {p1idFTAMans5}$ $\sage {p1idFTAMans1}$ $\sage {p1idFTAMans2}$

How many zeros (and what type) does the polynomial $\sage {p2idFTAMf1}$ have?

Since it is degree two it has two zeros, one of which is complex, the other one is real.According to FTA it has two real zeros and two complex zeros.It has no real zeros, but according to FTA is has 2 complex zeros.It has no real zeros, so it must have complex zeros; but we cannot determine how many.According to FTA it has two real zeros and no complex zeros.