Polynomials
Polynomial functions in which we have one independent variable. Example (1)
y = 4 + 3x4– 5x +2x3
y = 3x4+ 2x3– 5x +4 rewrite in descending order of powers Order or Degree = 4
Number of terms = 4 Leading term = 3x4
Coefficient of leading term = 3 Constant term = 4
Conditions:
All exponents are positive integers Coefficients can be any real numbers
Radicals, fractional, trigonometric, logarithmic expressions of independent variable are not included.
Example (2) Linear functions (or) polynomials of first degree.
y = mx + c famous equation of straight line
c = y-intercept of the line m = slope of the line =
value -in x change value -y in
change = constant for each straight line
a) Equation of the straight line passing the two points A(2, 3), and B(-2, 4) is needed. Choose any third point C on the line with coordinates (x, y).
Now we have three points: A(2, 3), B(-2, 4), and C(x, y). The relation between the x and y is the equation of the line.
Slope of line =
points two same e between th s
in x value difference points o between tw y values in difference
, (keep the order of subtraction) Slope = 2 -x 3
-y using points C and A
Slope = 4 1 2 2 1 -(-2) -2 4
-3
using points A and B
2 -x
3 -y =
4 1
4(y – 3) = -1(x – 2) 4y – 12 = -x + 2
4y = -x + 2 + 12 = -x + 14
If we know two points (x1, y1), and (x2, y2) on a straight line, then the equation of the straight line
is
Slope x
x y y x x
y y
1 2
1 2
1
1
If we know the y-intercept (0, c) and slope (m) of the straight line, then the equation of the straight line is
y = mx + c
Slope = Run Rise
Rise Go up (+), Go down (-)
Run Go right (+), Go left (-)
b) Using the below data, construct the linear relation between C (cost) and x (number of products).
C 200 250 300 350
x 0 20 40 60
C and n are linearly related the question implies Check for linear relation between C and x
When x is increased by 20 (0 to 20) C is increased by (250 - 200) = 50 When x is increased by 20 (20 to 40)C is increased by (300 - 250) = 50 When x is increased by 20 (40 to 60)C is increased by (350 - 300) = 50
The change of C when x changes by 20 units is the same. They are linearly related. Therefore,
C = mx + b use letter ‘b’ not to confuse with C
From the table, When x = 0, C = 200 200 = m(0) + b b = 200
When x = 40, C = 300 300 = m(40) + 200 300 – 200 = 40m 100 = 40m 100/40 = m 2.5 = m
c) Find the equation of the line given below.
Pick the two points where you can read the coordinates. I would choose A(-4, 6) and B(10, -2)
We have two points on the graph: A(-4, 6) and B(10, -2) Equation of the given straight line is
14 8 Slope )
4 ( x
6 y
7 4 4 x
6
y
7(y – 6) = -4(x + 4) 7y – 42 = -4x – 16
7y = -4x – 16 + 42 = -4x + 26
Parabolas or Quadratic functions (domain: set of real numbers unless specified) y = f(x) = ax2+ bx + c General form (or) standard form
y = f(x) = a(x-p)2+ q vertex form
y = f(x) = a(x – x1)(x – x2) Factor form
Where a 0
Polynomial of 2ndorder or 2nddrgree
x1, x2 “zeros” of the function y = f(x)
x1, x2 x-intercepts (x1, 0) & (x2, 0) of the graph of y = f(x)
x1, x2 “roots” or “solutions” of the equation y = f(x) = 0
Values of x1, x2 are given by quadratic formula:
a 2
ac 4 b b 2
ac 4
b2 = discriminant
(p, q) Vertex
a > 0 Parabola opens upward q = minimum value of y or f(x) a < 0 Parabola opens downwardq = maximum value of y or f(x)
ac 4
b2 > 0 Two x-intercepts ac
4
b2 = 0 One x-intercept Parabola touches x-axis ac
4
b2 < 0 No x-intercepts
Example (3) Investigate the function y =f(x) = (x-3)2+ 1
y = (x-3)2+ 1 y = x2– 6x + 9 + 1
y = x2– 6x + 10
a = 1 > 0 Open upward
(p, q) = (3, 1) Vertex or y-minimum (3, 1)
ac 4
If a point P(x, y) is on the graph, then these (x, y) values must satisfy the equation: y = (x-3)2+ 1
If there is a point R(x, y) not on the graph, these (x, y) values do not satisfy the equation: y = (x-3)2+ 1
For example R(8, 2) is not on the graph. Therefore (x-3)2+ 1 = (8-3)2+ 1 = (5)2+ 1 = 25 + 1 = 26 2
Example (4)
Completing the square to convert general form to vertex form
Graph the function: y = 2x2+ 4x + 3 and indicate its vertex, maximum or minimum y-value and x-intercepts and y-intercept if any.
y = 2x2+ 4x + 3 a = 2, b = 4, c = 3
y = 2[x2+ 2x] + 3 factor 2 out to make coefficient of x2one
Inside the brackets add and subtract the square of half of the coefficient of x y = 2[x2+ 2x + )2
2 2
( - )2
2 2 ( ] + 3
y = 2[x2+ 2x + (1)2– (1)2] + 3 y = 2[x2+ 2x + 1 – 1] + 3
y = 2[(x+1)2– 1] + 3 a2+ 2ab + b2= (a + b)2 y = 2(x+1)2– 2 + 3
y = 2(x+1)2+1 Vertex form: p = -1, q = 1
a = 2 > 0 Open up
(p, q) = vertex = (-1, 1) Minimum y-value = 1, above x-axis
at this point we know that the graph does not intersect xaxis; it stays completely positive
-b2– 4ac = 42– 4(2)(3) = 16 – 24 = -8 < 0 Parabola does not intersect x-axis
No x-intersect.
At y-intercept, x = 0 y = 2x2+ 4x + 3 y = 2(0)2+ 4(0) + 3 y = 3
y-intercept = (0, 3) Range: y 1
Example (5)
Find the equation of the parabola which crosses x-axis at (1, 0) and (6, 0) and passes through the point (2, 4).
x1= 1 and x2= 6 are zeros of the parabola.
So the equation of the parabola must be y = a(x – 1)(x – 6) and we must find a. (2, 4) is on the parabola
y = a(x – 1)(x – 6) 4 = a(2 – 1)(2 – 6) 4 = a(1)(-4) 4 = -4a -1 = a
So the parabola is y = -(x – 1)(x – 6)
y = -(x2– 6x – x + 6) = -(x2– 7x + 6) = -x2+ 7x - 6
Example (6)
Famous polynomials with odd degrees (power functions containing single term) y = x
y = x3 y = x5
When x is (+), y is also (+) check by plugging in few x values When x is (-), y is also (-)
Their domains: {x | x is a real number}, their ranges: {y | y is a real number} They are all increasing functions. The graphs rise when x value is increased.
Graphs of y = -x, y = -x3, and y = - x5 y = - x
y = - x3
y = - x5
When x is (+), y is (-) check by plugging few x values When x is (-), y is (+)
When x = 0, y is 0
The above six functions are called monotonic functionsbecause they are either always rising or falling.
Example (7) Famous polynomials with even degrees (power functions containing single term) y = x2
y = x4
When x = (+), y is (+) check by plugging in few x values When x = (-), y is still (+)
When x = 0 , y is also 0
Domains: {x | x is a real number} Ranges: {y | y 0}
These functions are not monotonic functions. They fall for x < 0, and rise for x > 0.
At x = 0, they both reach their minimum point.
Graphs of y = - x2, and y = - x4
y = - x2 y = - x4
When x is (+), y is (-) check by plugging few x values When x is (-), y is (+)
Domains: {x | x is a real number} Ranges: {y | y 0}
These functions rise for x < 0, and fall for x > 0 They have their maximums when x = 0
Properties of polynomial functions
y = f(x) = n 1 0
nx a x a
a Polynomial function
f(x) = 1 0
n
nx a x a
a = 0 Corresponding polynomial equation f(x) = 0 represents x-intersect points of the graph of y = f(x)
All powers or exponents of x are positive integers. All coefficients are real numbers.
Leading term = n
nx a
Order or degree = n
Leading coefficient = an0
Constant term = a0
Zeros of {y = f(x)} x-values for which f(x) = 0 Zero = it is not 0, it is the x-value at which f(x) = 0
If c is a zero of y = f(x) This paragraph is factor theorem
f(c) = 0
(c, 0) is the x-intercept of the graph of y = f(x). (x – c) is a factor of f(x)
f(x) = (x – c)q(x), where q(x) is a polynomial with 1 degree less than degree of f(x) If the constant term a0= 0, then x = 0 is a zero of y = f(x)
Number of “zeros” or x-intercepts of {y = f(x)} n OR
Number of solutions or roots of {f(x) = 0} n If c is a zero of multiplicity k (k = positive integer) of y = f(x),
(x - c)kis a factor of f(x)
(x - c)k+1is not the factor of f(x)
If k = even y = f(x) graph touches x-axis at x = c and turns around If k = odd y = f(x) graph crosses the x-axis at x = c
Turning point = a point at which the graph changes direction = falls then rises (or) rises then falls
No of turning points n – 1
Leading coefficient test for end behaviors: this test does not say anything about the shape of the graph in the central area.
n
Odd
Even
a
n(+)
(-)
Example (8)
a) Investigate the following polynomial function for end behaviors, maximum number of zeros, maximum number of turning points.
b) Find all zeros, and c) sketch the function and write down the actual number of turning points. y = f(x) = -2x4+ 3x2– x
Leading coefficient = -2 Degree n = 4, even number
a)
End behavior
When x +, y goes down to - When x -, y goes down to -
Maximum number of zeros = n = 4
Maximum number of turning points = n – 1 = 4 – 1 = 3
b)
There is no constant term. Constant term is 0. When x = 0, y = f(x) = 0 x = 0 is a zero.
Look at 2x3- 3x + 1
When x = 1 2x3- 3x + 1 = 2(1)3- 3(1) + 1 = 2 – 3 + 1 = 0 [by Luck!]
(x – 1) is a factor of 2x3- 3x + 1
2x3- 3x + 1 = 2x3- 2x – x + 1 do this if you foresee = 2x(x2– 1) - (x - 1) the second step
= 2x(x + 1)(x – 1) - (x – 1) general methods in example (9) = (x – 1)[2x(x + 1) - 1]
= (x – 1)(2x2+2x – 1) y = -2x4+ 3x2– x = -x (2x3- 3x + 1)
= -x (x – 1)(2x2+2x – 1)
We have two zeros so far: x = 0, and x = 1
Now we look at 2x2+2x – 1
We will find x-values at which 2x2+2x – 1 = 0 These will be =
a 2
ac 4 b b 2
Quadratic formula
= ) 2 ( 2 ) 1 )( 2 ( 4 2
2 2
= 4 12 2 = 4 ) 3 )( 4 ( 2 = 4 3 2 2
= 0.50.5 3 two more zeros
The zeros are : 0, 1, 0.50.5 3 (about 0.366), 0.50.5 3(about – 1.366) x-intersects are : (0,0), (1, 0), (0.366, 0), (- 1.366, 0)
They are all zeros of single multiplicity. Multiplicities are all odd (= 1). The graph passes x-axis at these zeros. Together with the end behaviors we can graph y = f(x).
We will find few more points in addition to the four x-intersects.
A B C D E F G H I
X -1.4 -1.366 -1 0 0.2 0.366 0.8 1 1.2
y = -2x4+ 3x2–
x
From the graph, we see that it turns 3 times. From left to right,
1stturn occurs near C Rising then falling
2ndturn occurs near E Falling then rising 3rdturn occurs near G Rising then falling
Note:Only when degree of polynomial is 1 (linear functions) and 2 (quadratic functions) we have definite ways to find the zeros of polynomials.
For the degrees higher than two, some easy polynomials such as the one in example (8) can be found by plugging in some possible (guessed) values.
Division algorithm
7 divided by 3 = 3 7=
3 1 2 3
1 6 OR
7 = 3(2) + 1 Dividend = (Divisor)(Quotient) + (Remainder) 7 = dividend
Dividend = (Divisor)(Quotient) + (Remainder) OR
Divisor
Dividend= (Quotient) +
Divisor mainder Re
Remainder theorem
When a polynomial f(x) is divided by (x – c), the remainder is f(c) f(x) = (x – c)(quotient) + f(c)
Quotient is a polynomial with the degree one less than f(x)
Factor theorem
Let f(x) be a polynomial.
If f(c) = 0 x = c is a x-intersect (x – c) is a factor of f(x) If (x – c) is a factor of f(x) f(c) = 0 x = c is a x-intersect Then, f(x) = (x – c)(quotient) + no remainder
Rational zero theorem
f(x) = n 1 0
nx a x a a
All powers are positive integers. All coefficients are integers. a00
If p/q is a possible rational zero of f(x)
p =factors of a0 =factors of constant term
q =factors of an =factors of leading coefficient
If an= 1
p is a possible integral zero of f(x) =factors of a0
Example (9) f(x) = x3– 3x – 2
a) Determine the end behaviors of the function, and graph it. b) Find all zeros
c) Find the behavior of the graph at x-intersects, the number turns the graph makes, and graph the function.
Degree = n = 3 (odd number)
Leading coefficient = 1 (+), possibility of integral zero Constant term = -2
a) End behaviors
b) There are at most 3 zeros.
Possible integral zeros = factors of (-2) = 2, 1
Try the possible zeros
f(1) = x3– 3x – 2 = (1)3– 3(1) – 2 = 1 – 3 – 2 = 1 – 5 = -4 (no) f(-1) = (-1)3– 3(-1) – 2 = -1 +3 – 2 = 0
x = -1 is a zero of f(x) (-1, 0) is a x-intercept
To find another two possible zeros Method I
f(x) = x3– 3x – 2 = [x - (-1)] (Quotient) + (no remainder) x3– 3x – 2 = (x + 1)(Quotient)
x3– 3x – 2 = (x + 1)(ax2+ bx + c) Quotient must be one degree less than f(x) x3– 3x – 2 = ax3+ bx2+ cx + ax2+ bx + c
x3– 3x – 2 = ax3+ (b + a)x2+ (c + b)x + c
Equating the coefficients on both sides of equation a = 1
b + a = 0 b = -a = -1
c + b = -3 b = -3 =c = -3 –(-2) = -3 + 2 = -1 (Okay!) c = -2
f(x) = x3– 3x – 2 = (x + 1)(ax2+ bx + c)
f(x) = x3– 3x – 2 = (x + 1)(x2- x - 2)
Now we need to find factors or zeros of x2- x – 2 They are =
a 2
ac 4 b b 2
Quadratic formula
=
) 1 ( 2
) 2 )( 1 ( 4 ) 1 ( ) 1
( 2
= 2
3
1 = 2 or -1
The zeros are: -1, 2, -1
Method II
After getting the first zero which is -1, we continue to try more possible numbers. f(-2) = x3– 3x – 2
= (-2)3– 3(-2) – 2 = -8 + 6 – 2 = -4 (no) f(2) = x3– 3x – 2
= (2)3– 3(2) – 2 = 8 – 6 – 2 = 0
2 is another zero. So we now have two zeros: -1, 2 f(x) =
x3– 3x – 2 = (x +1)(x - 2)(Quotient) + (no remainder) x3– 3x – 2 = (x2– 2x + x – 2)(Quotient)
x3– 3x – 2 = (x2– x – 2)(Quotient)
This time instead of finding and equating coefficients on both sides of the equation, we will divide (x3– 3x – 2) by (x2– x – 2) to find the quotient.
x + 1
_________________ x2– x – 2 | x3+ 0 – 3x – 2
| x3– x2– 2x
0 + x2– x – 2 x3- x3= 0, 0 – (- x2) = x2, -3x – (-2x) = -x
x2– x - 2 x2- x2= 0, -x – (-x) = 0, -2 – (-2) = 0
---0 No Remainder
x3– 3x – 2 = (x2– x – 2)(x + 1)
x3– 3x – 2 = (x + 1)(x – 2)(x + 1) = (x – 2)(x + 1)2
Zeros are: x = 2 (multiplicity one) cross the x-axis x = -1 (multiplicity two) touch the x-axis
c) The graph
At (-1, 0) the graph touches the x-axis
At (2, 0) the graph crosses the x-axis
There are two number of turning points.
2 = n – 1 = 3 – 1 = 2
Which is the maximum possible number of turns
Polynomial functions whose zeros (or) x-intersects are known. n = degree of polynomial
If number of real zeros = m < n
m – n = even number because imaginary zeros come in pairs If the degree n = 3, and there are only two real zeros found on the graph, one zero must be repeated two times. The graph must touch the x-axis at that zero and turn around.
Quadratic function f(x) with two real zeros: x1and x2
x-intersects at (x1, 0) and (x2, 0)
f(x) = a(x - x1)(x – x2)
a = to be found from one more information such as one more point on the graph Quadratic function f(x) with one real zero x1(it must be repeated two times)
x-intersect at (x1, 0)
x = x1 is zero of multiplicity two
f(x) = a(x - x1)2
Quadratic function f(x) with no real zeros No x-intersects
f(x) = ax2+ bx + c
ac 4 b2 < 0
Polynomial of 3rddegree f(x) with three real zeros: x
1, x2, and x3
x-intersects: (x1, 0), (x2, 0), (x3, 0)
f(x) = a(x - x1)(x – x2)(x – x3)
Polynomial of 3rddegree f(x) with two real zeros: x1, x2(one of these must be repeated 2 times)
x-intersects: (x1, 0), (x2, 0)
f(x) = a(x - x1)(x – x2)2 graph touches x-axis at (x2, 0) and turn around
Polynomial of 3rddegree f(x) with one real zero x1and graph crosses x-axis at x = x1but x-axis is
not tangent to the graph at this point x-intersect (x1, 0)
f(x) = (x - x1)(ax2+ bx + c)
ac 4 b2 < 0
Polynomial of 4thdegree f(x) with four real zeros: x1, x2, x3, x4
x-intersects: (x1, 0), (x2, 0), (x3, 0), (x4, 0)
f(x) = a(x - x1)(x – x2)(x – x3)(x – x4)
Polynomial of 4thdegree f(x) with three real zeros: x1, x2, x3 (one of these repeats 2 times)
x-intersects: (x1, 0), (x2, 0), (x3, 0)
Multiplicity of real zeros
Practice Question (1)
State the degree and list the zeros of the polynomial function
f(x) = (x − 1)3(x + 2)2. State the multiplicity of each zero and whether the graph crosses the x-axis
at the corresponding x-intercept. Then sketch the graph of the polynomial function by hand.
Question (2)
Using only algebra, find a cubic function (degree of three) with zeros given by 1,1 + 2,1 − 2.
Question (3)
Sketch f(x) = (4x − 7)(9 − x)(13 − x)2
.
Plot the following functions: Question (4)
f (x) = (x - 3)2(x + 1) Question (5)
f (x) = (x + 4) (x + 1)2(x - 3)
Question (6)
Using algebraic method find a cubic function that satisfies the given table values.
x f(x)
-2 0
-1 24
1 0
Question (7)
Determine the exact functions given the following information:
a) 3rd degree function has zeros at -2, +2 and +3 and passes through point (1, 12)
b) 4th degree function touches the x-axis at -4 and crosses x-axis at 0 and +2. (-1, -9) is a point on function.
Question (8)
Find the 3rddegree polynomial function for the below graph.
Question (9)
Find the 4th degree polynomial function for the below graph.
Question (10)
Question (11)
Tickets to a school dance cost $4 and the projected attendance is 300 people. For every $0.10 increase in ticket price, the dance committee projects that attendance will decrease by 5. a) Determine the dance committee’s greatest possible revenue.
b) What ticket price will produce the greatest revenue?
Questions with answers Question (12)
The sum of the squares of two consecutive even integers is 452. Find the integers. (Answer: 14, 16 or -14, 16)
Question (13)
The width of a rectangle is 2 m less than the length. The area is 48 m2. Find the dimensions. (Answer: 6 m by 8 m)
Question (14)
One side of a right triangle is 2 cm shorter than the hypotenuse and 7 cm longer than the third side. Find the lengths of the sides of the triangle.
(Answer: 8 cm, 15 cm, 17 cm)
