Marc's Responses in WyzAnt Answers

Hi Nish, and welcome to the wonderful world of factoring polynomials of degree 3 and above. That wild beast is a quartic because it's degree 4, and here's how I would try to tame it by factoring it into nice domesticated polynomials of degree 1 and 2. These are the the linear and quadratic polynomials that you've worked with in the past.

The first step is to see if all four terms have a factor in common. They don't because that would just be way too easy. In algebra 1, all four terms would have had something in common.

The second step is to see if factoring by grouping will work, so take a look at (x⁴−x_³)+(4x²−8). What factors do the first two terms have in common? An x³ term. And what factors do the second two terms have in common? Just a little 4. Factor those expressions one at a time, and you're left with:

x³(x−1)+4(x_²−2).

Uh oh. In algebra 1 those expressions in parentheses would have been identical, and now their not. Factoring by grouping doesn't work either.

A third step is to see if a convenient substitution can work. You've probably been asked to factor a quartic like x⁴+x²−6 in the past. By setting y=x² , this expression becomes y²+y−6, and now your task has become a lot simpler. But for the quartic in your problem, this algebra 2 method doesn't work either. You must be taking precalculus!

In precalculus, you're expected to know certain facts about polynomials. Every precalculus course requires that you know the Fundamental Theorem of Algebra. If you include multiplicity then that fourth degree polynomial will have four roots. If you were looking to solve for x by setting that polynomial equal to zero, this theorem would let you know to expect four solutions at most. You should know how the roots of a polynomial set equal to zero relate to the factors of the polynomial, and that relationship leads to another polynomial theorem.

According to the Linear Factors Theorem, you should be able to factor that thing all the way to the form (x−c₁)(x−c₂)(x−c₃)(x−c₄) where the four constants are complex numbers. Those four constants might or might not be real, and they might or might not be repeated more than once. This theorem provides some guidance for you with this problem. If we completely factor that quartic, we'll have a maximum of four linear terms.

Descartes' Rule of Signs, the Conjugate Roots Theorem for Complex Roots, the Conjugate Roots Theorem for Irrational Roots, and theorems about upper and lower limits for roots also might help you, and your course might expect you to know most or all of these. Descartes' Rule of Signs can be usefully applied to your expression, but we're going to use the Rational Roots Theorem.

According to the Rational Roots Theorem, for polynomials with integer coefficients, every rational root of a polynomial, p/q, must have p as a factor of the constant term of the polynomial and q as a factor of the leading coefficient of the polynomial. This means you should be able to factor a term (x−p/q) out of the polynomial. You can understand why the Rational Roots Theorem works by applying it to quadratics. We'll start out by knowing the factored form of a quadratic equation with two rational roots:

(x−p₁/q₁)(x−p₂/q₂)=0

Multiplying both sides by q₁q₂ gives us:

q₁(x−p₁/q₁)q₂(x−p₂/q₂)=0 or

(q₁x−p1)(q₂x−p₂)=0. After expanding by using FOIL, we get:

q₁q₂x²−(p₁q₁+p₂q₂)x+ p₁p₂=0.

Notice how the x² coefficient has factors that are the denominators of the rational roots and the units coefficient has factors that are the numerators.

Of course, the rational roots theorem only works for polynomials with rational roots. But if you're given a quartic in a precalculus course that can't be factored by another method, you can bet that it will have rational roots. If it didn't, you wouldn't be able to answer the question without using a calculator or computer.

For your problem, we're looking for roots of the form p/q where p is a factor of −8 and q is a factor of 1. The possibilities are ±1, ±2, ±4, and ±8. Test these possibilities one at a time, starting with the easiest numbers, 1 and −1. You can just plug these numbers into the expression and see if it evaluates to zero. However, there can be advantages to using synthetic division while testing rational roots because you can apply theorems about the upper and lower limits of roots to gain information about whether an incorrect attempt was too large or too small. Descartes's Rule of Signs can help you narrow down the possibilities, but let's move on the fact that −1 will work!

After you find this first root, you'll need to use synthetic division to determine the quotient when your original polynomial is divided by the factor corresponding to that root.

(x⁴−3x³+4x²−8)/(x+1) = x³−4x²+8x−8.

Congratulations! Now instead of factoring a quartic, you only need to factor a cubic. Try using the same theorems and tests again. Here's what your final answer should be:

http://www.wolframalpha.com/input/?i=Factor+x^4-3x^3%2B4x^2-8

Hi Diane,

When you heat ice at −10.0 °C until it becomes superheated steam at 129.0 °C, you are proceeding through several steps, and you have to take the time and make the calculations to account for each of those steps. In a general sense, adding thermal energy to H₂O or any other pure substance can do one of two things. First, it can raise the temperature of the substance. In that case, the heat exchanged is called sensible heat. Second, it can cause a phase transition in the substance. If this happens, the heat exchanged is called latent heat. For your question, you need to add together the contributions of all of these steps for 49.0 g of H₂O:

step 1) Find the sensible heat needed to increase the temperature of H₂O(s) (ice) by 10.0 °C from −10.0 °C to 0.0 °C.

step 2) Find the latent heat needed to melt ice at 0 °C into liquid water at 0 °C. This is called the latent heat of fusion.

step 3) Find the sensible heat needed to increase the temperature of H₂O(l) (liquid water) by 100.0 °C from 0.0 °C to 100.0 °C.

step 4) Find the latent heat needed to boil water at 100 °C into steam at 100 °C. This is called the latent heat of vaporization.

step 5) Find the sensible heat needed to increase the temperature of H2O(g) (water vapor) by 29.0 °C from 100.0 °C to 129.0 °C.

The three steps involving sensible heat (#1, #3, and #5) all use the formula:

Q = c_p · m · ΔT

but solid ice, liquid water, and gaseous water vapor each have their own specific heat, so you'll need three different values of c_p. Fortunately, these values are easy to locate for H₂O. Also be sure to use the correct change in temperature that corresponds to that particular step in the process.

The two steps involving latent heat (#2 and #4)  both use the formula:

Q =  m · L

The latent heat of fusion is different from the latent heat of vaporization, so you'll need two different values of L. Again, these values for the latent heat of fusion and vaporization are easy to find for water.

Add the contributions together for each substep, and you'll have the energy required for the entire process: Q_total = Q₁ + Q₂ + Q₃ + Q₄ + Q₅.

Before solving this problem, it's useful to know how the three Pythagorean identities arise from the Pythagorean formula.

If you have a right triangle, select one of the non-right angles as x, and label the three sides as adj, opp, and hyp to represent the lengths of the sides adjacent to x, opposite from x, and the length of the hypotenuse. Then Pythagorean theorem states:

opp² + adj² = hyp²

Divide both sides by hyp² to obtain:

(opp/hyp)² + (adj/hyp)² = 1

By substituting sin x = opp/hyp and cos x = adj/hyp , we get the most important trigonometric identity:

sin² x + cos² x = 1.

If you've learned circular trigonometry then I recommend the website: http://www.mathsisfun.com/geometry/unit-circle.html for an interactive demonstration, and the site: http://www.mathsisfun.com/algebra/trig-interactive-unit-circle.html has a great interactive unit circle.

I don't even bother memorizing the other two Pythagorean identities. I just derive them when I need them by dividing both sides of that most important trigonometric identity by cos² x or sin² x. Dividing by cos² x gives us:

(sin x / cos x)² + 1 = (1 / cos x)²

Using the definitions for tangent and secant in terms of sine and cosine yields:

tan² x + 1 = sec² x

Of course, if you like memorization better than derivation then memorize all three identities, but you still should understand how to derive them. Using this identity with your problem, we get:

(2tan x) / (1+tan² x) = 2tan x / sec² x

= 2(sin x / cos x) / (1/cos² x)

= 2(sin x / cos x)(cos² x)

= 2 sin x cos x

That's a fairly simple expression, isn't it? But using other trigonometric identities that you're expected to know, you should be able to think of a way to simplify it even further.

Hi Ej,

A probability model is just a set of all possible outcomes. For this carnival game, I'm assuming that you're not permitted to keep playing after you win, so the two general classes of outcomes are you either win within 4 throws and then stop playing or you don't win within 4 throws. The five specific outcomes are that you can win on your first, second, third, or fourth throw or not win after 4 throws. Each outcome can be assigned a specific probability of taking place, and you can calculate the expected values from these probabilities.

Each individual throw consists of a Bernoulli trial with a probability of success p = 0.1 and a probability of failure q = 1−p = 0.9. For each of the five specific outcomes, you'll need to calculate the probability of achieving that outcome, the number of darts thrown to achieve that outcome, and the winnings for that outcome. For example, in order to win on the third throw, you need to have failed on the first two throws and succeeded on the third:

P(win on the third throw)=q×q×p=q²p=(0.9)²(0.1)=0.081

Darts(win on the third throw)=3

Winnings(win on the third throw)=$100−$5×3=$100-$15=$85

This means that there's an 8.1% chance that you'll throw exactly 3 darts and an 8.1% chance you'll have winnings of exactly $85. Do the same thing for the other four specific outcomes. Don't forget to use "negative winnings" for the case of missing with all four darts. If your five calculated P probabilities add up to one then you've probably found them correctly.

Finally, calculate weighted averages for the number of darts and for your winnings. For each of the five possible outcomes, multiply the values for "Darts" by the probability that the related outcome will occur, and do the same for "Winnings." For the outcome above of winning on the third throw, you would have:

3×0.081 = 0.243 for number of darts

$85×0.081 = $6.885 for winnings

Add these products together for all five outcomes to obtain a weighted average. These weighted averages are your expected values.

∫₋₁³ |e^x−4| dx is the area between the function and the x-axis.

With f(x) = e^x−4, the function being integrated, g(x)=|e^x−4|, is equal to f(x) when f(x)≥0, and g(x) = −f(x) = 4−e^x when f(x)≤0. So we need to take the extra pre-calculus step of determining the values of x that make f(x) positive or negative.

Solving e^x−4 = 0 results in x = ln 4. This is the value of x where f(x)=0. Then a "sign-chart" can be used just like in other calculus skills to determine that f(x)≤0 on the interval [-1, ln 4] and f(x)≥0 on the interval [ln 4, 3]. Using the additivity of integration on intervals, we can get rid of those nasty absolute value brackets by splitting apart the definite integral:

∫₋₁³ |e^x−4| dx = ∫₋₁^{ln 4} (−f(x)) dx + ∫_{ln 4}³ f(x) dx

=∫₋₁^{ln 4} (4−e^x) dx + ∫_{ln 4}³ (e^x−4) dx

= (4x-e^x)¦₋₁^{ln 4} + (e^x−4x)¦_{ln 4}³

=(4 ln 4 − 4) − (−4 − 1/e) + (e³ − 12) − (4 − 4 ln 4)

= e³ + 1/e + 8 ln 4 − 16 = 15.544...

Wolfram Alpha gives: http://www.wolframalpha.com/input/?i=Area+between+e^x-4+and+y%3D0+on+[-1%2C3]

The original question was well-written because it used the word between. If the question had been "find the area under f(x)" then we'd need a definition for the area under a function when it dips below the x-axis. Functions can have a signed area of less than zero along these regions, but signed areas are a mathematical convention. Actual areas in the real world can never be negative, so the question "find the area under f(x)" always must assume that f(x)≥0. If f(x) is negative with a poorly-written question that uses the word "under" then an otherwise rational calculus class can deteriorate into a discussion about whether Mauna Loa is taller than Mount Everest, and calculus teachers hate when that happens.