Solving Polynomial Equations

7 = 1/5t^3-2t^2+21/5t+7 Next, group all the terms on the same side of the equation. This can be done by subtracting 7 from both sides.

To get rid of the fractions, multiply both sides of the equation by 5.

Note that all the terms on the left-hand side have t as a common factor. Thus, it can be factored out. t^3-10t^2+21t = 0 ⇓ t(t^2-10t+21) = 0 By the Zero Product Property, either t=0 or the quadratic polynomial between the parentheses is equal to 0. t(t^2-10t+21) = 0 ↙ ↘ t=0 t^2-10t+21=0 To find the solutions to the quadratic equation, the Quadratic Formula can be used.

The solutions to the quadratic equation are t=7 and t=3. Since it was already stated that t=0 is a solution, and now two more solutions were found, the initial polynomial equation has three different solutions. 1/5t^3-2t^2+21/5t+7 = 7 ⇓ t_1 = 0 t_2 = 3 t_3 = 7 Notice that Tearrik is interested in the moments in which the car will be 7 meters above the ground after the ride starts. Thus, t_1=0 can be discarded. Consequently, based on the context, the car will be 7 meters above the ground twice — the first time will occur 3 seconds after the ride starts and the second time will occur 7 seconds after the ride starts.

8.8 = - 15t^3+ 65t^2- 15t+10 To get rid of the fractions and the decimal number, multiply both sides by 5. After that, group all the terms on the same side of the equation.

To solve the resulting polynomial equation, factor the polynomial on the left-hand side. This can be done by grouping.

According to the Zero Product Property, either the linear factor or the quadratic factor is equal to 0. (t-6)(t^2+1) = 0 ↙ ↘ t-6=0 t^2+1=0 Solving the left-hand side equation for t gives 6 as a solution. Because the quadratic equation is a sum of two squares, it can be factored using the following formula. a^2+b^2 = (a+bi)(a-bi) Here, i is the imaginary unit. For that reason, the quadratic equation can be rewritten as follows. t^2+1 = (t+i)(t-i) The solutions to the quadratic equation are i and -i — both imaginary numbers. While they are solutions to the polynomial equation, they do not make sense in the context of the ride. For this reason, during the first few seconds of the ride, the car will be 8.8 meters above the ground only once, and it will happen 6 seconds after the ride starts.

48x^9 &= 2* 2* 2* 2* 3 * x^7* x^2 [0.25em] 108x^7 &= 2* 2* 3* 3* 3 * x^7 Comparing the factors of each term, the GCF of the two terms is 2* 2* 3* x^7 which, when the coefficients are multiplied, is the same as 12x^7. Next, rewrite each term of the polynomial in terms of the GCF. 48x^9 &= 12x^7* 4x^2 [0.25em] 108x^7 &= 12x^7 * 9 Now, substitute these expressions into the polynomial equation and factor out the GCF.

The next task is factoring the quadratic expression. This can be done using the Quadratic Formula. However, notice that the terms in the parenthesis can be expressed differently. In fact, 4x^2 can be expressed as (2x)^2, and 9 as 3^2. Thus, the quadratic expression can be rewritten as a difference of two squares. 12x^7( 4x^2 - 9) = 0 ⇕ 12x^7 ((2x)^2 - 3^2) = 0 The difference of squares can be factored as the sum of the bases multiplied by the difference of the bases. 12x^7 (( 2x)^2 - 3^2) = 0 ⇕ 12x^7( 2x+ 3)( 2x- 3) = 0 Notice that each of the factors cannot be factored further.

90x^6 &= 2* 3* 3 * 5 * x^5* x [0.25em] 75x^5 &= 3* 5 * 5 * x^5 [0.25em] 27x^7 &= 3* 3* 3* x^5* x^2 Taking the underlined factors into consideration, the GCF of the polynomial is 3 x^5. Now, rewrite each term of the polynomial in terms of the GCF. 90x^6 &= 3 x^5* 30x 75x^5 &= 3 x^5* 25 27x^7 &= 3 x^5* 9x^2 Next, substitute these expressions into the given polynomial and factor out the GCF.

To factor the quadratic expression between the parentheses, first rearrange its terms so that it is in standard form. 3 x^5( 30x + 25 + 9x^2) = 0 ⇕ 3 x^5( 9x^2 + 30x + 25) = 0 Instead of using the Quadratic Formula, notice that 9x^2 equals (3x)^2, 30x equals 2* 3x* 5, and 25 equals 5^2. 3 x^5( 9x^2 + 30x + 25) = 0 ⇕ 3 x^5( (3x)^2 + 2* 3x * 5 + 5^2) = 0 As can be seen, the quadratic expression has the form of a perfect square trinomial. Therefore, it can be rewritten as the square of the sum of 3x and 5. 3 x^5( ( 3x)^2 + 2* 3x * 5 + 5^2) = 0 ⇕ 3 x^5( 3x + 5)^2 = 0 The given polynomial has been factored as the product of unfactorable factors with integer coefficients.

V_(Rubik) = y^3 V_(Die) = 2^3 ↘ ↙ V = y^3+2^3 Notice that V has the form of the sum of two cubes. Therefore, it can be factored using the following formula. a^3+b^3 = (a+b)(a^2 - ab + b^2) In this case, a=y and b=2. y^3+2^3 &= (y+2)(y^2 - y* 2 + 2^2) &⇓ y^3+2^3 &= (y+2)(y^2 - 2y + 4) Now, before factoring the quadratic expression, its discriminant will be examined.

Since the discriminant is negative, the quadratic expression has no real solutions. Therefore, it cannot be factored using integer coefficients. Consequently, the factored form of V looks as follows. V = (y+2)(y^2 - 2y + 4)

Using the fact that 8 is equal to 2^3 and 343 is equal to 7^3, the left-hand side expression can be rewritten.

The left-hand side expression is a sum of cubes and can be factored using the same formula used in Part A. In this case, a=2x and b=7.

So far, one solution to the equation has been found — namely, x=- 72. Next, use the Quadratic Formula to solve the second equation.

Using the positive and negative signs, the two solutions to the quadratic equation are obtained. All the solution are written in the following table.

Now that all the solutions are known, their sum can be calculated.

The sum of the solutions to the given equation is 0.

V_(Large) &= x^3 V_(Small) &= 3^3 = 27 Since tío Angelito has to remove the small cube from the large one, the volume of the resulting object is equal to the volume of the large cube minus the volume of the small one. V_(Object) &= V_(Large) - V_(Small) &⇓ V_(Object) &= x^3-27 On the other hand, tío Angelito said that this object should have a volume of 37 cubic inches. Therefore, equate the previous expression to 37. x^3 - 27 = 37 This equation models the situation described by tío Angelito.

Notice that 64 can be written as 4^3. Doing this will allow seeing the equation as a difference of two cubes.

By the Zero Product Property, either the linear factor or the quadratic factor is equal to zero. (x-4)(x^2 + 4x + 16) = 0 ↙ ↘ x-4=0 x^2 + 4x + 16 = 0 Solving the left-hand side equation gives x=4 as a solution. Next, use the Quadratic Formula to solve the right-hand side equation.

The solutions to the quadratic equation are x=-2-2sqrt(3)i and x=-2+2sqrt(3)i.

The solutions to the previous equation represent the moments when tío Angelito was 90 meters underwater. These solutions can be found graphically. Begin by considering the function defined by the polynomial on the left. f(t) = t^4 - 25t^3 + 210t^2 -680t+704 The next step is to graph y=f(t) which can be done using a table of values. Since t represents time, negative values do not make sense. Also, only values within the domain of D(t) will be considered.

Use the values in the table to make a graph of y=f(t).

The t-intercepts represent the solutions to the equation t^4- 25t^3+ 210t^2- 680t+ 704=0. As the graph shows, there are four t-intercepts.

Given those four intercepts, it can be determined that there were four moments in which tío Angelito was exactly 90 meters underwater — namely, at 2, 4, 8, and 11 minutes after he began his dive.

D(t) = t^4-25t^3+210t^2−680t+614 ⇓ -200 = t^4 - 25t^3 + 210t^2 - 680t + 614 If this equation has at least one real solution, it means that tío Angelito reached 200 meters underwater. As in Part A, in this context, imaginary solutions do not make sense, the equation will be solved graphically, and solving for the solution can begin by grouping all the terms on the left-hand side.

Next, consider the function defined by the polynomial expression. g(t) = t^4 - 25t^3 + 210t^2 - 680t + 814 To draw the graph of y=g(t), a table of values can be used. Again, only positive values of t within the domain of D(t) are going to be considered.

At first glance, it can be seen that all the values are positive, which means that the graph does not intersect the t-axis. Nevertheless, it is best to graph the function to gain a better understanding.

As expected, the graph does not have t-intercepts. Therefore, the equation t^4- 25t^3+ 210t^2- 680t+ 814 = 0 has no real solutions. This implies that tío Angelito never reached 200 meters underwater.

D(t) = -150 ⇓ -150 =t^4 - 25t^3 + 210t^2 - 680t + 614 As in Part B, if the previous equation has at least one real solution, it means that tío Angelito, indeed, reached at least 150 meters below sea level. To determine this, begin by grouping all the terms on the left-hand side.

Consider the function defined by the polynomial on the left. h(t) = t^4 - 25t^3 + 210t^2 - 680t + 764 The graph of y=h(t) can be made by making a table of values. Once more, only positive values of t within the domain of D(t) will be considered.

By using the points in the table, graph y=h(t).

The graph of y=h(t) intersects the t-axis twice, which means that tío Angelito reached 150 meters below sea level twice — once going down and the other going up.

It can be seen that the t-intercepts occur around 9 and around 10.5. The difference between these approximations represents the amount of time tío Angelito was below the 150-meter mark. 10.5 - 9 ≈ 1.5 Given this value, it can be determined that tío Angelito was under the 150-meter mark for between 1 and 2 minutes long.