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Exercises
1. Given that when x2−3x+5k is divided by x+4 the remainder is 9, nd the value of k by choosing Compute > Polynomials >
Divide and then choosing Compute > Solve > Exact.
2. De ne functions f (x) = x3+x ln xand g(x) = x+ex. Evaluate f (g(x)), g( f (x)), f (x)g(x), and f (x) + g(x).
3. Find the equation of the line passing through the two points (x1, y1), (x2, y2).
4. Find the equation of the line passing through the two points (2, 5), (3,−7).
5. Find the equation of the line passing through the two points (1, 2), (2, 4).
6. Find the slope of the line given by the equation sx + ty = c.
7. Factor the difference of powers xn− ynfor several values of n, and deduce a general formula.
8. Applying Factor to x2+(√
5− 3) x− 3√
5gives the factor-ization
x2+(√
5− 3) x− 3√
5 = (
x +√ 5
) (x− 3) showing that the system can factor some polynomials with irra-tional roots. However, applying Factor to x2− 3 and x3+ 3x2− 5x + 1does not do anything. Find a way to factor these poly-nomials.
9. Find the standard form for the equation of the circle x2− 6x + 18 + y2+ 10y = 0by “completing the square.” Determine the center and radius of this circle.
Solutions
1. Choose Compute > Polynomials > Divide to get x2− 3x + 5k
x + 4 = x +(5k + 28) x + 4 − 7
us, the remainder is 5k + 28. Solve the equation 5k + 28 = 9 to get k =−195.
2. De ning functions f (x) = x3+ x ln xand g(x) = x + exand evaluating gives
f (g(x)) = (x + ex)3+ ln (x + ex) (x + ex) g( f (x)) = ex ln x+x3+ x ln x + x3
f (x)g(x) = (x + ex)(
x ln x + x3) f (x) + g(x) = x + ex+ x ln x + x3
3. For any two distinct points (x1, y1)and (x2, y2)in the plane, there is a unique line ax + by + c = 0 through these two points.
Substituting these points in the equation for the line gives the two equations ax1+ by1+ c = 0and ax2+ by2+ c = 0. Set Tools > Preferences > Computation, Engine page, to Principal Value Only and Ignore Special Cases, then choose Compute >
Solve > Exact with the insert point in this system ax1+ by1+ c = 0 ax2+ by2+ c = 0
of linear equations, solving for the variables a, b.
Solution:
[
a = cy1− cy2
x1y2− x2y1, b =−cx1− cx2
x1y2−y1x2 ]
Consequently, the equation for the line is cy1− cy2
x1y2− x2y1x− cx1− cx2
x2y1− x1y2y + c = 0 or, clearing fractions and collecting coefficients,
(y1− y2) x− (x1− x2) y + (x1y2− y1x2) = 0
Note
An interesting method for nding the equation of a line through two speci ed points using determinants is described in a matrix algebra exercise on page 337.
4. For the points (2, 5), (3,−7), the system of equations is 2a + 5b + c = 0
3a− 7b + c = 0 Choose Compute > Solve > Exact to get
Solution: [
a =−1229c, b =−291c] . Consequently, the equation for the line is−1229cx−(
−291) cy + c = 0, or, clearing fractions and simplifying,
−12x + y + 29 = 0.
5. Since the point (0, 0) lies on the line, you do not get a unique so-lution to the system of equations for the pair a, b. us, choos-ing Solve + Exact and specifychoos-ing a, b for the variables gives no response. However, specifying a, c for Variable(s) to Solve for gives the solution
[a =−2b,c = 0]
us, the equation for the line is
−2bx + by = 0 or, dividing by b and applying Simplify,
(−2bx + by)1
b =−2x + y = 0
6. e slope-intercept form of the equation for a line is y = mx + b, where m is the slope and b the y-intercept. If a line is given as a linear equation in the form sx + ty = c, you can nd the slope by solving the equation for y. Expand the solution y =−sx−ct to get y =c
t −s
tx, revealing the slope to be−s t. 7. Apply Factor to several differences:
x2− y2= (x− y)(x + y) A er looking at only these few examples, you might nd it rea-sonable to conjecture that, for n odd,
xn− yn= (x− y)n
∑
−1k=0
xn−k−1yk
We leave the general conjecture for you. Experiment.
8. Using the clue from the example that the system will factor over roots that appear as coefficients, factor the product√
3( divide out the extraneous√
3to get x2−3 =( For the polynomial x3+3x2−5x+1, choose Compute > Poly-nomials > Roots to nd the roots:
[ 1,√
5− 2,−√ 5− 2]
. You can multiply by√
5to factor this polynomial:
√5( en, dividing out the extraneous factor of√
5, you have
9. To nd the center and radius of the circle x2− 6x + 18 + y2+ 10y = 0,
rst subtract the constant term 18 from both sides of the equa-tion to get
x2− 6x + 18 + y2+ 10y− 18 = 0 − 18
Simplify each side of the equation. is gives the equation x2− 6x + y2+ 10y =−18. Add parentheses to put the equation in
To complete the squares, add the square of one-half the coeffi-cient of x to both sides. Do the same for the coefficoeffi-cient of y.
(
Simplify the right side of the equation to get (x− 3)2+ (y + 5)2= 16.
You can read the solution to this problem from this form of the equation. e center of the circle is (3,−5) and the radius is
√16= 4.
Trigonometry 4
Since you are now studying geometry and trigonometry, I will give you a problem. A ship sails the ocean. It left Boston with a cargo of wool. It grosses 200 tons. It is bound for Le Havre. The main mast is broken, the cabin boy is on deck, there are 12 passengers aboard, the wind is blowing east-north-east, the clock points to a quarter past three in the afternoon. It is the month of May. How old is the captain?
Gustave Flaubert (1821–1880)