# Honors Algebra 2: Assignment 33 -- Unit 4 Review

• Class: Honors Algebra 2
• Author: Peter Atlas
• Text: Algebra and Trigonometry: Structure and Method, Brown

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1. How many terms are there in a quadratic polynomial?
2. Solution The $$ANSWER$$ is not 3. There is no $$ANSWER$$ to this question. Quadratic refers to the degree of the polynomial. $$x^2 = 0$$ is a quadratic equation where the quadratic only has one term.
3. Simplify: $$\displaystyle \left( -3x^2 \right) ^3 \cdot \left( -\frac{2}{5x^2} \right) ^4$$
4. Solution $$\displaystyle \frac{-432}{625x^{2}}$$
5. Given $$f(x)$$ is a cubic function, give two other words for the solutions to $$f(x) = 0$$.
6. Solution "roots of the equation" and "zeros of the function"
7. Multiply: $$(x + 1)(x - 3)(x - 1)$$
8. Solution Multiply the difference of squares first: $$(x^2 - 1)(x - 3) = x^3 -3x^2 - x + 3$$
9. Factor: $$6x^2 - 7x - 20$$
10. Solution $$(3x + 4)(2x - 5)$$
11. Factor: $$-3x^3+ 12x$$
12. Solution $$-3x(x + 2)(x - 2)$$
13. Factor: $$3(x + 2)^2 - 11(x + 2) - 4$$
14. Solution $$(3x + 7)(x - 2)$$
15. Factor: $$x^2 + y^2 + 2xy - 4$$
16. Solution $$(x + y + 2)(x + y - 2)$$
17. Factor: $$p^2 - (p + q)^2$$
18. Solution $$(p + (p + q))(p - (p + q) = -q(2p + q)$$
19. Factor: $$3x^4 - 243$$
20. Solution $$3(x^2 + 9)(x + 3)(x - 3)$$
21. Factor: $$x^{2n + 1} + 3x^{n + 1} + 2x$$ given $$n$$ is a positive integer
22. Solution Factor out an $$x$$: $$x(x^{2n} + 3x^n + 2)$$. Now let $$u = x^n$$. $$x(u^2 + 3u + 2) = x(u + 2)(u + 1) = x(x^n + 2)(x^n + 1)$$
23. Solve analytically: $$3m^2 - 22m = 16$$
24. Solution $$\displaystyle m = \left\{ -\frac{2}{3}, 8 \right\}$$
25. Solve analytically: $$(5x + 3)(x - 4) = 11x$$
26. Solution $$\displaystyle x = \left\{ -\frac{2}{5}, 6 \right\}$$
27. A rectangle that is 5 cm longer than it is wide has an area of 84 $$\text{cm}^2$$. What is its width?
28. Solution $$7$$ cm
29. Solve and graph: $$x^3 + 10x^2 +25x < 0$$
30. Solution $$\left\{ \left. x \right| x < 0 \land x \ne -5 \right\}$$ The graph has an open ciricle at 0, extends to the left, and has a hole at -5.
31. Factor completely given $$m$$ is a positive integer: $$x^{2m + 7} + 2x^{m + 4} - 24x$$
32. Solution $$x(x^{2m + 6} + 2x^{m + 3} - 24)$$. Let $$u = x^{m + 3}$$. $$x(u^{2} + 2u - 24) = x(u + 6)(u - 4) = x(x^{m + 3} + 6)(x^{m + 3} - 4)$$
33. Solve analytically: $$-2x^{7} + 50x^{5} + 2x^3 - 50x = 0$$
34. Solution Factor out $$-2x$$: $$-2x(x^{6} - 25x^{4} - x^{2} + 25) = 0$$. Factor the 6th degree polynomial by grouping: $$x^{4}(x^{2} - 25) - 1(x^{2} -25) = (x^{4} - 1)(x^{2} -25)$$. So the equation becomes: $$-2x(x^{4} - 1)(x^{2} -25) = 0$$. Factor the differences of squares: $$-2x(x^{2} + 1)(x^{2} - 1)(x + 5)(x - 5) = 0$$. Factor the remaining difference of squares:$$-2x(x^{2} + 1)(x + 1)(x - 1)(x + 5)(x - 5) = 0$$. $$x^{2} + 1$$ is never 0. The solution is $$x = \{ 0, \pm 1, \pm 5 \}$$
35. Multiply: $$(a + b + c)^2$$. Look at the solution before you go on to the next problem.
36. Solution $$a^{2} + b^{2} + c^{2} + 2ab + 2ac + 2bc$$. Hmmm. It seems there's a pattern here. Each term gets squared, and you get twice the linear combination of each of the terms. $$(a + b)^{2} = a^{2} + b^{2} + 2ab. (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2ac + 2bc$$. I bet $$(a + b + c + d)^{2}$$ would be $$a^{2} + b^{2} + c^{2} + d^{2} + 2ab + 2ac + 2ad + 2bd + 2cd.$$
37. $$Multiply: (a - b - c)^2$$.
38. Solution This would be the same as the last problem, but with $$-b$$ substituted for $$b$$, and $$-c$$ substituted for $$c$$. $$a^{2} + b^{2} + c^{2} - 2ab - 2ac + 2bc$$. That last term is still positive, since the product of $$-b$$ and $$-c$$ is still $$+bc$$.
39. Factor: $$x^4 + y^4 + z^4 - 2 \left( x^2y^2 + y^2z^2 + x^2z^2 \right)$$
40. Solution Distribute the -2: $$x^{4} + y^{4} + z^{4} - 2x^{2}y^{2} - 2x^{2}z^{2} - 2y^{2}z^{2}$$. If only that last term were positive, this would be the square of $$x^{2} - y^{2} - z^{2}$$. (If you don't see that, look more carefully at the last two problems.) So I'll add and subtract $$4y^{2}z^{2}$$ to make that term positive: $$x^{4} + y^{4} + z^{4} - 2x^{2}y^{2} - 2x^{2}z^{2} + 2y^{2}z^{2} - 4y^{2}z^{2}$$. We can now factor the first 6 terms: $$\left( x^{2} - y^{2} - z^{2} \right) ^{2} - 4y^{2}z^{2}$$. And now we have a difference of squares! $$\left( x^{2} - y^{2} - z^{2} + 2yz \right) \left(x^{2} - y^{2} - z^{2} - 2yz \right)$$. Now each of those are factorable. Consider $$x^{2} - y^{2} - z^{2} + 2yz$$. Rearrange the terms to bring out the perfect square: $$x^{2} - \left( y^{2} - 2yz + z^{2} \right)$$. Factor that: $$x^{2} - (y - z)^{2}$$. And that, again, is a difference of squares: $$\left( x + (y - z) \right) \left( x - (y - z) \right) = (x + y - z)(x - y + z)$$. We can play the same game with the second factor. The final answer is: $$(x + y - z)(x - y + z)(x - y - z)(x + y + z)$$.
41. A very difficult question (a true honors challenge problem). If you like challenging problems, this one's for you! Work it every which way you can. Do your best, but don't sweat buckets if you have to look at the solution: Three times Ed's age plus Tom's age equals twice Harry's age. Double the cube of Harry's age is equal to three times the cube of Ed's age added to the cube of Tom's age. Their respective ages are integers, and relativly prime (that is, they have no common factors.) Find their ages.
42. Solution Let $$e$$ be Ed's age, $$t$$ be Tom's age, and $$h$$ be Harry's age. Translating, I get two equations: $(1): 3e + t = 2h$ $(2): 2h^{3} = 3e^{3} + t^{3}$ Consider equation (2). I want something I can factor on both sides, so I'm going to break up the $$3e^{3}$$ into $$2e^{3} + e^{3}$$, and subtract $$2e^{3}$$ from both sides to get $2h^{3} - 2e^{3} = e^{3} + t^{3}$ I'll factor out the 2 on the left, and factor the difference and sum of cubes: $(3): 2(h - e) \left( h^{2} + he + e^{2} \right) = (e + t)\left(e^{2} - et + t^{2}\right)$ Looking at equation (1), I see that I can get an expression for both $$2(h - e)$$ and $$e + t$$ by doing the same trick: write $$3e$$ as $$2e + e$$, and subtract $$2e$$ from both sides to get $e + t = 2h - 2e$$e + t = 2(h - e)$ Now I can divide both sides of equation (3) by the same number. On the left, I'll divide by $$2(h - e)$$ and on the right I'll divide by $$e + t$$ (and make a note that $$h \ne e$$, and $$e \ne -t$$, or I'd be dividing by 0. The latter isn't true, since you can't have a negative age. I'll have to check to make sure the former isn't true at the end.) I get cancellation on both sides, and I'm left with $h^{2} + he + e^{2} = e^{2} - et + t^{2}$ I can subtract $$e^{2}$$ from both sides to get $h^{2} + he = - et + t^{2}$ Now I'm going to move the the terms around so I get a difference of squares: $t^{2} - h^{2} = he + et$ I'll factor both sides: $(t - h)(t + h) = e(t + h)$ I can divide both sides by $$t + h$$ (I'm not dividing by 0 since $$t \ne -h$$. That would mean one of them would be a negative age.) Now I have $e = t - h$ $(4): e + h = t$ Substituting equation (4) into equation (1) I get: $3e + e + h = 2h$ $4e = h$ Let's look at another form of equation (4): $h = t - e$ Substituting that into equathion (1) gives: $3e + t = 2(t - e) = 2t - 2e$ $t = 5e$ Now we have $$h$$ and $$t$$ both in terms of $$e$$. Since they're relatively prime and integral, I'll pick the smallest integer I can for $$e$$. Let $$e = 1$$. Then $$h = 4e = 4$$, and $$t = 5e = 5$$. Their ages are 1, 4, and 5. Above, I had the requirement that $$h \ne e$$, and they're not, so the answer is valid. Nifty, huh? Sometimes, just being able to follow a difficult solution is a skill in and of itself. No, there won't be a question this difficult on the test.