1. §5.7 PolyNomial Eqns & Apps
Chabot Mathematics
§5.7 PolyNomial Eqns & Apps
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. 5.6 Review § Any QUESTIONS About Any QUESTIONS About HomeWork
MTH 55
Review §
Any QUESTIONS About
§5.6 → Factoring Strategies
Any QUESTIONS About HomeWork
§5.6 → HW-16

3. §5.7 Solving PolyNomial Eqns
The Principle of Zero Products
Factoring to Solve Equations
Algebraic-Graphical Connection

4. Quadratic Equations
Second degree equations such as 9t2 – 4 = 0 and x2 + 6x + 9 = 0 are called quadratic equations
A quadratic equation is an equation equivalent to one of the form
ax2 + bx + c = 0
where a, b, and c are constants, with a ≠ 0.

5. Principle of Zero Products
An equation AB = 0 is true if and only if A = 0 or B = 0, or both = 0.
That is, a product is 0 if and only if at LEAST ONE factor in the multiplication-chain is 0
i.e.; Need a Zero-FACTOR to create a Zero-PRODUCT

6. Example  Solve (x + 4)(x − 3) = 0
In order for a product to be 0, at least one factor must be 0. Therefore, either
x + 4 = 0 or x − 3 = 0
Solving each equation:
x + 4 = or x − 3 = 0
x = −4 or x = 3
Both −4 and 3 should be checked in the original equation.

7. Check for (x + 4)(x − 3) = 0 The solutions are −4 and 3.
For x = −4: For x = 3:
(x + 4)(x − 3) = 0 (x + 4)(x − 3) = 0
(−4 + 4)(−4 − 3) (3 + 4)(3 − 3)
0(−7) (0)
0 = = 0
True True
The solutions are −4 and 3.

8. Solve  4(3x + 1)(x − 4) = 0
Since the factor 4 is constant, the only way for 4(3x + 1)(x − 4) to be 0 is for one of the other factors to be 0. That is,
3x + 1 = or x − 4 = 0
3x = − or x = 4
x = −1/3
So the solutions to the Equation are
x = −1/3 and x = 4
{−1/3 , 4}
Can divide both sides by 4 first to reveal Zero Products condition

9. Check 4(3x + 1)(x – 4) = 0 For −1/3: 4(3x + 1)(x − 4) = 0
4(3•[−1/3] + 1)([−1/3] − 4) = 0
4(−1 + 1)(−1/3 − 12/3) = 0
4(0)(−13/3 ) = 0
0 = 0
For 4: 4(3x + 1)(x − 4) = 0
4((3(4) + 1)(4 − 4) = 0
4(13)(0) = 0
The solutions are −1/3 and 4

10. Solve  3y(y − 7) = 0 SOLUTION 3  y (y − 7) = 0 y = 0 or y − 7 = 0
The solutions are 0 and 7
The Check is Left to the Student
Should be easily “EyeBalled”

11. Factoring to Solve Equations
By factoring and using the principle of zero products, we can now solve a variety of quadratic equations.
Example: Solve x2 + 9x + 14 = 0
SOLUTION: This equation requires us to FIRST factor the polynomial since there are no like terms to combine and there is a squared term. THEN we use the principle of zero products

12. Solve  x2 + 9x + 14 = 0
Factor the Left Hand Side (LHS) by Educated Guessing (FOIL Factoring):
x2 + 9x + 14 = 0
(x + 7)(x + 2) = 0
x + 7 = or x + 2 = 0
x = − or x = −2.
The Tentative Solutions are −7 and −2
Let’s Check

13. Check  x = −7 & x = −2 For −7: For −2:
x2 + 9x + 14 = 0 x2 + 9x + 14 = 0
(−7)2 + 9(−7) (–2)2 + 9(–2) + 14
49 − −
− −
0 = = 0
True True
Thus −7 and −2 are VERIFIED as Solutions to x2 + 9x + 14 = 0

14. Example  Solve x2 + 9x = 0
SOLUTION: Although there is no constant term, because of the x2-term, the equation is still quadratic. Try factoring:
x2 + 9x = 0 → see GCF = x
x(x + 9) = 0
x = or x + 9 = 0
x = or x = −9
The solutions are 0 and −9.

15. Caveat Mathematicus
CAUTION: We must have 0 on one side of the equation before the principle of zero products can be used.
Get all nonzero terms on one side of the equation and 0 on the other
Example: Solve: x2 − 12x = −36

16. Solve  x2 − 12x = −36
SOLUTION: We first add 36 to BOTH Sides to get 0 on one side:
x2 − 12x = −36
x2 − 12x = − = 0
(x − 6)(x − 6) = 0
x − 6 = 0 or x − 6 = 0
x = 6 or x = 6
There is only one solution, 6.

17. Standard Form
To solve a quadratic equation using the principle of zero products, we first write it in standard form: with 0 on one side of the equation and the leading coefficient POSITIVE.
We then factor and determine when each factor is 0.

18. Example  Functional Eval
Given f(x) = x2 + 10x Find a such that f(a) = 1.
SOLUTION
f (a) = a2 + 10a + 26 = 1
Set f (a) = 1
a2 + 10a + 25 = 0
(a + 5)(a + 5) = 0
a + 5 = 0 so a = −5
The check is left for you & I to do Later

19. Example  Solve 9x2 = 49 SOLUTION: 9x2 = 49 9x2 − 49 = 0
► Diff of Sqs: (3x)2 & 72
(3x − 7)(3x + 7) = 0
3x − 7 = or 3x + 7 = 0
3x = or 3x = −7 x = 7/3 or x = −7/3
The solutions are 7/3 & −7/3

20. Solve: 14x2 + 9x + 2 = 10x + 6
SOLUTION: Be careful with an equation like this! Since we need 0 on one side, we subtract 10x and 6 from Both Sides to get the RHS = 0
14x2 + 9x + 2 = 10x + 6
14x2 + 9x − 10x + 2 − 6 = 0
14x2 − x − 4 = 0
(7x − 4)(2x + 1) = 0
7x − 4 = 0 or 2x + 1 = 0
7x = 4 or x = −1
x = 4/7 or x = −1/2
The solutions are 4/7 and −1/2.

21. Solve Eqns by Zero Products
Write an equivalent equation with 0 on one side, using the addition principle.
Factor the nonzero side of the equation
Set each factor that is not a constant equal to 0
Solve the resulting equations

22. Parabola intercepts
Find the x-intercepts for the graph of the equation shown.
Parabola
The x-intercepts occur where the plot crosses y = 0
Thus at the x-intercepts
y = 0 = x2 + 2x − 8
So Can use the principle of zero products
y = x2 + 2x  8

23. Parabola intercepts cont.
SOLUTION: To find the intercepts, let y = 0 and solve for x.
Parabola
0 = x2 + 2x − 8
0 = (x + 4)(x − 2)
x + 4 = or x − 2 = 0
x = −4 or x = 2
The x-intercepts are (−4, 0) and (2, 0).
y = x2 + 2x  8

24. −x2 − x + 6 = 0  Solve with Graph
Recall from Graphing that the x-axis is the Location where y = 0
Thus on Graph find x for y = 0
Solns: x = −3 and x = 2

25. Example  Find x-Intercepts
Find the x-intercepts for graph (at Left) of Equation
At intercepts y = 0; so Use ZERO Products:

26. Example  Find x-Intercepts
At Intercepts y = 0, so
y = 0 = 2x2 + 3x − 9
FOIL-factor the Quadratic expression
0 = 2x2 + 3x − 9 = (2x − 3)(x + 3) = 0
By ZERO PRODUCTS
(2x − 3) = 0 or (x + 3) = 0
Solving for x (the intercept values):
x = 3/2 or x = −3

27. Example  x-Intercepts

28. PolyNomial Fcns and Graphs
Consider, for example the eqn, x2 − 2x = 8. One way to begin to solve this equation is to graph the function f (x) = x2 − 2x. Then look for any x-value that is paired with 8, as shown at Right y
y = 8 8 7 6 5 4 3 2 1 x -1

29. PolyNomial Fcns and Graphs
Equivalently, we could graph the function given by g(x) = x2 − 2x − 8 and look for the values of x for which g(x) = 0. These values are what we call the roots, or zeros, of a polynomial function
Root-1
Root-2

30. Problem Solving
Some problems can be translated to quadratic equations, which we can now solve.
The problem-solving process is the same as for other kinds of problems.

31. The Pythagorean Theorem
Recall Pythagorus’ Great Discovery:
In any right triangle, if a and b are the lengths of the legs and c is the length of the hypotenuse, then
a2 + b2 = c2 c a b

32. Example  Screen Diagonal
A computer screen has the dimensions (in inches) shown below.
Find the length of the diagonal of the screen.

33. Example  Screen Diagonal
Familiarize. A right triangle is formed using the diagonal and sides of the screen. x + 6 is the hypotenuse with x and x + 3 as legs.
Translate. Applying the Pythagorean Theorem, Use the Diagram to translate as follows:
a2 + b2 = c2
x2 + (x + 3)2 = (x + 6)2

34. Example  Screen Diagonal
Carry out. Solve the equation by:
x2 + (x2 + 6x + 9) = x2 + 12x + 36
2x2 + 6x + 9 = x2 + 12x + 36
x 2 – 6x – 27 = 0
(x + 3)(x – 9) = 0
x + 3 = 0 or x – 9 = 0
x = –3 or x = 9

35. Example  Screen Diagonal
Check. The integer −3 cannot be the length of a side because it is negative. For x = 9, we have x + 3 = 12 and x + 6 = 15. Since = 225, the lengths determine a right triangle.
Thus 9, 12,and 15 check.
State. The length of the diagonal of the screen is 15 inches

36. Example  Area Allotment
A LandScape Architect designs a flower bed of uniform width around a small reflecting pool. The pool if 6ft by 10ft. The plans call for 36 ft2 of plant coverage. How WIDE should the Border be?
Familiarize with Diagram

37. Example  Area Allotment
Now LET x ≡ the Border Width
Translate using Diagram
The OverAll
Width is 6ft + 2x
Length is 10ft + 2x

38. Example  Area Allotment

39. Example  Area Allotment
Carry Out
Zero Products
Since a Length can Never be Negative, Discard −9 as solution

40. Example  Area Allotment
State: The reflecting pool border width should be 1 ft
Check: 2·(12ft·1ft) + 2·(6ft·1ft) = 24ft2 + 12ft2 = 36ft2 
12 ft
8 ft

41. Example  Rocket Ballistics
A Model Rocket is fired UpWards from the Ground. The Height, h, in feet of the rocket can be found from this equation:
Where t is time in seconds
Find the time that it takes for the Rocket to reach a height of 48 feet

42. Example  Rocket Ballistics
Familiarize: We must find t such that h(t) = 48.
Thus Substitute 48 for h(t) in the ballistics equation
Carry Out:
Subtract 48 from both sides

43. Example  Rocket Ballistics
Divide Both Sides by −16
Write in Standard form
Use QUADRATIC Formula with a = 1, b = −5, and c = 3

44. Example  Rocket Ballistics
Approximate the Sq-Roots
Since “What goes UP must come DOWN” The Rocket will reach 48ft TWICE; Once while blasting-UP and again while Free-Falling DOWN

45. Example  Rocket Ballistics
State: The rocket will climb to 48ft in about 0.7 seconds, continue its climb, and then it will descend (fall) to a height of 48ft after a total flite time of 4.3 seconds as it continues its FreeFall to the Ground

46. WhiteBoard Work Problems From §5.7 Exercise Set Model Rockets
74, 80, 82
Model Rockets

47. All Done for Today
Blast Off!

48. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Chabot Mathematics
Appendix
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer –

49. Graph y = |x|
Make T-table

51. Quadratic Formula