Polynomial Functions EVALUATING POLYNOMIAL FUNCTIONS A polynomial function

Polynomial Functions

EVALUATING POLYNOMIAL FUNCTIONS A polynomial function is a function of the form f (x) = an x nn + an – 1 x nn – 11 +· · ·+ a 1 x + aa 0 0 Where aann >0 0 and the exponents are all whole numbers. coefficient For this polynomial function, aann is the leading coefficient, constant term, term and n is the degree aa 00 is the constant A polynomial function is in standard form if its terms are descending order of of exponents from left to to right. written in descending

EVALUATING POLYNOMIAL FUNCTIONS You are already familiar with some types of polynomial functions. Here is a summary of common types of polynomial functions. Degree Type Standard Form 0 Constant f (x) = a 0 1 Linear f (x) = a 1 x + a 0 2 Quadratic f (x) = a 2 x 2 + a 1 x + a 0 3 Cubic f (x) = a 3 x 3 + a 2 x 2 + a 1 x + a 0 4 Quartic f (x) = a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0

Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. f (x) = 1 x 2 – 3 x 4 – 7 2 SOLUTION The function is a polynomial function. Its standard form is f (x) = – 3 x 4 + 1 2 x – 7. 2 It has degree 4, so it is a quartic function. The leading coefficient is – 3.

Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. f (x) = x 3 + 3 x SOLUTION The function is not a polynomial function because the x term 3 does not have a variable base and an exponent that is a whole number.

Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. – f (x) = 6 x 2 + 2 x 1 + x SOLUTION The function is not a polynomial function because the term 2 x – 1 has an exponent that is not a whole number.

Identifying Polynomial Functions Polynomial function? f (x) = 12 x 2 – 3 x 4 – 7 f (x) = x 3 + 3 x f (x) = 6 x 2 + 2 x– 1 + x

Using Synthetic Substitution One way to evaluate polynomial functions is to use direct substitution. Another way to evaluate a polynomial is to use synthetic substitution. Use synthetic division to evaluate f (x) = 2 x 4 + -8 x 2 + 5 x - 7 when x = 3.

Using Synthetic Substitution SOLUTION 2 x 4 + 0 x 3 + (– 8 x 2) + 5 x + (– 7) Polynomialinin Polynomial standardform standard 33 • 2 0 – 8 5 – 7 Coefficients 6 18 30 105 6 10 35 98 x-value 2 The value of f (3) is the last number you write, In the bottom right-hand corner.

GRAPHING POLYNOMIAL FUNCTIONS The end behavior of a polynomial function’s graph is the behavior of the graph as x approaches infinity (+ ∞) or negative infinity (– ∞). The expression x + ∞ is read as “x approaches positive infinity. ”

GRAPHING POLYNOMIAL FUNCTIONS END BEHAVIOR

GRAPHING POLYNOMIAL FUNCTIONS CONCEPT SUMMARY END BEHAVIOR FOR POLYNOMIAL FUNCTIONS x +∞ +∞ f (x) –∞ f (x) +∞ even f (x) –∞ odd f (x) +∞ f (x) –∞ an n x >0 even f (x) >0 odd <0 <0 –∞

Graphing Polynomial Functions Graph f (x) = x 3 + x 2 – 4 x – 1. SOLUTION To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. – 3 and– 2 – 1 coefficient 0 1 is positive, 2 3 x The degree is odd the leading 3 and f (x) – 1 + – 3 as x 3 + 23. so f (x) f(x)– – 7 as x 3 –

Graphing Polynomial Functions Graph f (x) = –x 4 – 2 x 3 + 2 x 2 + 4 x. SOLUTION To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. – 2 the leading – 1 0 coefficient 1 is negative, 2 3 x The degree is– 3 even and – 16 +. as x – 105 so f (x) – – 21 as x 0 – – 1 and f 0(x) 3–

Adding, Subtracting, & Multiplying Polynomials

To + or - , + or – the coeff. of like terms! Vertical format : • Add 3 x 3+2 x 2 -x-7 and x 3 -10 x 2+8. • 3 x 3 + 2 x 2 – x – 7 + x 3 – 10 x 2 + 8 Line up like terms • 4 x 3 – 8 x 2 – x + 1

Horizontal format : Combine like terms • (8 x 3 – 3 x 2 – 2 x + 9) – (2 x 3 + 6 x 2 – x + 1)= • (8 x 3 – 2 x 3)+(-3 x 2 – 6 x 2)+(-2 x + x) + (9 – 1)= • 6 x 3 + -9 x 2 + -x + 8= • 6 x 3 – 9 x 2 – x + 8

Examples: Adding & Subtracting • (9 x 3 – 2 x + 1) + (5 x 2 + 12 x -4) = • 9 x 3 + 5 x 2 – 2 x + 1 – 4 = • 9 x 3 + 5 x 2 + 10 x – 3 • • (2 x 2 + 3 x) – (3 x 2 + x – 4)= 2 x 2 + 3 x – 3 x 2 – x + 4 = 2 x 2 - 3 x 2 + 3 x – x + 4 = -x 2 + 2 x + 4

Multiplying Polynomials: Vertically • (-x 2 + 2 x + 4)(x – 3)= -x 2 + 2 x + 4 * x– 3 3 x 2 – 6 x – 12 -x 3 + 2 x 2 + 4 x -x 3 + 5 x 2 – 2 x – 12

Multiplying Polynomials : Horizontally • • (x – 3)(3 x 2 – 2 x – 4)= (x – 3)(3 x 2) + (x – 3)(-2 x) + (x – 3)(-4) = (3 x 3 – 9 x 2) + (-2 x 2 + 6 x) + (-4 x + 12) = 3 x 3 – 9 x 2 – 2 x 2 + 6 x – 4 x +12 = 3 x 3 – 11 x 2 + 2 x + 12

Multiplying 3 Binomials : • • (x – 1)(x + 4)(x + 3) = FOIL the first two: (x 2 – x +4 x – 4)(x + 3) = (x 2 + 3 x – 4)(x + 3) = Then multiply the trinomial by the binomial (x 2 + 3 x – 4)(x) + (x 2 + 3 x – 4)(3) = (x 3 + 3 x 2 – 4 x) + (3 x 2 + 9 x – 12) = x 3 + 6 x 2 + 5 x - 12

Some binomial products appear so much we need to recognize the patterns! • Sum & Difference (S&D): • (a + b)(a – b) = a 2 – b 2 • Example: (x + 3)(x – 3) = x 2 – 9 • Square of Binomial: • (a + b)2 = a 2 + 2 ab + b 2 • (a - b)2 = a 2 – 2 ab + b 2

Last Pattern • Cube of a Binomial • (a + b)3 = a 3 + 3 a 2 b + 3 ab 2 + b 3 • (a – b)3 = a 3 - 3 a 2 b + 3 ab 2 – b 3

Example: • (x + 5)3 = a = x and b = 5 x 3 + 3(x)2(5) + 3(x)(5)2 + (5)3 = x 3 + 15 x 2 + 75 x + 125

Factoring and Solving Polynomial Expressions

Types of Factoring: – GCF : 6 x 2 + 15 x = 3 x (2 x + 5) – POS : x 2 + 10 x + 25 = (x + 5)2 – DOS : 4 x 2 – 9 = (2 x + 3)(2 x – 3) – “Bustin’ da B” = 2 x 2 – 5 x – 12 = » (2 x 2 - 8 x) + (3 x – 12) = » 2 x(x – 4) + 3(x – 4)= » (x – 4)(2 x + 3)

Now we will use Sum of Cubes: a 3 + b 3 = (a + b)(a 2 – ab + b 2) x 3 + 8 = (x)3 + (2)3 = (x + 2)(x 2 – 2 x + 4)

Difference of Cubes a 3 – b 3 = (a – b)(a 2 + ab + b 2) 8 x 3 – 1 = (2 x)3 – 13 = (2 x – 1)((2 x)2 + 2 x*1 + 12) (2 x – 1)(4 x 2 + 2 x + 1)

When there are more than 3 terms – use GROUPING x 3 – 2 x 2 – 9 x + 18 = (x 3 – 2 x 2) + (-9 x + 18) = x 2(x – 2) - 9(x – 2) = (x – 2)(x 2 – 9) = (x – 2)(x + 3)(x – 3) Group in two’s with a ‘+’ in the middle GCF each group Factor all that can be factored

Factoring in Quad form: 81 x 4 – 16 = (9 x 2)2 – 42 = (9 x 2 + 4)(9 x 2 – 4)= Can anything be factored still? ? ? (9 x 2 + 4)(3 x – 2)(3 x +2) Keep factoring ‘till you can’t factor any more!!

You try this one! 4 x 6 – 20 x 4 + 24 x 2 = 4 x 2 (x 4 - 5 x 2 +6) = 4 x 2 (x 2 – 2)(x 2 – 3)

Solve: • 2 x 5 + 24 x = 14 x 3 2 x 5 - 14 x 3 + 24 x = 0 2 x (x 4 – 7 x 2 +12) = 0 2 x (x 2 – 3)(x 2 – 4) = 0 2 x (x 2 – 3)(x + 2)(x – 2) = 0 2 x=0 x 2 -3=0 x+2=0 x=±√ 3 x=-2 Put in standard form GCF Bustin’ da ‘b’ Factor EVERYTHING x-2=0 set all to zero x=2

Now, you try one! • 2 y 5 – 18 y = 0 • y=0 y=±√ 3 y=±i√ 3